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taco | verifiable_code | https://codeforces.com/problemset/problem/1013/B | Solve the following coding problem using the programming language python:
There is an array with n elements a_1, a_2, ..., a_{n} and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element a_{i} with a_{i} & x, where & denotes the bitwise and operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that a_{i} = a_{j}. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
-----Input-----
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers a_{i} (1 ≤ a_{i} ≤ 100 000), the elements of the array.
-----Output-----
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
-----Examples-----
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
-----Note-----
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(n, x) = [int(i) for i in input().split()]
nums = [int(i) for i in input().split()]
nums1 = nums[:]
nums = set(nums)
if len(set(nums)) < n:
print(0)
exit(0)
for i in range(len(nums)):
if nums1[i] & x in nums and nums1[i] != nums1[i] & x:
print(1)
exit(0)
for i in range(len(nums)):
nums1[i] = nums1[i] & x
if len(set(nums1)) < n:
print(2)
exit(0)
print(-1)
``` | vfc_134280 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1013/B",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4 3\n1 2 3 7\n",
"output": "1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2 228\n1 1\n",
"output": "0\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3 7\n1 2 3\n",
"output": "-1\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/IARCSJUD/problems/ENDCORR | Solve the following coding problem using the programming language python:
A despotic king decided that his kingdom needed to be rid of corruption and disparity. He called his prime minister and ordered that all corrupt citizens be put to death. Moreover, he wanted this done quickly.
The wily prime minister realised that investigating every citizen to decide who was corrupt and who was not was rather difficult. So he decided on the following plan: He ordered all the citizens to appear in the court one by one and declare their wealth.
The king does not sit in the court all the time (he has other important business to attend to - for instance, meet dignitaries from neighbouring kingdoms, spend time with his family …) Whenever the king walks into the court, the prime minister pulls out the richest man who has appeared before the court so far and is still alive and beheads him for being corrupt. Since the rich are more likely to be corrupt, he hopes to get rid of most of the corrupt and the king is happy as he sees his policy being implemented enthusiastically.
Suppose the wealth of the citizens trooping into the court is
1376518911241376518911241\; 3\; 7\; 6\; 5\; 18\; 9\; 11\; 2\; 4
and the king walked in three times: the first time after the first four persons have seen the minister, the second time after the first five persons have seen the minister and, finally after the first nine persons have seen the minister.
At the king's first visit the richest person to have met the minister has wealth $7$ and he would be beheaded. At the second visit, the wealth of the richest person who has met the minister and is still alive has wealth $6$ and so he would be beheaded. At the third visit the richest person to have met the minister who is still alive has wealth $18$ and so he would be beheaded.
You may assume that the input is such that whenever the king walks in, it is always possible to behead someone.
Your aim is to write a program that will enable the prime minister to identify the richest man to have met the minister and who is still alive quickly. You may assume that no two citizens have the same wealth.
-----Input:-----
The first line of the input consists of two numbers $N$ and $M$, where $N$ is the number of citizens in the kingdom and M is the number of visits to the court by the king.
The next $N+M$ lines describe the order in which the $N$ citizens' appearances are interleaved with the $M$ visits by the king. A citizen's visit is denoted by a positive integer, signifying his wealth. You may assume that no two citizens have the same wealth. A visit by the king is denoted by $-1$.
-----Output:-----
Your output should consist of $M$ lines, where the $i^{th}$ line contains the wealth of the citizen who is beheaded at the $i^{th}$ visit of the king.
-----Constraints:-----
- $1 \leq M \leq 10000$.
- $1 \leq N \leq 100000$.
- You may assume that in $50 \%$ of the inputs $1 \leq M \leq 1000$ and $1 \leq N \leq 8000$.
-----Sample Input-----
10 3
1
3
7
6
-1
5
-1
18
9
11
2
-1
4
-----Sample Output-----
7
6
18
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(n, m) = map(int, input().split())
l = []
leng = 0
for i in range(n + m):
w = int(input())
if w == -1:
cm = 0
mi = 0
for j in range(leng):
if l[j] > cm:
cm = l[j]
mi = j
print(cm)
l[mi] = -1
else:
l.append(w)
leng += 1
``` | vfc_134296 | {
"difficulty": "unknown_difficulty",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/IARCSJUD/problems/ENDCORR",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "10 3\n1\n3\n7\n6\n-1\n5\n-1\n18\n9\n11\n2\n-1\n4\n",
"output": "7\n6\n18\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/SUMPARITY | Solve the following coding problem using the programming language python:
Vedant has a hidden array of N elements. Given the [bitwise AND] of all elements of the array, determine whether the total sum of the elements is odd, even or cannot be determined.
------ Input Format ------
- The first line of each input contains T - the number of test cases. The test cases then follow.
- The only line of each test case contains two space-separated integers N and A - the number of elements in the hidden array and the bitwise AND of all the elements in the hidden array.
------ Output Format ------
For each test case, output Even if the sum of all elements in the hidden array is even, Odd if the sum of all elements in the hidden array is odd, or Impossible if the parity cannot be determined.
Note: You may print each character of the string in uppercase or lowercase (for example, the strings eVen, EvEn, even and EVEN will all be treated as identical).
------ Constraints ------
$1 ≤ T ≤ 10000$
$1 ≤ N ≤ 10^{9}$
$0 ≤ A ≤ 10^{9}$
----- Sample Input 1 ------
3
1 11
1 2
74120341 829182732
----- Sample Output 1 ------
Odd
Even
Impossible
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for _ in range(int(input())):
(n, a) = map(int, input().split())
if a % 2:
if n % 2:
print('Odd')
else:
print('Even')
elif n == 1:
print('Even')
else:
print('Impossible')
``` | vfc_134300 | {
"difficulty": "medium",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/SUMPARITY",
"time_limit": "0.5 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n1 11\n1 2\n74120341 829182732",
"output": "Odd\nEven\nImpossible",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/gambling-stark3304/1 | Solve the following coding problem using the programming language python:
Given a number N, you have to say if it is adjacent to two prime numbers. Two numbers are adjacent if the absolute difference between them is 1.
Example 1:
Input: N = 4
Output: Y
Explaination: The adjacent numbers 3 and
5 are both prime.
Eaxmple 2:
Input: 7
Output: N
Explaination: The numbers 6 and 8 are
not prime.
Your Task:
You do not need to read input or print anything. Your task is to complete the function primeAdjacent() which takes N as input parameter and reutrns 1 if it has two adjacent prime numbers, otherwise, returns 0.
Expected Time Complexity: O(sqrt(N))
Expected Auxiliary Space: O(1)
Constraints:
2 ≤ N ≤ 5000
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def primeAdjacent(self, N):
# code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def is_prime(self, n):
if n <= 1:
return 0
for i in range(2, n):
if n % i == 0:
return 0
return 1
def primeAdjacent(self, N):
ans1 = self.is_prime(N + 1)
ans2 = self.is_prime(N - 1)
return ans1 and ans2
``` | vfc_134304 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/gambling-stark3304/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 4",
"output": "Y",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "7",
"output": "N",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/MINIL | Solve the following coding problem using the programming language python:
Read problems statements in [Hindi], [Mandarin Chinese], [Russian], [Vietnamese], and [Bengali] as well.
You are given a grid with $N$ rows (numbered $1$ through $N$) and $M$ columns (numbered $1$ through $M$). Each cell of this grid contains either water or land. The initial state of the grid is described by strings $S_{1}, S_{2}, \ldots, S_{N}$; for each valid $i$ and $j$, the $j$-th character of $S_{i}$ is '.' if the cell in the $i$-th row and $j$-th column initially contains water or '*' if this cell initially contains land.
A maximal group of connected land cells (by moving between side-adjacent land cells) is called an *island*.
You may perform any number of operations (including zero). In one operation, you should choose one cell of the grid and convert it either to a land cell or to a water cell. Find the smallest number of operations required to maximise the number of islands in the grid.
------ Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first line of each test case contains two space-separated integers $N$ and $M$.
$N$ lines follow. For each valid $i$, the $i$-th of these lines contains a single string $S_{i}$ with length $M$.
------ Output ------
For each test case, print a single line containing one integer ― the smallest number of operations required to maximise the number of islands.
------ Constraints ------
$1 ≤ T ≤ 1,000$
$2 ≤ N, M ≤ 10$
$S_{1}, S_{2}, \ldots, S_{N}$ contain only characters '.' and '*'
----- Sample Input 1 ------
2
2 2
..
*.
2 4
*..*
*.*.
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Example case 1: The largest number of islands is $2$ and it can be achieved by converting one cell from water to land. The resulting grid is
.*
*.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
t = int(input())
for _ in range(t):
(n, m) = map(int, input().split())
grid = []
ans1 = 0
ans2 = 0
for i in range(n):
grid.append(input())
if not (n % 2 == 1 and m % 2 == 1):
for i in range(n):
for j in range(m):
if (i + j) % 2 == 0:
cmp = '*'
else:
cmp = '.'
if grid[i][j] != cmp:
ans1 += 1
else:
ans2 += 1
print(min(ans1, ans2))
else:
for i in range(n):
for j in range(m):
if (i + j) % 2 == 0:
cmp = '*'
else:
cmp = '.'
if grid[i][j] != cmp:
ans1 += 1
print(ans1)
``` | vfc_134305 | {
"difficulty": "medium_hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/MINIL",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n2 2\n..\n*.\n2 4\n*..*\n*.*.",
"output": "1\n2",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/CHEFTWOS | Solve the following coding problem using the programming language python:
Chef's loves his dog so much! Once his dog created two strings a and b each of length n consisting of digits 1 and 2, and even a problem about them!
Chef's Dog will tell by barking if a string x (also containing only digits 1 and 2 and with length N) is good or not by performing the following actions.
- It starts at first digit of the string, i.e. at i = 1.
- It can move from digit i to either i - 1 or i + 1 if xi equals 1 and the corresponding digits exist.
- It can move from digit i to either i - 2 or i + 2 if xi equals 2 and the corresponding digits exist.
- It must visit each digit exactly once.
- It must finish at the last digit (XN).
Chef's dog wants to make both the strings a and b good by choosing some subset S (possibly empty) of indices of set {1, 2, ..., n} and swapping each index i ϵ S between string a and b, i.e. swapping ai and bi. Can you find how many such subsets S exist out there? As the answer could be large, output it modulo 109 + 7.
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line contains string a.
The second line contains string b.
-----Output-----
For each test case, output a single line containing answer of the problem.
-----Constraints-----
- 1 ≤ T ≤ 20
- 1 ≤ |a| = |b| ≤ 105
- '1' ≤ ai, bi ≤ '2'
-----Subtasks-----
- Subtask #1 (30 points) |a|, |b| ≤ 10
- Subtask #2 (70 points) original constraints
-----Example-----
Input:
2
1111
2211
222
111
Output:
8
0
-----Explanation-----
Test case 1.
Possible subsets are:
{}, {1, 2}, {1, 2, 3}, {1, 2, 4}, {1, 2, 3, 4}, {3}, {4}, {3, 4}.
Test case 2. There are no possible sets S which can make both the strings good.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def fun(a):
n = len(a)
i = 0
b = []
while i < n:
c = 0
while i < n - 1 and a[i] == a[i + 1]:
c += 1
i += 1
b.append([a[i], c + 1])
i += 1
return b
def tatti(h):
n = len(h)
i = 0
b = []
while i < n:
c = 0
while i < n - 1 and h[i][0] == h[i + 1][0]:
c += 1
i += 1
b.append([h[i][0], h[i][1], c + 1])
i += 1
return b
def mod(a, b, m):
x = 1
y = a
while b > 0:
if b % 2:
x = x * y % m
y = y * y % m
b //= 2
return x % m
def rem(s):
s = list(s)
l = len(s)
i = l - 1
ans = []
c = 0
while i >= 0:
if s[i] == 1 and s[i - 1] == 2 and (s[i - 2] == 1) and (i >= 2):
i -= 3
c += 1
else:
i -= 1
ans.append(c)
c = 0
ans.append(c)
count = 0
flag = 0
for i in range(len(ans)):
if ans[i] != 0:
count += 1
flag = max(flag, ans[i])
return (count, flag)
for _ in range(int(input())):
md = 10 ** 9 + 7
s1 = input()
s2 = input()
x = []
l = len(s1)
for i in range(l):
if s1[i] == '2' and s2[i] == '2':
x.append(2)
elif s1[i] == '1' and s2[i] == '1':
x.append(0)
else:
x.append(1)
if l == 1:
ans = 2
elif l == 2:
if s1[0] == '2' or s2[0] == '2':
ans = 0
else:
ans = 4
else:
x.pop()
x.pop()
h = fun(x)
for i in range(1, len(h) - 1):
if h[i][0] == 2 and h[i][1] == 1:
if h[i - 1][0] == 1 and h[i + 1][0] == 1:
h[i - 1][1] -= 1
h[i + 1][1] -= 1
h[i][0] = -1
hn = []
for i in range(len(h)):
if h[i][1] != 0:
hn.append(h[i])
hnn = tatti(hn)
flag = 0
ans = 1
for i in range(len(hnn)):
if hnn[i][0] == 0:
ans = ans * mod(2, hnn[i][1], md) % md
elif hnn[i][0] == 1:
if hnn[i][1] % 2 == 0:
ans *= 2
ans = ans % md
else:
ans *= 0
elif hnn[i][0] == 2:
if hnn[i][1] == 2:
if i - 1 >= 0 and hnn[i - 1][0] != 0 or (i + 1 < len(hnn) and hnn[i + 1][0] != 0):
ans *= 0
else:
ans *= 4
ans = ans % md
else:
ans *= 0
elif (i - 1 >= 0 and hnn[i - 1][0] == 1) and (i + 1 <= len(hnn) - 1 and hnn[i + 1][0] == 1):
if hnn[i][2] - 1 >= 0:
ans = ans * mod(2, hnn[i][2] - 1, md)
ans = ans % md
elif i - 1 >= 0 and hnn[i - 1][0] == 1 or (i + 1 <= len(hnn) - 1 and hnn[i + 1][0] == 1):
ans = ans * mod(2, hnn[i][2], md)
ans = ans % md
else:
ans = ans * mod(2, hnn[i][2] + 1, md)
ans = ans % md
if s1[l - 2] == '2' or s2[l - 2] == '2':
ans *= 0
else:
ans *= 4
ans = ans % md
print(ans % md)
``` | vfc_134309 | {
"difficulty": "hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/CHEFTWOS",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n1111\n2211\n222\n111\n",
"output": "8\n0\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/ARRANGE | Solve the following coding problem using the programming language python:
The Chef has prepared the appetizers in the shapes of letters to spell a special message for the guests. There are n appetizers numbered from 0 to n-1 such that if the appetizers are arrayed in this order, they will display the message. The Chef plans to display them in this order on a table that can be viewed by all guests as they enter. The appetizers will only be served once all guests are seated.
The appetizers are not necessarily finished in the same order as they are numbered. So, when an appetizer is finished the Chef will write the number on a piece of paper and place it beside the appetizer on a counter between the kitchen and the restaurant. A server will retrieve this appetizer and place it in the proper location according to the number written beside it.
The Chef has a penchant for binary numbers. The number of appetizers created is a power of 2, say n = 2^{k}. Furthermore, he has written the number of the appetizer in binary with exactly k bits. That is, binary numbers with fewer than k bits are padded on the left with zeros so they are written with exactly k bits.
Unfortunately, this has unforseen complications. A binary number still "looks" binary when it is written upside down. For example, the binary number "0101" looks like "1010" when read upside down and the binary number "110" looks like "011" (the Chef uses simple vertical lines to denote a 1 bit). The Chef didn't realize that the servers would read the numbers upside down so he doesn't rotate the paper when he places it on the counter. Thus, when the server picks up an appetizer they place it the location indexed by the binary number when it is read upside down.
You are given the message the chef intended to display and you are to display the message that will be displayed after the servers move all appetizers to their locations based on the binary numbers they read.
------ Input ------
The first line consists of a single integer T ≤ 25 indicating the number of test cases to follow. Each test case consists of a single line beginning with an integer 1 ≤ k ≤ 16 followed by a string of precisely 2^{k} characters. The integer and the string are separated by a single space. The string has no spaces and is composed only of lower case letters from a to z.
------ Output ------
For each test case you are to output the scrambled message on a single line.
----- Sample Input 1 ------
2
2 chef
4 enjoyourapplepie
----- Sample Output 1 ------
cehf
eayejpuinpopolre
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def get_binary(num, bits):
ans = []
for i in range(bits):
ans.append('0')
i = bits - 1
while i >= 0:
ans[i] = str(num % 2)
num //= 2
i -= 1
return ''.join(ans)
for t in range(int(input())):
temp = list(input().split())
k = int(temp[0])
s = temp[1]
ans = []
for i in range(2 ** k):
ans.append('0')
for i in range(2 ** k):
bi = get_binary(i, k)
bi = bi[::-1]
x = eval('0b' + bi)
ans[x] = s[i]
print(''.join(ans))
``` | vfc_134313 | {
"difficulty": "medium",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/ARRANGE",
"time_limit": "1.43314 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n2 chef\n4 enjoyourapplepie",
"output": "cehf\neayejpuinpopolre",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/SGARDEN | Solve the following coding problem using the programming language python:
Read problems statements in Mandarin Chinese and Russian.
All bandits are afraid of Sheriff. Sheriff constantly fights crime, but when bandits lay low, he gets bored and starts to entertain himself.
This time Sheriff gathered all the bandits in his garden and ordered them to line up. After the whistle all bandits should change the order in which they stand.
Sheriff gave all the bandits numbers from 1 to N. For each place i he determined the unique position j. After whistling the bandit staying on position i should run to the j-th position. Sheriff loved seeing how the bandits move around, and he continued whistling until the evening. He finished the game only when he noticed that the bandits are in the same order in which they were standing originally.
Now the Sheriff asks the question: How many times has he whistled?
------ Input ------
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains a single integer N denoting the number of bandits. The second line contains N space-separated integers A_{1}, A_{2}, ..., A_{N} denoting that the bandit staying on position i should run to the A_{i}-th position after the whistle.
------ Output ------
For each test case, output a single line containing number of times the sheriff had to whistle, print it modulo 10^{9} + 7.
------ Constraints ------
$1 ≤ T ≤ 5$
$1 ≤ N ≤ 100000$
$All A_{i} are distinct numbers from 1 to N$
----- Sample Input 1 ------
2
3
1 2 3
5
2 3 1 5 4
----- Sample Output 1 ------
1
6
----- explanation 1 ------
Example case 2.
the bandits positions are:
0. 1 2 3 4 5
1. 3 1 2 5 4
2. 2 3 1 4 5
3. 1 2 3 5 4
4. 3 1 2 4 5
5. 2 3 1 5 4
6. 1 2 3 4 5.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import fractions
test = int(input())
for i in range(test):
n = int(input())
array = [int(x) for x in input().split()]
array.insert(0, 0)
count = []
seen = {}
toSee = 0
for i in range(1, n + 1):
j = 1
x = array[i]
if x not in seen:
seen[x] = None
while x != i:
x = array[x]
seen[x] = None
j += 1
count.append(j)
toSee += 1
lcm = 1
for i in range(toSee):
lcm = lcm * count[i] // fractions.gcd(lcm, count[i])
print(lcm % (10 ** 9 + 7))
``` | vfc_134321 | {
"difficulty": "medium_hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/SGARDEN",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n3\n1 2 3\n5\n2 3 1 5 4",
"output": "1\n6",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n2\n1 2 3\n5\n2 3 1 5 4",
"output": "1\n6\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1585/C | Solve the following coding problem using the programming language python:
A total of $n$ depots are located on a number line. Depot $i$ lies at the point $x_i$ for $1 \le i \le n$.
You are a salesman with $n$ bags of goods, attempting to deliver one bag to each of the $n$ depots. You and the $n$ bags are initially at the origin $0$. You can carry up to $k$ bags at a time. You must collect the required number of goods from the origin, deliver them to the respective depots, and then return to the origin to collect your next batch of goods.
Calculate the minimum distance you need to cover to deliver all the bags of goods to the depots. You do not have to return to the origin after you have delivered all the bags.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10\,500$). Description of the test cases follows.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The second line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($-10^9 \le x_i \le 10^9$). It is possible that some depots share the same position.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, output a single integer denoting the minimum distance you need to cover to deliver all the bags of goods to the depots.
-----Examples-----
Input
4
5 1
1 2 3 4 5
9 3
-5 -10 -15 6 5 8 3 7 4
5 3
2 2 3 3 3
4 2
1000000000 1000000000 1000000000 1000000000
Output
25
41
7
3000000000
-----Note-----
In the first test case, you can carry only one bag at a time. Thus, the following is a solution sequence that gives a minimum travel distance: $0 \to 2 \to 0 \to 4 \to 0 \to 3 \to 0 \to 1 \to 0 \to 5$, where each $0$ means you go the origin and grab one bag, and each positive integer means you deliver the bag to a depot at this coordinate, giving a total distance of $25$ units. It must be noted that there are other sequences that give the same distance.
In the second test case, you can follow the following sequence, among multiple such sequences, to travel minimum distance: $0 \to 6 \to 8 \to 7 \to 0 \to 5 \to 4 \to 3 \to 0 \to (-5) \to (-10) \to (-15)$, with distance $41$. It can be shown that $41$ is the optimal distance for this test case.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
input = sys.stdin.readline
def positive_dist(x, n, k, last):
cur = n - 1
total = 0
while cur >= 0 and x[cur] > 0:
total += 2 * x[cur]
cur -= k
if last and x[-1] > 0:
total -= x[-1]
return total
def negative_dist(x, n, k, last):
cur = 0
total = 0
while cur < n and x[cur] < 0:
total -= 2 * x[cur]
cur += k
if last and x[0] < 0:
total += x[0]
return total
t = int(input())
for _ in range(t):
(n, k) = map(int, input().split())
x = [int(i) for i in input().split()]
x.sort()
if x[0] >= 0:
print(positive_dist(x, n, k, True))
elif x[-1] <= 0:
print(negative_dist(x, n, k, True))
elif abs(x[0]) > abs(x[-1]):
print(positive_dist(x, n, k, False) + negative_dist(x, n, k, True))
else:
print(positive_dist(x, n, k, True) + negative_dist(x, n, k, False))
``` | vfc_134325 | {
"difficulty": "easy",
"memory_limit": "256 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1585/C",
"time_limit": "1 second"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n5 1\n1 2 3 4 5\n9 3\n-5 -10 -15 6 5 8 3 7 4\n5 3\n2 2 3 3 3\n4 2\n1000000000 1000000000 1000000000 1000000000\n",
"output": "25\n41\n7\n3000000000\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1272/B | Solve the following coding problem using the programming language python:
Recently you have bought a snow walking robot and brought it home. Suppose your home is a cell $(0, 0)$ on an infinite grid.
You also have the sequence of instructions of this robot. It is written as the string $s$ consisting of characters 'L', 'R', 'U' and 'D'. If the robot is in the cell $(x, y)$ right now, he can move to one of the adjacent cells (depending on the current instruction). If the current instruction is 'L', then the robot can move to the left to $(x - 1, y)$; if the current instruction is 'R', then the robot can move to the right to $(x + 1, y)$; if the current instruction is 'U', then the robot can move to the top to $(x, y + 1)$; if the current instruction is 'D', then the robot can move to the bottom to $(x, y - 1)$.
You've noticed the warning on the last page of the manual: if the robot visits some cell (except $(0, 0)$) twice then it breaks.
So the sequence of instructions is valid if the robot starts in the cell $(0, 0)$, performs the given instructions, visits no cell other than $(0, 0)$ two or more times and ends the path in the cell $(0, 0)$. Also cell $(0, 0)$ should be visited at most two times: at the beginning and at the end (if the path is empty then it is visited only once). For example, the following sequences of instructions are considered valid: "UD", "RL", "UUURULLDDDDLDDRRUU", and the following are considered invalid: "U" (the endpoint is not $(0, 0)$) and "UUDD" (the cell $(0, 1)$ is visited twice).
The initial sequence of instructions, however, might be not valid. You don't want your robot to break so you decided to reprogram it in the following way: you will remove some (possibly, all or none) instructions from the initial sequence of instructions, then rearrange the remaining instructions as you wish and turn on your robot to move.
Your task is to remove as few instructions from the initial sequence as possible and rearrange the remaining ones so that the sequence is valid. Report the valid sequence of the maximum length you can obtain.
Note that you can choose any order of remaining instructions (you don't need to minimize the number of swaps or any other similar metric).
You have to answer $q$ independent test cases.
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 2 \cdot 10^4$) — the number of test cases.
The next $q$ lines contain test cases. The $i$-th test case is given as the string $s$ consisting of at least $1$ and no more than $10^5$ characters 'L', 'R', 'U' and 'D' — the initial sequence of instructions.
It is guaranteed that the sum of $|s|$ (where $|s|$ is the length of $s$) does not exceed $10^5$ over all test cases ($\sum |s| \le 10^5$).
-----Output-----
For each test case print the answer on it. In the first line print the maximum number of remaining instructions. In the second line print the valid sequence of remaining instructions $t$ the robot has to perform. The moves are performed from left to right in the order of the printed sequence. If there are several answers, you can print any. If the answer is $0$, you are allowed to print an empty line (but you can don't print it).
-----Example-----
Input
6
LRU
DURLDRUDRULRDURDDL
LRUDDLRUDRUL
LLLLRRRR
URDUR
LLL
Output
2
LR
14
RUURDDDDLLLUUR
12
ULDDDRRRUULL
2
LR
2
UD
0
-----Note-----
There are only two possible answers in the first test case: "LR" and "RL".
The picture corresponding to the second test case: [Image] Note that the direction of traverse does not matter
Another correct answer to the third test case: "URDDLLLUURDR".
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
t = int(input())
while t:
s = input()
(l, r, u, d) = (0, 0, 0, 0)
for i in s:
if i == 'U':
u += 1
if i == 'D':
d += 1
if i == 'R':
r += 1
if i == 'L':
l += 1
mlr = min(l, r)
mud = min(u, d)
l = mlr
r = mlr
u = mud
d = mud
if mud != 0 and mlr == 0:
print(2)
print('UD')
elif mlr != 0 and mud == 0:
print(2)
print('LR')
elif mlr == 0 and mud == 0:
print(0)
print()
else:
print(l + r + u + d)
print('L' * l + 'U' * u + 'R' * r + 'D' * d)
t -= 1
``` | vfc_134333 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1272/B",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "6\nLRU\nDURLDRUDRULRDURDDL\nLRUDDLRUDRUL\nLLLLRRRR\nURDUR\nLLL\n",
"output": "2\nLR\n14\nLLLUUUURRRDDDD\n12\nLLLUUURRRDDD\n2\nLR\n2\nUD\n0\n\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\nLRU\nDURLDRUDRULRDURDDL\nLRUDDLRUDRUL\nLLLLRRRR\nURDUR\nLLL\n",
"output": "2\nLR\n14\nLLLUUUURRRDDDD\n12\nLLLUUURRRDDD\n2\nLR\n2\nUD\n0\n\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\nLRU\nDURLDRUDRULRDURDDL\nLURDURLDDURL\nLLLLRRRR\nURDUR\nLLL\n",
"output": "2\nLR\n14\nLLLUUUURRRDDDD\n12\nLLLUUURRRDDD\n2\nLR\n2\nUD\n0\n\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\nLRU\nLDDRUDRLURDURDLRUD\nLURDURLDDURL\nLLLLRRRR\nURDUR\nLLL\n",
"output": "2\nLR\n14\nLLLUUUURRRDDDD\n12\nLLLUUURRRDDD\n2\nLR\n2\nUD\n0\n\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/MGCRNK | Solve the following coding problem using the programming language python:
Everybody loves magic, especially magicians who compete for glory on the Byteland Magic Tournament. Magician Cyael is one such magician.
Cyael has been having some issues with her last performances and today she’ll have to perform for an audience of some judges, who will change her tournament ranking, possibly increasing it. As she is a great magician she managed to gather a description of the fixed judges’ disposition on the room (which is represented as an N × N square matrix), such that she knows in advance the fixed points each judge will provide. She also knows that the room is divided into several parallel corridors, such that we will denote the j-th cell on corridor i, as [i][j]. Note that some judges can award Cyael, zero points or negative points, as they are never pleased with her performance. There is just one judge at each cell of the matrix, except the cells [1][1] and [N][N].
To complete her evaluation, she must start on the top leftmost corner of the room (cell [1][1]), and finish on the bottom right corner (cell [N][N]), moving either to the cell directly in front of her on the same corridor (that is, moving from cell [r][c] to cell [r][c+1], where c+1 ≤ N) or to the cell in the next corridor directly in front of where she is (that is, moving from cell [r][c] to cell [r+1][c], where r+1 ≤ N). She will keep doing this until she reaches the end point of the room, i.e. last cell [N][N] on the last corridor. Cyael will be judged at all visited cells with a judge.
Cyael wants to maximize her average score at end of her performance. More specifically, if she passes K judges, each being on cell [i_{1}][j_{1}], cell [i_{2}][j_{2}], ..., cell [i_{K}][j_{K}] respectively, then she wants to maximize (S[i_{1}][j_{1}] + S[i_{2}][j_{2}] + ... + S[i_{K}][j_{K}]) / K, where S[i][j] denotes the points that the judge will give her on the cell [i][j].
Help her determine the best path she has to follow in order to maximize her average points.
------ Input ------
The first line contains a single integer T denoting the number of test cases. The description for T test cases follows. For each test case, the first line contains a single integer N. Each of the next N lines contains N space-separated integers.
The j-th integer S[i][j] in i-th line denotes the points awarded by the judge at cell [i][j].
Note that the cells [1][1] and [N][N] have no judges, so S[1][1] and S[N][N] will be 0.
------ Output ------
For each test case, if the maximum possible average points Cyael can obtain is negative, output a single line containing "Bad Judges" (quotes for clarity). Otherwise, output the maximum possible average points. The answer will be considered correct if it has an absolute error no more than 10^{-6}.
------ Constraints ------
1 ≤ T ≤ 20
2 ≤ N ≤ 100
-2500 ≤ S[i][j] ≤ 2500
S[1][1] = S[N][N] = 0
Your code will be judged against several input files.
------ Example ------
----- Sample Input 1 ------
2
2
0 -4
8 0
2
0 -45
-3 0
----- Sample Output 1 ------
8.000000
Bad Judges
----- explanation 1 ------
Test case $1$: An optimal path for Cyael would be $(1,1)\rightarrow (2,1)\rightarrow (2,2)$. This way Cyael faces $1$ judge and gets a total score of $8$. Thus, the average score is $\frac{8}{1} = 8$.
Test case $2$: No matter what path Cyael chooses, the final score would be less than $0$.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import math
import sys
import string
t = int(input())
while t:
n = int(input())
arr = [0] * n
arr[0] = [int(x) for x in input().split()]
for i in range(1, n):
arr[0][i] += arr[0][i - 1]
for i in range(1, n):
arr[i] = [int(x) for x in input().split()]
arr[i][0] += arr[i - 1][0]
for j in range(1, n):
arr[i][j] += max(arr[i][j - 1], arr[i - 1][j])
if arr[n - 1][n - 1] < 0:
print('Bad Judges')
else:
print('%.6f' % (arr[n - 1][n - 1] / (2 * n - 3)))
t -= 1
``` | vfc_134337 | {
"difficulty": "medium",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/MGCRNK",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n2\n0 -4\n8 0\n2\n0 -45\n-3 0",
"output": "8.000000\nBad Judges",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n2\n0 -4\n8 0\n2\n0 -45\n-2 0",
"output": "8.0000000\nBad Judges\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/largest-number-in-one-swap1520/1 | Solve the following coding problem using the programming language python:
Given a non-negative number N in the form of a string.Find the largest number that can be formed by swapping two characters at most once.
Example 1:
Input:
N=768
Output:
867
Explanation:
Swapping the 1st and 3rd
characters(7 and 8 respectively),
gives the largest possible number.
Example 2:
Input:
N=333
Output:
333
Explanation:
Performing any swaps gives the
same result i.e 333.
Your Task:
You don't need to read input or print anything. Your task is to complete the function LargestSwap() which takes the string N as input parameter and returns the maximum possible string possible with at most one swap.
Expected Time Complexity:O(|N|)
Expected Auxillary Space:O(1)
Constraints:
1<=|N|<=10^{5}
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def LargestSwap(self,S):
#code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def LargestSwap(self, S):
n = len(S)
arr = list(S)
right = [0] * n
right[n - 1] = n - 1
rmax = n - 1
for i in range(n - 2, -1, -1):
right[i] = rmax
if arr[i] > arr[rmax]:
rmax = i
for i in range(0, n):
if arr[right[i]] > arr[i]:
(arr[right[i]], arr[i]) = (arr[i], arr[right[i]])
break
return ''.join(arr)
``` | vfc_134341 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/largest-number-in-one-swap1520/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N=768",
"output": "867",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N=333",
"output": "333",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/261/B | Solve the following coding problem using the programming language python:
Maxim has opened his own restaurant! The restaurant has got a huge table, the table's length is p meters.
Maxim has got a dinner party tonight, n guests will come to him. Let's index the guests of Maxim's restaurant from 1 to n. Maxim knows the sizes of all guests that are going to come to him. The i-th guest's size (a_{i}) represents the number of meters the guest is going to take up if he sits at the restaurant table.
Long before the dinner, the guests line up in a queue in front of the restaurant in some order. Then Maxim lets the guests in, one by one. Maxim stops letting the guests in when there is no place at the restaurant table for another guest in the queue. There is no place at the restaurant table for another guest in the queue, if the sum of sizes of all guests in the restaurant plus the size of this guest from the queue is larger than p. In this case, not to offend the guest who has no place at the table, Maxim doesn't let any other guest in the restaurant, even if one of the following guests in the queue would have fit in at the table.
Maxim is now wondering, what is the average number of visitors who have come to the restaurant for all possible n! orders of guests in the queue. Help Maxim, calculate this number.
-----Input-----
The first line contains integer n (1 ≤ n ≤ 50) — the number of guests in the restaurant. The next line contains integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 50) — the guests' sizes in meters. The third line contains integer p (1 ≤ p ≤ 50) — the table's length in meters.
The numbers in the lines are separated by single spaces.
-----Output-----
In a single line print a real number — the answer to the problem. The answer will be considered correct, if the absolute or relative error doesn't exceed 10^{ - 4}.
-----Examples-----
Input
3
1 2 3
3
Output
1.3333333333
-----Note-----
In the first sample the people will come in the following orders: (1, 2, 3) — there will be two people in the restaurant; (1, 3, 2) — there will be one person in the restaurant; (2, 1, 3) — there will be two people in the restaurant; (2, 3, 1) — there will be one person in the restaurant; (3, 1, 2) — there will be one person in the restaurant; (3, 2, 1) — there will be one person in the restaurant.
In total we get (2 + 1 + 2 + 1 + 1 + 1) / 6 = 8 / 6 = 1.(3).
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
a = list(map(int, input().split()))
p = int(input())
fact = [1]
for i in range(1, 51):
fact.append(fact[-1] * i)
if sum(a) <= p:
print(n)
else:
dp = [[0] * 56 for _ in range(56)]
dp[0][0] = 1
for i in range(n):
for j in range(n, -1, -1):
for k in range(p):
if a[i] + k <= p:
dp[j + 1][a[i] + k] += dp[j][k]
tot = sum((dp[i][j] * fact[i] * fact[n - i] for j in range(1, p + 1) for i in range(1, n + 1)))
print(f'{tot / fact[n]:.9f}')
``` | vfc_134342 | {
"difficulty": "hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/261/B",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n1 2 3\n3\n",
"output": "1.3333333333\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "9\n2 2 2 2 2 2 2 1 2\n9\n",
"output": "4.5555555556\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "7\n2 1 1 2 1 1 2\n2\n",
"output": "1.2857142857\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "8\n3 1 5 6 1 5 4 4\n7\n",
"output": "1.6250000000\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n1 3\n3\n",
"output": "1.0000000000\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n1 2\n2\n",
"output": "1.0000000000\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/26/E | Solve the following coding problem using the programming language python:
You are given the following concurrent program. There are N processes and the i-th process has the following pseudocode:
repeat ni times
yi := y
y := yi + 1
end repeat
Here y is a shared variable. Everything else is local for the process. All actions on a given row are atomic, i.e. when the process starts executing a row it is never interrupted. Beyond that all interleavings are possible, i.e. every process that has yet work to do can be granted the rights to execute its next row. In the beginning y = 0. You will be given an integer W and ni, for i = 1, ... , N. Determine if it is possible that after all processes terminate, y = W, and if it is possible output an arbitrary schedule that will produce this final value.
Input
In the first line of the input you will be given two space separated integers N (1 ≤ N ≤ 100) and W ( - 109 ≤ W ≤ 109). In the second line there are N space separated integers ni (1 ≤ ni ≤ 1000).
Output
On the first line of the output write Yes if it is possible that at the end y = W, or No otherwise. If the answer is No then there is no second line, but if the answer is Yes, then on the second line output a space separated list of integers representing some schedule that leads to the desired result. For more information see note.
Examples
Input
1 10
11
Output
No
Input
2 3
4 4
Output
Yes
1 1 2 1 2 2 2 2 2 1 2 1 1 1 1 2
Input
3 6
1 2 3
Output
Yes
1 1 2 2 2 2 3 3 3 3 3 3
Note
For simplicity, assume that there is no repeat statement in the code of the processes, but the code from the loop is written the correct amount of times. The processes are numbered starting from 1. The list of integers represent which process works on its next instruction at a given step. For example, consider the schedule 1 2 2 1 3. First process 1 executes its first instruction, then process 2 executes its first two instructions, after that process 1 executes its second instruction, and finally process 3 executes its first instruction. The list must consists of exactly 2·Σ i = 1...N ni numbers.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
from array import array
def input():
return sys.stdin.buffer.readline().decode('utf-8')
(n, w) = map(int, input().split())
a = [0] + list(map(int, input().split()))
total = sum(a)
def ng():
print('No')
exit()
def ok(a):
print('Yes')
print(*a)
exit()
if w < 1 or total < w:
ng()
if n == 1:
if w == a[1]:
ok([1] * (a[1] * 2))
else:
ng()
if w == 1:
if min(a[1:]) > 1:
ng()
min_i = a.index(1)
ans = []
for i in range(1, n + 1):
if i == min_i:
continue
ans += [i] * (a[i] * 2)
ok([min_i] + ans + [min_i])
(ans1, ans2, ans3) = ([], [], [])
w -= 2
if w:
for i in range(1, 3):
x = min(a[i] - 1, w)
w -= x
a[i] -= x
ans3 += [i] * (2 * x)
for i in range(3, n + 1):
x = min(a[i], w)
w -= x
a[i] -= x
ans3 += [i] * (2 * x)
ans1 = [2] * ((a[2] - 1) * 2)
for i in range(3, n + 1):
ans1 += [i] * (a[i] * 2)
ans1 = [1] + ans1 + [1]
a[1] -= 1
ans2 = [2] + [1] * (a[1] * 2) + [2]
if w == 0:
ok(ans1 + ans2 + ans3)
else:
ng()
``` | vfc_134346 | {
"difficulty": "very_hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/26/E",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5 1\n2 3 4 5 6\n",
"output": "No\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/TRUESCORE | Solve the following coding problem using the programming language python:
Chef is watching a football match. The current score is A:B, that is, team 1 has scored A goals and team 2 has scored B goals. Chef wonders if it is possible for the score to become C:D at a later point in the game (i.e. team 1 has scored C goals and team 2 has scored D goals). Can you help Chef by answering his question?
------ Input Format ------
- The first line contains a single integer T - the number of test cases. Then the test cases follow.
- The first line of each test case contains two integers A and B - the intial number of goals team 1 and team 2 have scored respectively.
- The second line of each test case contains two integers C and D - the final number of goals team 1 and team 2 must be able to score respectively.
------ Output Format ------
For each testcase, output POSSIBLE if it is possible for the score to become C:D at a later point in the game, IMPOSSIBLE otherwise.
You may print each character of POSSIBLE and IMPOSSIBLE in uppercase or lowercase (for example, possible, pOSsiBLe, Possible will be considered identical).
------ Constraints ------
$1 ≤ T ≤ 1000$
$0 ≤ A,B,C,D ≤ 10$
----- Sample Input 1 ------
3
1 5
3 5
3 4
2 6
2 2
2 2
----- Sample Output 1 ------
POSSIBLE
IMPOSSIBLE
POSSIBLE
----- explanation 1 ------
Test case 1: The current score is $1:5$. If team $1$ scores $2$ more goals, the score will become $3:5$. Thus $3:5$ is a possible score.
Test case 2: The current score is $3:4$. It can be proven that no non-negative pair of integers $(x, y)$ exists such that if team $1$ scores $x$ more goals and team $2$ scores $y$ more goals the score becomes $2:6$ from $3:4$. Thus in this case $2:6$ is an impossible score.
Test case 3: The current score is already $2:2$. Hence it is a possible score.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for _ in range(int(input())):
(a, b) = map(int, input().split())
(c, d) = map(int, input().split())
if c >= a and d >= b:
print('POSSIBLE')
else:
print('IMPOSSIBLE')
``` | vfc_134350 | {
"difficulty": "easy",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/TRUESCORE",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n1 5\n3 5\n3 4\n2 6\n2 2\n2 2\n",
"output": "POSSIBLE\nIMPOSSIBLE\nPOSSIBLE\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/180/E | Solve the following coding problem using the programming language python:
Let's imagine that you're playing the following simple computer game. The screen displays n lined-up cubes. Each cube is painted one of m colors. You are allowed to delete not more than k cubes (that do not necessarily go one after another). After that, the remaining cubes join together (so that the gaps are closed) and the system counts the score. The number of points you score equals to the length of the maximum sequence of cubes of the same color that follow consecutively. Write a program that determines the maximum possible number of points you can score.
Remember, you may delete no more than k any cubes. It is allowed not to delete cubes at all.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 2·105, 1 ≤ m ≤ 105, 0 ≤ k < n). The second line contains n integers from 1 to m — the numbers of cube colors. The numbers of colors are separated by single spaces.
Output
Print the maximum possible number of points you can score.
Examples
Input
10 3 2
1 2 1 1 3 2 1 1 2 2
Output
4
Input
10 2 2
1 2 1 2 1 1 2 1 1 2
Output
5
Input
3 1 2
1 1 1
Output
3
Note
In the first sample you should delete the fifth and the sixth cubes.
In the second sample you should delete the fourth and the seventh cubes.
In the third sample you shouldn't delete any cubes.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
(self.buffer.truncate(0), self.buffer.seek(0))
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda : self.buffer.read().decode('ascii')
self.readline = lambda : self.buffer.readline().decode('ascii')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
input = lambda : sys.stdin.readline().rstrip('\r\n')
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError('Out of ranges')
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
avl = AvlTree()
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a, b)):
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return 'SegmentTree({0})'.format(self.data)
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return 'SegmentTree({0})'.format(self.data)
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print('Invalid argument to calculate n!')
print('n must be non-negative value. But the argument was ' + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print('Invalid argument to calculate n^(-1)')
print('n must be non-negative value. But the argument was ' + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print('Invalid argument to calculate (n^(-1))!')
print('n must be non-negative value. But the argument was ' + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n)
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * self.invModulos[i % p] % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for (_, x) in sorted(zipped_pairs)]
return z
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while left <= right:
mid = int((right + left) / 2)
if arr[mid] <= key:
count = mid + 1
left = mid + 1
else:
right = mid - 1
return count
def countdig(n):
c = 0
while n > 0:
n //= 10
c += 1
return c
def countGreater(arr, n, k):
l = 0
r = n - 1
leftGreater = n
while l <= r:
m = int(l + (r - l) / 2)
if arr[m] >= k:
leftGreater = m
r = m - 1
else:
l = m + 1
return n - leftGreater
(n, m, k) = map(int, input().split())
l = list(map(int, input().split()))
ans = 1
ind = defaultdict(list)
for i in range(n):
ind[l[i]].append(i)
cop = k
for i in ind:
if len(ind[i]) == 1:
continue
st = 0
r = 0
k = cop
while r < len(ind[i]) - 1:
if k >= 0:
ans = max(ans, r - st + 1)
r += 1
k -= ind[i][r] - ind[i][r - 1] - 1
else:
st += 1
k += ind[i][st] - ind[i][st - 1] - 1
if k >= 0:
ans = max(ans, r - st + 1)
print(ans)
``` | vfc_134354 | {
"difficulty": "medium_hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/180/E",
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "30 5 6\n4 2 2 1 3 4 2 3 2 4 3 1 1 4 4 3 5 1 4 5 5 1 2 2 1 2 4 4 1 2\n",
"output": "4",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "20 6 3\n4 1 2 6 3 3 2 5 2 5 2 1 1 4 1 2 2 1 1 4\n",
"output": "5",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "100 5 10\n4 4 4 4 4 4 4 4 4 4 4 4 2 4 4 4 4 4 4 4 4 4 2 2 2 3 3 3 3 3 3 3 4 4 4 3 3 2 1 1 1 2 3 3 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 4 4 4 4 5 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 5 5 5 3 3 4 3 3 3\n",
"output": "21",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 2 2\n1 1 1 2 1 2 1 2 1 1\n",
"output": "5",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1158/C | Solve the following coding problem using the programming language python:
Vasya has written some permutation $p_1, p_2, \ldots, p_n$ of integers from $1$ to $n$, so for all $1 \leq i \leq n$ it is true that $1 \leq p_i \leq n$ and all $p_1, p_2, \ldots, p_n$ are different. After that he wrote $n$ numbers $next_1, next_2, \ldots, next_n$. The number $next_i$ is equal to the minimal index $i < j \leq n$, such that $p_j > p_i$. If there is no such $j$ let's let's define as $next_i = n + 1$.
In the evening Vasya went home from school and due to rain, his notebook got wet. Now it is impossible to read some written numbers. Permutation and some values $next_i$ are completely lost! If for some $i$ the value $next_i$ is lost, let's say that $next_i = -1$.
You are given numbers $next_1, next_2, \ldots, next_n$ (maybe some of them are equal to $-1$). Help Vasya to find such permutation $p_1, p_2, \ldots, p_n$ of integers from $1$ to $n$, that he can write it to the notebook and all numbers $next_i$, which are not equal to $-1$, will be correct.
-----Input-----
The first line contains one integer $t$ — the number of test cases ($1 \leq t \leq 100\,000$).
Next $2 \cdot t$ lines contains the description of test cases,two lines for each. The first line contains one integer $n$ — the length of the permutation, written by Vasya ($1 \leq n \leq 500\,000$). The second line contains $n$ integers $next_1, next_2, \ldots, next_n$, separated by spaces ($next_i = -1$ or $i < next_i \leq n + 1$).
It is guaranteed, that the sum of $n$ in all test cases doesn't exceed $500\,000$.
In hacks you can only use one test case, so $T = 1$.
-----Output-----
Print $T$ lines, in $i$-th of them answer to the $i$-th test case.
If there is no such permutations $p_1, p_2, \ldots, p_n$ of integers from $1$ to $n$, that Vasya could write, print the only number $-1$.
In the other case print $n$ different integers $p_1, p_2, \ldots, p_n$, separated by spaces ($1 \leq p_i \leq n$). All defined values of $next_i$ which are not equal to $-1$ should be computed correctly $p_1, p_2, \ldots, p_n$ using defenition given in the statement of the problem. If there exists more than one solution you can find any of them.
-----Example-----
Input
6
3
2 3 4
2
3 3
3
-1 -1 -1
3
3 4 -1
1
2
4
4 -1 4 5
Output
1 2 3
2 1
2 1 3
-1
1
3 2 1 4
-----Note-----
In the first test case for permutation $p = [1, 2, 3]$ Vasya should write $next = [2, 3, 4]$, because each number in permutation is less than next. It's easy to see, that it is the only satisfying permutation.
In the third test case, any permutation can be the answer because all numbers $next_i$ are lost.
In the fourth test case, there is no satisfying permutation, so the answer is $-1$.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
input = sys.stdin.readline
T = int(input())
for _ in range(T):
n = int(input())
l = list(map(int, input().split()))
stack = []
out = [-1] * n
curr = 0
works = True
for i in range(n):
while stack and stack[-1][0] == i:
(_, j) = stack.pop()
curr += 1
out[j] = curr
nex = l[i] - 1
if nex == -2:
curr += 1
out[i] = curr
elif stack and nex > stack[-1][0]:
works = False
else:
stack.append((nex, i))
while stack:
(_, j) = stack.pop()
curr += 1
out[j] = curr
if works:
print(*out)
else:
print(-1)
``` | vfc_134358 | {
"difficulty": "hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1158/C",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "6\n3\n2 3 4\n2\n3 3\n3\n-1 -1 -1\n3\n3 4 -1\n1\n2\n4\n4 -1 4 5\n",
"output": "1 2 3 \n2 1 \n1 2 3 \n-1\n1 \n3 1 2 4 \n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/modified-fibonacci4449/1 | Solve the following coding problem using the programming language python:
Given 4 integers A, B, C and N, find the value of F(N) such that
F(1) = A + B
F(2) = B + C
F(N) = F(N-1) - F(N-2), for N > 2.
Example 1:
Input: N = 2, A = 2, B = 3, C = 4
Output: 7
Explaination: F(2) = B+C = 3+4 = 7
Example 2:
Input: N = 3, A = 2, B = 3, C = 4
Output: 2
Explaination: F(3) = F(2)- F(1) = 7-5 = 2
Your Task:
You do not need to read input or print anything. Your task is to complete the function modifiedFib() which takes the values N, A, B and C as input parameters and returns F(N). Since F(N) modulo (10^{9}+7).
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{12}
1 ≤ A, B, C ≤ 10^{9}
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def modifiedFib(self, N, A, B, C):
# code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def modifiedFib(self, N, A, B, C):
(a, b, c) = (A + B, B + C, C - A)
n = N % 6
if n == 1:
return a % 1000000007
if n == 2:
return b % 1000000007
if n == 3:
return c % 1000000007
if n == 4:
return -a % 1000000007
if n == 5:
return -b % 1000000007
if n == 0:
return -c % 1000000007
``` | vfc_134368 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/modified-fibonacci4449/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 2, A = 2, B = 3, C = 4",
"output": "7",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N = 3, A = 2, B = 3, C = 4",
"output": "2",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/CHEFHOME | Solve the following coding problem using the programming language python:
Our hardworking chef is bored of sleeping in his restaurants. He has decided to settle down. The first thing he must do is to find a suitable location to build a palatial home.
Think of the city as a two-dimensional grid. There are N restaurants in the city. Each of the chef's restaurant is a point denoted by (X , Y). A house can be located at a grid point (R, S) if the sum of the distances between this point and each of the restaurants is as small as possible. Find the number of possible house locations in the city to help out chef build a home.
More than one restaurant can be located at the same point.
Houses and restaurants can be located at the same point.
Every house must have integer co-ordinates. In other words, R and S are integers.
The distance between two points (A,B) and (C,D) is |A-C| + |B-D|. Here |X| is the absolute function.
------ Input ------
First line in the input contains T, number of test cases.
First line of each test case contains N, number of restaurants.
Each of the next N lines contain two integers X and Y separated by a space.
T ≤ 100
N ≤ 10^{3}
-10^{8} ≤ X ≤10^{8}
-10^{8} ≤ Y ≤10^{8}
------ Output ------
The number of possible locations (grid points) where houses can be built.
----- Sample Input 1 ------
3
5
0 0
-1 0
1 0
0 1
0 -1
5
31 11
30 -41
20 14
25 18
25 38
2
0 0
1 1
----- Sample Output 1 ------
1
1
4
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def findHouseLocations():
t = int(input())
for j in range(t):
n = int(input())
x = []
y = []
for i in range(n):
m = list(map(int, input().split()))
x.append(m[0])
y.append(m[1])
x.sort()
y.sort()
x1 = x[n // 2] - x[(n - 1) // 2] + 1
y1 = y[n // 2] - y[(n - 1) // 2] + 1
print(x1 * y1)
findHouseLocations()
``` | vfc_134369 | {
"difficulty": "medium_hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/CHEFHOME",
"time_limit": "0.352542 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n5\n0 0\n-1 0\n1 0\n0 1\n0 -1\n5\n31 11\n30 -41\n20 14\n25 18\n25 38\n2\n0 0\n1 1",
"output": "1\n1\n4",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n5\n0 0\n-1 0\n1 0\n0 1\n0 -1\n5\n31 11\n30 -41\n20 15\n25 18\n25 38\n2\n0 0\n1 1",
"output": "1\n1\n4\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n5\n0 0\n0 0\n1 0\n0 1\n0 -1\n5\n31 3\n30 -41\n20 15\n25 18\n25 38\n1\n0 0\n1 1",
"output": "1\n1\n1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n5\n0 0\n-1 0\n1 1\n0 1\n0 -1\n5\n31 11\n52 -41\n20 4\n25 18\n25 60\n2\n0 0\n1 2",
"output": "1\n1\n6\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n5\n-1 0\n-1 0\n1 2\n0 1\n0 -1\n5\n31 11\n52 -41\n18 1\n25 18\n25 60\n2\n0 0\n1 3",
"output": "1\n1\n8\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n5\n0 0\n-1 0\n1 1\n0 1\n0 -1\n5\n31 11\n30 -41\n20 8\n25 18\n25 38\n2\n0 0\n1 0",
"output": "1\n1\n2\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1419/D2 | Solve the following coding problem using the programming language python:
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
arr = list(map(int, input().split()))
flag = 1
if n <= 2:
print(0)
print(*arr)
flag = 0
arr.sort(reverse=True)
final = ['0']
pair = 0
if flag == 1:
k = 0
j = 1
if n % 2 != 0:
temp1 = arr[:n // 2 + 1]
temp2 = arr[n // 2 + 1:]
else:
temp1 = arr[:n // 2]
temp2 = arr[n // 2:]
rem_val = temp2.pop(0)
for i in range(0, len(temp2)):
if temp1[k] == temp2[i] or temp1[j] == temp2[i]:
pass
else:
pair += 1
final.pop(-1)
final.append(temp1[k])
final.append(temp2[i])
final.append(temp1[j])
k += 1
j += 1
if n % 2 == 0:
final.append(rem_val)
if flag == 1:
print(pair)
print(*final)
``` | vfc_134373 | {
"difficulty": "medium",
"memory_limit": "256 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1419/D2",
"time_limit": "1 second"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "7\n1 3 2 2 4 5 4\n",
"output": "3\n3 1 4 2 4 2 5 ",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1\n1\n",
"output": "0\n1 ",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n1000000000 1000000000\n",
"output": "0\n1000000000 1000000000 ",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5\n1 2 2 4 5\n",
"output": "2\n2 1 4 2 5 ",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/812/E | Solve the following coding problem using the programming language python:
Sagheer is playing a game with his best friend Soliman. He brought a tree with n nodes numbered from 1 to n and rooted at node 1. The i-th node has a_{i} apples. This tree has a special property: the lengths of all paths from the root to any leaf have the same parity (i.e. all paths have even length or all paths have odd length).
Sagheer and Soliman will take turns to play. Soliman will make the first move. The player who can't make a move loses.
In each move, the current player will pick a single node, take a non-empty subset of apples from it and do one of the following two things: eat the apples, if the node is a leaf. move the apples to one of the children, if the node is non-leaf.
Before Soliman comes to start playing, Sagheer will make exactly one change to the tree. He will pick two different nodes u and v and swap the apples of u with the apples of v.
Can you help Sagheer count the number of ways to make the swap (i.e. to choose u and v) after which he will win the game if both players play optimally? (u, v) and (v, u) are considered to be the same pair.
-----Input-----
The first line will contain one integer n (2 ≤ n ≤ 10^5) — the number of nodes in the apple tree.
The second line will contain n integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^7) — the number of apples on each node of the tree.
The third line will contain n - 1 integers p_2, p_3, ..., p_{n} (1 ≤ p_{i} ≤ n) — the parent of each node of the tree. Node i has parent p_{i} (for 2 ≤ i ≤ n). Node 1 is the root of the tree.
It is guaranteed that the input describes a valid tree, and the lengths of all paths from the root to any leaf will have the same parity.
-----Output-----
On a single line, print the number of different pairs of nodes (u, v), u ≠ v such that if they start playing after swapping the apples of both nodes, Sagheer will win the game. (u, v) and (v, u) are considered to be the same pair.
-----Examples-----
Input
3
2 2 3
1 1
Output
1
Input
3
1 2 3
1 1
Output
0
Input
8
7 2 2 5 4 3 1 1
1 1 1 4 4 5 6
Output
4
-----Note-----
In the first sample, Sagheer can only win if he swapped node 1 with node 3. In this case, both leaves will have 2 apples. If Soliman makes a move in a leaf node, Sagheer can make the same move in the other leaf. If Soliman moved some apples from a root to a leaf, Sagheer will eat those moved apples. Eventually, Soliman will not find a move.
In the second sample, There is no swap that will make Sagheer win the game.
Note that Sagheer must make the swap even if he can win with the initial tree.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def coloring(i, ancestors, color):
while i != 0 and color[ancestors[i - 1]] is None:
color[ancestors[i - 1]] = not color[i]
i = ancestors[i - 1]
def main():
n = int(input())
a = list(map(int, input().split()))
ancestors = list(map(lambda x: int(x) - 1, input().split()))
descendants = [[] for i in range(n)]
for i in range(n - 1):
descendants[ancestors[i]].append(i + 1)
color = [None for i in range(n)]
for i in range(n):
if not descendants[i]:
color[i] = True
coloring(i, ancestors, color)
reds = 0
blues = 0
xor = 0
count_red = dict()
count_blue = dict()
for i in range(n):
if color[i]:
blues += 1
xor ^= a[i]
if str(a[i]) in count_blue:
count_blue[str(a[i])] += 1
else:
count_blue[str(a[i])] = 1
else:
reds += 1
if str(a[i]) in count_red:
count_red[str(a[i])] += 1
else:
count_red[str(a[i])] = 1
res = 0
if xor == 0:
res += (blues - 1) * blues // 2
res += (reds - 1) * reds // 2
for i in count_blue.items():
if i[0] in count_red:
res += i[1] * count_red[i[0]]
else:
for i in count_blue.items():
if str(xor ^ int(i[0])) in count_red:
res += i[1] * count_red[str(xor ^ int(i[0]))]
print(res)
main()
``` | vfc_134377 | {
"difficulty": "hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/812/E",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n2 2 3\n1 1\n",
"output": "1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n1 2 3\n1 1\n",
"output": "0\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "8\n7 2 2 5 4 3 1 1\n1 1 1 4 4 5 6\n",
"output": "4\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\n7 7 7 7 7 7\n1 1 1 1 1\n",
"output": "0\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/687/A | Solve the following coding problem using the programming language python:
Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.
Suppose the graph G is given. Subset A of its vertices is called a vertex cover of this graph, if for each edge uv there is at least one endpoint of it in this set, i.e. $u \in A$ or $v \in A$ (or both).
Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.
They have agreed to give you their graph and you need to find two disjoint subsets of its vertices A and B, such that both A and B are vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).
-----Input-----
The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — the number of vertices and the number of edges in the prize graph, respectively.
Each of the next m lines contains a pair of integers u_{i} and v_{i} (1 ≤ u_{i}, v_{i} ≤ n), denoting an undirected edge between u_{i} and v_{i}. It's guaranteed the graph won't contain any self-loops or multiple edges.
-----Output-----
If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).
If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer k denoting the number of vertices in that vertex cover, and the second line contains k integers — the indices of vertices. Note that because of m ≥ 1, vertex cover cannot be empty.
-----Examples-----
Input
4 2
1 2
2 3
Output
1
2
2
1 3
Input
3 3
1 2
2 3
1 3
Output
-1
-----Note-----
In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to Pari and keep vertex number 4 for yourself (or give it someone, if you wish).
In the second sample, there is no way to satisfy both Pari and Arya.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from collections import defaultdict
(n, m) = list(map(int, input().strip().split()))
g = defaultdict(list)
for _ in range(m):
(u, v) = list(map(int, input().strip().split()))
g[u].append(v)
g[v].append(u)
v1 = []
v2 = []
mark = [-1] * (n + 1)
def bfs(start):
q = []
q.append(start)
mark[start] = 1
v2.append(start)
while q:
v = q.pop(0)
for u in g[v]:
if mark[u] == -1:
mark[u] = 1 - mark[v]
q.append(u)
if mark[u] == 0:
v1.append(u)
else:
v2.append(u)
elif mark[u] == mark[v]:
return -1
return 0
sw = False
for i in range(1, n + 1):
if mark[i] == -1 and len(g[i]):
res = bfs(i)
try:
if res == -1:
sw = True
break
except:
continue
if sw:
print(-1)
else:
print(len(v1))
print(' '.join((str(x) for x in v1)))
print(len(v2))
print(' '.join((str(x) for x in v2)))
``` | vfc_134381 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/687/A",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4 2\n1 2\n2 3\n",
"output": "1\n2 \n2\n1 3 \n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3 3\n1 2\n2 3\n1 3\n",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5 7\n3 2\n5 4\n3 4\n1 3\n1 5\n1 4\n2 5\n",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 11\n4 10\n8 10\n2 3\n2 4\n7 1\n8 5\n2 8\n7 2\n1 2\n2 9\n6 8\n",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 9\n2 5\n2 4\n2 7\n2 9\n2 3\n2 8\n2 6\n2 10\n2 1\n",
"output": "1\n2 \n9\n1 5 4 7 9 3 8 6 10 \n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 16\n6 10\n5 2\n6 4\n6 8\n5 3\n5 4\n6 2\n5 9\n5 7\n5 1\n6 9\n5 8\n5 10\n6 1\n6 7\n6 3\n",
"output": "2\n5 6 \n8\n1 2 10 4 8 9 7 3 \n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.hackerrank.com/challenges/rirb/problem | Solve the following coding problem using the programming language python:
Given an integer range [A,B],
What’s the probability to get a 1-bit if we first randomly choose a number x in the range and then randomly choose a bit from x?
What’s the expected number of bit 1s if we randomly choose a number x in the range?
Input Format
The first line of input is the number of test cases $\mathbf{T}$
Each test cases is a line contains 2 integers $\mbox{A}$ and $\mbox{B}$ separated by a space.
Output Format
For each test case output a line containing 2 float numbers separated by a space. The first one is the probability and the second one is the expected number. You should output the number accurate to 5 fractional digits.
Constraints
$1\leq T\leq200$
$1\leq A\leq B\leq10^{10}$
Sample Input
1
2 4
Sample Output
0.61111 1.33333
Explanation
(10) (11) (100)
(1) So we got a one in $\frac{1}{3}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{1}+\frac{1}{3}\times\frac{1}{3}=\frac{11}{18}$
(2) The expected 1 we have is : $1\times\frac{1}{3}+2\times\frac{1}{3}+1\times\frac{1}{3}=\frac{4}{3}$
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from functools import lru_cache
@lru_cache(maxsize=None)
def s1(n):
if n < 3:
return n
if n % 2 == 0:
return s1(n // 2) + s1(n // 2 - 1) + n // 2
return 2 * s1(n // 2) + n // 2 + 1
def s2(n):
res = 0
for i in range(1, 40):
if 2 ** i > n:
return res + (s1(n) - s1(2 ** (i - 1) - 1)) / i
res += (1 + 1 / i) / 2 * 2 ** (i - 1)
for _ in range(int(input())):
(a, b) = map(int, input().split())
ans1 = (s2(b) - s2(a - 1)) / (b - a + 1)
ans2 = (s1(b) - s1(a - 1)) / (b - a + 1)
print('%.5f %.5f' % (ans1, ans2))
``` | vfc_134389 | {
"difficulty": "medium_hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.hackerrank.com/challenges/rirb/problem",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "1\n2 4\n",
"output": "0.61111 1.33333\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/MIXCOLOR | Solve the following coding problem using the programming language python:
Read problems statements in Mandarin chinese, Russian and Vietnamese as well.
Chef likes to mix colors a lot. Recently, she was gifted N colors A_{1}, A_{2}, ..., A_{N} by her friend.
Chef wants to make the values of all her colors pairwise distinct, as they will then look stunningly beautiful. In order to achieve that, she can perform the following mixing operation zero or more times:
Choose any two colors. Let's denote their values by x and y.
Mix the color with value x into the color with value y. After the mixing, the value of the first color remains x, but the value of the second color changes to x + y.
Find the minimum number of mixing operations Chef needs to perform to make her colors beautiful.
------ Input ------
The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains a single integer N denoting the number of colors.
The second line contains N space-separated integers A_{1}, A_{2}, ..., A_{N} denoting Chef's initial colors.
------ Output ------
For each test case, print a single line containing one integer — the minimum number of required mixing operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ A_{i} ≤ 10^{9} for each valid i$
------ Subtasks ------
Subtask #1 (30 points): 1 ≤ N ≤ 100
Subtask #2 (70 points): original constraints
----- Sample Input 1 ------
2
3
1 2 3
3
2 1 2
----- Sample Output 1 ------
0
1
----- explanation 1 ------
Example case 1: No two colors have the same value. Hence, they are already beautiful. No mixing is required.
Example case 2: Take the second color (with value x = 1) and mix it into the first color (with value x = 2). After the mixing, the colors will be 3 1 2. Now the colors are beautiful. This is the minimum number of mixing operations that are required. Hence, the answer is 1.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
print(n - len(set(arr)))
``` | vfc_134393 | {
"difficulty": "medium",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/MIXCOLOR",
"time_limit": "2 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n3\n1 2 3\n3\n2 1 2",
"output": "0\n1",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/CHEFAPPS | Solve the following coding problem using the programming language python:
Chef's phone has a total storage of S MB. Also, Chef has 2 apps already installed on his phone which occupy X MB and Y MB respectively.
He wants to install another app on his phone whose memory requirement is Z MB. For this, he might have to delete the apps already installed on his phone. Determine the minimum number of apps he has to delete from his phone so that he has enough memory to install the third app.
------ Input Format ------
- The first line contains a single integer T — the number of test cases. Then the test cases follow.
- The first and only line of each test case contains four integers S, X, Y and Z — the total memory of Chef's phone, the memory occupied by the two already installed apps and the memory required by the third app.
------ Output Format ------
For each test case, output the minimum number of apps Chef has to delete from his phone so that he can install the third app.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ S ≤ 500$
$1 ≤X ≤Y ≤S$
$X + Y ≤S$
$Z ≤S$
----- Sample Input 1 ------
4
10 1 2 3
9 4 5 1
15 5 10 15
100 20 30 75
----- Sample Output 1 ------
0
1
2
1
----- explanation 1 ------
Test Case 1: The unused memory in the phone is $7$ MB. Therefore Chef can install the $3$ MB app without deleting any app.
Test Case 2: There is no unused memory in the phone. Chef has to first delete one of the apps from the phone and then only he can install the $1$ MB app.
Test Case 3: There is no unused memory in the phone. Chef has to first delete both the apps from the phone and then only he can install the $15$ MB app.
Test Case 4: The unused memory in the phone is $50$ MB. Chef has to first delete the $30$ MB app from the phone and then only he can install the $75$ MB app.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
t = int(input())
for i in range(t):
(s, x, y, z) = map(int, input().split())
a = x + y
if s - a >= z:
print(0)
elif s - a < z and s - x >= z or s - y >= z:
print(1)
else:
print(2)
``` | vfc_134401 | {
"difficulty": "easy",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/CHEFAPPS",
"time_limit": "0.5 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n10 1 2 3\n9 4 5 1\n15 5 10 15\n100 20 30 75\n",
"output": "0\n1\n2\n1\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Navi is a counter strike pro. He always say how good he is at counter strike. After being tired of Navi, his friends decided to test his skills at shooting. They put M targets on a X-Y plane, each target is denoted by (X, Y) where X is x-coordinate and Y is y-coordinate. His friends also gave him N locations on X-Y plane from where Navi can shoot the targets. Navi knows that he can shoot a target if
Manhattan distance between his location and target is ≤ D. If Navi can shoot more than half of the targets (for odd values of M check only for the integral part of half of M, say M = 3, \frac{2}{3} = 1) only then his friends believe that he is a pro at counter strike otherwise he is not.
Input
First line of input contains an integers T denoting the number of test cases. First line of each test case consists of three integers N, M and D separated by a space. Next N lines of each test case contains a pair of integers denoting the X -co-ordinate and Y - co-ordinate. Next M lines of each test case contains a pair of integers denoting X -co-ordinate and Y - co-ordinate of each of the target.
N is number of shooting locations for Navi.
M is number of targets.
D is shooting range of Navi.
Output
For each test case , Print "YES" if he is a pro else "NO" (without quotes) .
Constraints
1 ≤ T ≤ 6
1 ≤ N ≤ 1000
1 ≤ M ≤ 1000
-50000 ≤ X, Y ≤ 50000
0 ≤ D ≤ 50000
SAMPLE INPUT
1
3 3 5
1 1
2 2
3 3
1 0
0 0
2 0
SAMPLE OUTPUT
YES
Explanation
First location is at (1, 1) , we can shot any of the given targets from here so count > \frac{3}{2} , therefore YES.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
'''
# Read input from stdin and provide input before running code
name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''
test = int(input())
for i in range(test):
mat = input()
[n,m,d] = list(map(int, mat.split()))
shoot_loc = []
for j in range(n):
loc = input()
[x,y] = list(map(int,loc.split()))
shoot_loc.append([x,y])
tar_loc = []
for k in range(m):
loc = input()
[x,y] = list(map(int,loc.split()))
tar_loc.append([x,y])
summ = 0
for loc in tar_loc:
dist = []
for sho in shoot_loc:
dist.append(abs(loc[0]-sho[0])+abs(loc[1]-sho[1]))
if min(dist) <= d:
summ = summ+1
if summ > m/2:
print("YES")
else:
print("NO")
``` | vfc_134410 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
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28239\n-23132 17028\n9621 -2169\n-24436 -12222\n672 13415\n-32448 9175\n32412 -19686\n-20521 -7720\n1603 2904\n-26845 23349\n12260 -2401\n-9786 -1013\n32100 2787\n-8173 26477\n32644 -9261\n-1371 -29316\n",
"output": "NO\nYES\nNO\nYES\nYES\nNO\n",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/distinct-occurrences/1 | Solve the following coding problem using the programming language python:
Given two strings S and T of length n and m respectively. Find count of distinct occurrences of T in S as a sub-sequence.
Example 1:
Input:
S = "banana" , T = "ban"
Output: 3
Explanation: There are 3 sub-sequences:
[ban], [ba n], [b an].
Example 2:
Input:
S = "geeksforgeeks" , T = "ge"
Output: 6
Explanation: There are 6 sub-sequences:
[ge], [ ge], [g e], [g e] [g e] and [ g e].
Your Task:
You don't need to read input or print anything.Your task is to complete the function subsequenceCount() which takes two strings as argument S and T and returns the count of the sub-sequences modulo 10^{9} + 7.
Expected Time Complexity: O(n*m).
Expected Auxiliary Space: O(n*m).
Constraints:
1 ≤ length of(T) ≤ 100
1 ≤ length of(S) ≤ 8000
Write your solution by modifying this code:
```python
# Your task is to complete this function
# Finction should return Integer
class Solution:
def sequenceCount(self,arr1, arr2):
# Code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def sequenceCount(self, arr1, arr2):
n = len(arr1)
m = len(arr2)
if n < m:
return 0
dp = [[0] * (m + 1) for _ in range(n + 1)]
for i in range(n + 1):
dp[i][0] = 1
for i in range(1, n + 1):
for j in range(1, m + 1):
p1 = dp[i - 1][j]
p2 = 0
if arr1[i - 1] == arr2[j - 1]:
p2 = dp[i - 1][j - 1]
dp[i][j] = (p1 + p2) % 1000000007
return dp[n][m]
``` | vfc_134418 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/distinct-occurrences/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "S = \"banana\" , T = \"ban\"",
"output": "3",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "S = \"geeksforgeeks\" , T = \"ge\"",
"output": "6",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Compute A + B.
Constraints
* -1000 ≤ A, B ≤ 1000
Input
The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF.
Output
For each pair of input integers A and B, you must output the sum of A and B in one line.
Example
Input
1 2
10 5
100 20
Output
3
15
120
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
ans = ''
while True:
try:
info = input()
except:
break
if ans != '':
ans += '\n'
info_2 = info.split()
ans += str(int(info_2[0]) + int(info_2[1]))
print(ans)
``` | vfc_134420 | {
"difficulty": "unknown_difficulty",
"memory_limit": "134.217728 megabytes",
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "1 2\n10 5\n101 20",
"output": "3\n15\n121\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 2\n10 4\n101 20",
"output": "3\n14\n121\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 2\n10 4\n101 37",
"output": "3\n14\n138\n",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/sub-array-sum-divisible-by-k2617/1 | Solve the following coding problem using the programming language python:
You are given an array A of N positive and/or negative integers and a value K. The task is to find the count of all sub-arrays whose sum is divisible by K.
Example 1:
Input: N = 6, K = 5
arr[] = {4, 5, 0, -2, -3, 1}
Output: 7
Explanation: There are 7 sub-arrays whose
is divisible by K {4, 5, 0, -2, -3, 1}, {5},
{5, 0}, {5, 0, -2, -3}, {0}, {0, -2, -3}
and {-2, -3}
Example 2:
Input: N = 6, K = 2
arr[] = {2, 2, 2, 2, 2, 2}
Output: 21
Explanation: All subarray sums are
divisible by 7
Your Task:
This is a function problem. You don't need to take any input, as it is already accomplished by the driver code. You just need to complete the function subCount() that takes array arr, integer N, and integer K as parameters and returns the desired output.
Expected Time Complexity: O(N+K).
Expected Auxiliary Space: O(K).
Constraints:
2 ≤ N ≤ 10^{5}
Write your solution by modifying this code:
```python
class Solution:
def subCount(self,arr, n, k):
# Your code goes here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def subCount(self, arr, n, k):
hashmap = {}
currSum = 0
for i in range(n):
currSum += arr[i]
arr[i] = currSum
arr[i] %= k
hashmap[arr[i]] = hashmap.get(arr[i], 0) + 1
ans = 0
for x in hashmap.keys():
y = hashmap[x]
if x == 0:
ans += y * (y + 1) // 2
else:
ans += y * (y - 1) // 2
return ans
``` | vfc_134424 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/sub-array-sum-divisible-by-k2617/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 6, K = 5\n arr[] = {4, 5, 0, -2, -3, 1}",
"output": "7",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N = 6, K = 2\n arr[] = {2, 2, 2, 2, 2, 2}",
"output": "21",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/DEC15/problems/CHEFST | Solve the following coding problem using the programming language python:
Chef has two piles of stones with him, one has n1 stones and the other has n2 stones. Fired up by boredom, he invented a game with the two piles.
Before the start of the game Chef chooses an integer m.
In the j-th move:
- He chooses a number xj such that 1 ≤ xj ≤ m, and removes xj stones from both the piles (this is only possible when both the piles have ≥ xj stones).
- The number chosen must be unique over all the moves in the game. That is, for all k < j, xj ≠ xk.
The game stops when Chef is unable to make any more moves.
Chef wants to make the moves in such a way that the sum of the number of stones remaining in the two piles is minimized. Please help Chef find this.
-----Input-----
- The first line of input contains an integer T denoting the number of test cases.
- Each test case consists of 1 line with three integers — n1, n2 and m — separated by single spaces.
-----Output-----
For each test case, output a single line containing the minimum sum of the number of stones of two piles.
-----Constraints-----
Subtask 1 : (5 pts)
- 1 ≤ T ≤ 100
- 0 ≤ m ≤ 18
- 0 ≤ n1, n2 ≤ 100
Subtask 2 : (25 pts)
- 1 ≤ T ≤ 1000
- 0 ≤ m ≤ 10000
- 0 ≤ n1, n2 ≤ 10000
Subtask 3 : (70 pts)
- 1 ≤ T ≤ 105
- 0 ≤ m ≤ 109
- 0 ≤ n1, n2 ≤ 1018
-----Example-----
Input:3
1 1 1
1 2 1
4 5 2
Output:0
1
3
-----Explanation-----
Example case 1. : Remove 1 stone from each of the piles. Now 0 stones are remaining, so chef cannot remove any more stones from the piles. Hence, answer is 0+0 = 0
Example case 2. : Again, remove 1 stone from both the piles to get (0,1) stones. Now chef cannot remove any more stones from pile 1, so he stops. Hence, answer is 0+1 = 1.
Example case 3. : First remove 1 stone from both the piles to get (3,4) stones. Now, remove 2 stones from both the piles so that (1,2) stones are remaining. Now chef cannot remove any more stones owing to the condition that he cannot remove the same number of stones twice. So, the answer is 1+2 = 3.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
n = int(sys.stdin.readline())
for _ in range(n):
(p1, p2, m) = list(map(int, sys.stdin.readline().split()))
l = min(p1, p2)
q = min(p1, p2)
d = min(m * (m + 1) / 2, q)
print(p1 - d + p2 - d)
``` | vfc_134425 | {
"difficulty": "unknown_difficulty",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/DEC15/problems/CHEFST",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n1 1 1\n1 2 1\n4 5 2\n",
"output": "0\n1\n3\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1204/A | Solve the following coding problem using the programming language python:
In the city of Saint Petersburg, a day lasts for $2^{100}$ minutes. From the main station of Saint Petersburg, a train departs after $1$ minute, $4$ minutes, $16$ minutes, and so on; in other words, the train departs at time $4^k$ for each integer $k \geq 0$. Team BowWow has arrived at the station at the time $s$ and it is trying to count how many trains have they missed; in other words, the number of trains that have departed strictly before time $s$. For example if $s = 20$, then they missed trains which have departed at $1$, $4$ and $16$. As you are the only one who knows the time, help them!
Note that the number $s$ will be given you in a binary representation without leading zeroes.
-----Input-----
The first line contains a single binary number $s$ ($0 \leq s < 2^{100}$) without leading zeroes.
-----Output-----
Output a single number — the number of trains which have departed strictly before the time $s$.
-----Examples-----
Input
100000000
Output
4
Input
101
Output
2
Input
10100
Output
3
-----Note-----
In the first example $100000000_2 = 256_{10}$, missed trains have departed at $1$, $4$, $16$ and $64$.
In the second example $101_2 = 5_{10}$, trains have departed at $1$ and $4$.
The third example is explained in the statements.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
s = input()
if s == '0' or s == '1':
print(0)
else:
x = len(s)
if x % 2 != 0 and '1' not in s[1:]:
print((x - 1) // 2)
else:
print((x + 1) // 2)
``` | vfc_134429 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1204/A",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "100000000\n",
"output": "4\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1328/E | Solve the following coding problem using the programming language python:
You are given a rooted tree consisting of $n$ vertices numbered from $1$ to $n$. The root of the tree is a vertex number $1$.
A tree is a connected undirected graph with $n-1$ edges.
You are given $m$ queries. The $i$-th query consists of the set of $k_i$ distinct vertices $v_i[1], v_i[2], \dots, v_i[k_i]$. Your task is to say if there is a path from the root to some vertex $u$ such that each of the given $k$ vertices is either belongs to this path or has the distance $1$ to some vertex of this path.
-----Input-----
The first line of the input contains two integers $n$ and $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.
Each of the next $n-1$ lines describes an edge of the tree. Edge $i$ is denoted by two integers $u_i$ and $v_i$, the labels of vertices it connects $(1 \le u_i, v_i \le n, u_i \ne v_i$).
It is guaranteed that the given edges form a tree.
The next $m$ lines describe queries. The $i$-th line describes the $i$-th query and starts with the integer $k_i$ ($1 \le k_i \le n$) — the number of vertices in the current query. Then $k_i$ integers follow: $v_i[1], v_i[2], \dots, v_i[k_i]$ ($1 \le v_i[j] \le n$), where $v_i[j]$ is the $j$-th vertex of the $i$-th query.
It is guaranteed that all vertices in a single query are distinct.
It is guaranteed that the sum of $k_i$ does not exceed $2 \cdot 10^5$ ($\sum\limits_{i=1}^{m} k_i \le 2 \cdot 10^5$).
-----Output-----
For each query, print the answer — "YES", if there is a path from the root to some vertex $u$ such that each of the given $k$ vertices is either belongs to this path or has the distance $1$ to some vertex of this path and "NO" otherwise.
-----Example-----
Input
10 6
1 2
1 3
1 4
2 5
2 6
3 7
7 8
7 9
9 10
4 3 8 9 10
3 2 4 6
3 2 1 5
3 4 8 2
2 6 10
3 5 4 7
Output
YES
YES
YES
YES
NO
NO
-----Note-----
The picture corresponding to the example:
[Image]
Consider the queries.
The first query is $[3, 8, 9, 10]$. The answer is "YES" as you can choose the path from the root $1$ to the vertex $u=10$. Then vertices $[3, 9, 10]$ belong to the path from $1$ to $10$ and the vertex $8$ has distance $1$ to the vertex $7$ which also belongs to this path.
The second query is $[2, 4, 6]$. The answer is "YES" as you can choose the path to the vertex $u=2$. Then the vertex $4$ has distance $1$ to the vertex $1$ which belongs to this path and the vertex $6$ has distance $1$ to the vertex $2$ which belongs to this path.
The third query is $[2, 1, 5]$. The answer is "YES" as you can choose the path to the vertex $u=5$ and all vertices of the query belong to this path.
The fourth query is $[4, 8, 2]$. The answer is "YES" as you can choose the path to the vertex $u=9$ so vertices $2$ and $4$ both have distance $1$ to the vertex $1$ which belongs to this path and the vertex $8$ has distance $1$ to the vertex $7$ which belongs to this path.
The fifth and the sixth queries both have answer "NO" because you cannot choose suitable vertex $u$.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
input = sys.stdin.readline
(n, m) = [int(i) for i in input().split()]
graph = [[] for i in range(n)]
for j in range(n - 1):
(v, u) = [int(i) - 1 for i in input().split()]
graph[v].append(u)
graph[u].append(v)
parent = [0] * n
tin = [0] * n
tout = [0] * n
time = 0
queue = [(0, 0)]
while queue:
(current, visit) = queue.pop()
if visit == 0:
queue.append((current, 1))
time += 1
tin[current] = time
for child in graph[current]:
if child != parent[current]:
parent[child] = current
queue.append((child, 0))
else:
tout[current] = time
for j in range(m):
q = [int(i) - 1 for i in input().split()]
q = q[1:]
for i in range(len(q)):
q[i] = parent[q[i]]
cin = 1
cout = n
for i in q:
cin = max(cin, tin[i])
cout = min(cout, tout[i])
if cin > cout:
print('NO')
else:
print('YES')
``` | vfc_134433 | {
"difficulty": "hard",
"memory_limit": "256 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1328/E",
"time_limit": "2 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "10 6\n1 2\n1 3\n1 4\n2 5\n2 6\n3 7\n7 8\n7 9\n9 10\n4 3 8 9 10\n3 2 4 6\n3 2 1 5\n3 4 8 2\n2 6 10\n3 5 4 7\n",
"output": "YES\nYES\nYES\nYES\nNO\nNO\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1155/A | Solve the following coding problem using the programming language python:
You are given a string $s$ consisting of $n$ lowercase Latin letters.
Let's define a substring as a contiguous subsegment of a string. For example, "acab" is a substring of "abacaba" (it starts in position $3$ and ends in position $6$), but "aa" or "d" aren't substrings of this string. So the substring of the string $s$ from position $l$ to position $r$ is $s[l; r] = s_l s_{l + 1} \dots s_r$.
You have to choose exactly one of the substrings of the given string and reverse it (i. e. make $s[l; r] = s_r s_{r - 1} \dots s_l$) to obtain a string that is less lexicographically. Note that it is not necessary to obtain the minimum possible string.
If it is impossible to reverse some substring of the given string to obtain a string that is less, print "NO". Otherwise print "YES" and any suitable substring.
String $x$ is lexicographically less than string $y$, if either $x$ is a prefix of $y$ (and $x \ne y$), or there exists such $i$ ($1 \le i \le min(|x|, |y|)$), that $x_i < y_i$, and for any $j$ ($1 \le j < i$) $x_j = y_j$. Here $|a|$ denotes the length of the string $a$. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
-----Input-----
The first line of the input contains one integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the length of $s$.
The second line of the input contains the string $s$ of length $n$ consisting only of lowercase Latin letters.
-----Output-----
If it is impossible to reverse some substring of the given string to obtain a string which is lexicographically less, print "NO". Otherwise print "YES" and two indices $l$ and $r$ ($1 \le l < r \le n$) denoting the substring you have to reverse. If there are multiple answers, you can print any.
-----Examples-----
Input
7
abacaba
Output
YES
2 5
Input
6
aabcfg
Output
NO
-----Note-----
In the first testcase the resulting string is "aacabba".
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
sen = input()
for i in range(0, n - 1, 1):
if sen[i] > sen[i + 1]:
print('YES')
print(i + 1, i + 2)
exit(0)
print('NO')
``` | vfc_134442 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1155/A",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "7\nabacaba\n",
"output": "YES\n2 3\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/859/E | Solve the following coding problem using the programming language python:
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 10^9 + 7.
-----Input-----
Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.
N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.
-----Output-----
Print the number of possible assignments, modulo 1000000007 = 10^9 + 7.
-----Examples-----
Input
4
1 5
5 2
3 7
7 3
Output
6
Input
5
1 10
2 10
3 10
4 10
5 5
Output
5
-----Note-----
These are the possible assignments for the first example: 1 5 3 7 1 2 3 7 5 2 3 7 1 5 7 3 1 2 7 3 5 2 7 3
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
m = 2 * n + 1
u = [[] for i in range(m)]
v = [0] * m
s = [0] * m
d = 10 ** 9 + 7
y = 1
for j in range(n):
(a, b) = map(int, input().split())
v[a] = b
if a != b:
s[b] += 1
u[b].append(a)
for b in range(m):
if not v[b]:
x = 0
p = [b]
while p:
x += 1
a = p.pop()
s[a] = -1
p += u[a]
y = x * y % d
for a in range(m):
if s[a] == 0:
b = v[a]
while s[b] == 1:
s[b] = -1
b = v[b]
s[b] -= 1
for a in range(m):
if s[a] == 1:
y = 2 * y % d
while s[a]:
s[a] = 0
a = v[a]
print(y)
``` | vfc_134446 | {
"difficulty": "hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/859/E",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n1 5\n5 2\n3 7\n7 3\n",
"output": "6\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5\n1 10\n2 10\n3 10\n4 10\n5 5\n",
"output": "5\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1\n1 2\n",
"output": "2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "30\n22 37\n12 37\n37 58\n29 57\n43 57\n57 58\n58 53\n45 4\n1 4\n4 51\n35 31\n21 31\n31 51\n51 53\n53 48\n60 55\n52 55\n55 33\n36 9\n10 9\n9 33\n33 19\n5 23\n47 23\n23 32\n50 44\n26 44\n44 32\n32 19\n19 48\n",
"output": "31\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "50\n73 1\n65 73\n16 65\n57 65\n33 16\n34 57\n98 16\n84 98\n55 34\n64 84\n80 55\n75 64\n28 75\n20 75\n42 75\n88 42\n50 20\n48 28\n32 48\n58 88\n92 76\n76 53\n53 15\n15 1\n1 10\n10 71\n71 37\n37 95\n95 63\n63 92\n45 97\n97 51\n51 96\n96 12\n12 62\n62 31\n31 5\n5 29\n29 19\n19 49\n49 6\n6 40\n40 18\n18 22\n22 17\n17 46\n46 72\n72 82\n82 14\n14 14\n",
"output": "2\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/627/A | Solve the following coding problem using the programming language python:
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
-----Input-----
The first line of the input contains two integers s and x (2 ≤ s ≤ 10^12, 0 ≤ x ≤ 10^12), the sum and bitwise xor of the pair of positive integers, respectively.
-----Output-----
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
-----Examples-----
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
-----Note-----
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(s, x) = map(int, input().split())
b = s - x
b = list(bin(s - x))
b = b[2:]
a = list(bin(x))
a = a[2:]
al = len(a)
bl = len(b)
c = []
if bl > al:
for i in range(bl - al):
c.append('0')
c.extend(a)
a = c
else:
for i in range(al - bl):
c.append('0')
c.extend(b)
b = c
p = 0
bl = len(b)
for i in range(bl - 1):
if b[i] == '1':
if a[i + 1] == '1':
p = 1
bl = len(b)
if b[bl - 1] == '1':
p = 1
if x > s:
print('0')
elif p == 1:
print('0')
elif s == x:
print(2 ** a.count('1') - 2)
elif x == 0:
print('1')
else:
print(2 ** a.count('1'))
``` | vfc_134450 | {
"difficulty": "medium_hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/627/A",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "9 5\n",
"output": "4\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3 3\n",
"output": "2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5 2\n",
"output": "0\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def main():
while 1:
(d, e) = map(int, input().split())
if d == e == 0:
break
x = d
y = 0
ans = abs(d - e)
while y != d:
a = abs((x ** 2 + y ** 2) ** 0.5 - e)
ans = min(a, ans)
x -= 1
y += 1
print(ans)
main()
``` | vfc_134454 | {
"difficulty": "unknown_difficulty",
"memory_limit": "268.435456 megabytes",
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "8.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2 1\n7 6\n7 6\n7 7\n76 5\n8 41\n0 0",
"output": "0.414213562373\n0.0827625302982\n0.0827625302982\n0.0\n48.7401153702\n33.0\n",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://www.hackerrank.com/challenges/extra-long-factorials/problem | Solve the following coding problem using the programming language python:
The factorial of the integer $n$, written $n!$, is defined as:
$n!=n\times(n-1)\times(n-2)\times\text{}\cdots\times3\times2\times1$
Calculate and print the factorial of a given integer.
For example, if $n=30$, we calculate $30\times29\times28\times\text{...}\times2\times1$ and get $265252859812191058636308480000000$.
Function Description
Complete the extraLongFactorials function in the editor below. It should print the result and return.
extraLongFactorials has the following parameter(s):
n: an integer
Note: Factorials of $n>20$ can't be stored even in a $64-bit$ long long variable. Big integers must be used for such calculations. Languages like Java, Python, Ruby etc. can handle big integers, but we need to write additional code in C/C++ to handle huge values.
We recommend solving this challenge using BigIntegers.
Input Format
Input consists of a single integer $n$
Constraints
$1\leq n\leq100$
Output Format
Print the factorial of $n$.
Sample Input
$25$
Sample Output
$1551121004330985984000000$
Explanation
$25!=25\times24\times23\times\text{...}\times3\times2\times1$
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from math import factorial
print(factorial(int(input())))
``` | vfc_134458 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.hackerrank.com/challenges/extra-long-factorials/problem",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "25\n",
"output": "15511210043330985984000000\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1566/B | Solve the following coding problem using the programming language python:
A binary string is a string that consists of characters $0$ and $1$.
Let $\operatorname{MEX}$ of a binary string be the smallest digit among $0$, $1$, or $2$ that does not occur in the string. For example, $\operatorname{MEX}$ of $001011$ is $2$, because $0$ and $1$ occur in the string at least once, $\operatorname{MEX}$ of $1111$ is $0$, because $0$ and $2$ do not occur in the string and $0 < 2$.
A binary string $s$ is given. You should cut it into any number of substrings such that each character is in exactly one substring. It is possible to cut the string into a single substring — the whole string.
A string $a$ is a substring of a string $b$ if $a$ can be obtained from $b$ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
What is the minimal sum of $\operatorname{MEX}$ of all substrings pieces can be?
-----Input-----
The input consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases. Description of the test cases follows.
Each test case contains a single binary string $s$ ($1 \le |s| \le 10^5$).
It's guaranteed that the sum of lengths of $s$ over all test cases does not exceed $10^5$.
-----Output-----
For each test case print a single integer — the minimal sum of $\operatorname{MEX}$ of all substrings that it is possible to get by cutting $s$ optimally.
-----Examples-----
Input
6
01
1111
01100
101
0000
01010
Output
1
0
2
1
1
2
-----Note-----
In the first test case the minimal sum is $\operatorname{MEX}(0) + \operatorname{MEX}(1) = 1 + 0 = 1$.
In the second test case the minimal sum is $\operatorname{MEX}(1111) = 0$.
In the third test case the minimal sum is $\operatorname{MEX}(01100) = 2$.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
t = int(input())
for test in range(t):
s = input()
zeroes = s.count('0')
if zeroes == 0:
print(0)
continue
first = s.find('0')
last = s.rfind('0')
if last - first + 1 == zeroes:
print(1)
else:
print(2)
``` | vfc_134467 | {
"difficulty": "easy",
"memory_limit": "256 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1566/B",
"time_limit": "1 second"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "6\n01\n1111\n01100\n101\n0000\n01010\n",
"output": "1\n0\n2\n1\n1\n2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\n01\n1111\n01100\n100\n0000\n01010\n",
"output": "1\n0\n2\n1\n1\n2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\n01\n1011\n01100\n100\n0000\n01010\n",
"output": "1\n1\n2\n1\n1\n2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\n01\n1001\n01100\n111\n0000\n01010\n",
"output": "1\n1\n2\n0\n1\n2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\n1\n1001\n01100\n111\n0000\n01010\n",
"output": "0\n1\n2\n0\n1\n2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6\n01\n1010\n01100\n100\n0000\n01010\n",
"output": "1\n2\n2\n1\n1\n2\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/largest-product-pallindrome3753/1 | Solve the following coding problem using the programming language python:
Given a value N, find out the largest palindrome number which is product of two N-digit numbers.
Example 1:
Input: N = 2
Output: 9009
Explanation: 9009 is the largest number
which is product of two 2-digit numbers.
9009 = 91*99.
Example 2:
Input: N = 1
Output: 9
Explanation: 9 is the largest number
which is product of two 1-digit numbers.
3 * 3 = 9.
Your Task:
You don't need to read or print anyhting. Your task is to complete the function LargestProductPalin() which takes N as input parameter and returns largest palindromic product of 2 N-digit numbers.
Expected Time Complexity: O(10^{n+1})
Expected Space Complexity: O(1)
Constraints:
1 <= N <= 6
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def LargestProductPalin(self, N):
# Code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def LargestProductPalin(self, n):
upper_limit = 10 ** n - 1
lower_limit = 1 + upper_limit // 10
max_product = 0
for i in range(upper_limit, lower_limit - 1, -1):
for j in range(i, lower_limit - 1, -1):
product = i * j
if product < max_product:
break
number = product
reverse = 0
while number != 0:
reverse = reverse * 10 + number % 10
number = number // 10
if product == reverse and product > max_product:
max_product = product
return max_product
``` | vfc_134475 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/largest-product-pallindrome3753/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 2",
"output": "9009",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N = 1",
"output": "9",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/103/A | Solve the following coding problem using the programming language python:
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys, math
s = int(input())
arr = list(map(int, input().split(' ')))
m = 0
for i in range(0, s):
m += (arr[i] - 1) * (i + 1) + 1
print(m)
``` | vfc_134476 | {
"difficulty": "easy",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/103/A",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "100\n1 3 2 1 1 2 1 3 2 2 3 1 1 1 2 2 1 3 3 1 1 2 2 3 2 1 3 1 3 2 1 1 3 3 2 1 2 2 2 3 2 2 3 2 2 3 2 1 3 1 1 2 1 3 2 2 1 1 1 1 1 1 3 1 2 3 1 1 1 1 1 2 3 3 1 1 1 1 2 3 3 1 3 2 2 3 2 1 3 2 2 3 1 1 3 2 3 2 3 1\n",
"output": "4667",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1\n84355694\n",
"output": "84355694",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n2 4 1\n",
"output": "10",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/check-if-divisible-by-114724/1 | Solve the following coding problem using the programming language python:
Given a number S. Check whether it is divisble by 11 or not.
Example 1:
Input:
S = 76945
Output: 1
Explanation: The number is divisible by 11
as 76945 % 11 = 0.
ââ¬â¹Example 2:
Input:
S = 12
Output: 0
Explanation: The number is not divisible
by 11 as 12 % 11 = 1.
Your Task:
You don't need to read input or print anything. Your task is to complete the function divisibleBy11() which takes the number in the form of string S as input and returns 1 if the number is divisible by 11. Else, it returns 0.
Expected Time Complexity: O(Log S) where S is the input number.
Expected Auxiliary Space: O(1).
Constraints:
1<=S<=10^{1000}+5
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def divisibleBy11(self, S):
# Your Code Here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def divisibleBy11(self, S):
if int(S) % 11 == 0:
return 1
else:
return 0
``` | vfc_134480 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/check-if-divisible-by-114724/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "S = 76945",
"output": "1",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "S = 12",
"output": "0",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1064/D | Solve the following coding problem using the programming language python:
You are playing some computer game. One of its levels puts you in a maze consisting of n lines, each of which contains m cells. Each cell either is free or is occupied by an obstacle. The starting cell is in the row r and column c. In one step you can move one square up, left, down or right, if the target cell is not occupied by an obstacle. You can't move beyond the boundaries of the labyrinth.
Unfortunately, your keyboard is about to break, so you can move left no more than x times and move right no more than y times. There are no restrictions on the number of moves up and down since the keys used to move up and down are in perfect condition.
Now you would like to determine for each cell whether there exists a sequence of moves that will put you from the starting cell to this particular one. How many cells of the board have this property?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2000) — the number of rows and the number columns in the labyrinth respectively.
The second line contains two integers r, c (1 ≤ r ≤ n, 1 ≤ c ≤ m) — index of the row and index of the column that define the starting cell.
The third line contains two integers x, y (0 ≤ x, y ≤ 109) — the maximum allowed number of movements to the left and to the right respectively.
The next n lines describe the labyrinth. Each of them has length of m and consists only of symbols '.' and '*'. The j-th character of the i-th line corresponds to the cell of labyrinth at row i and column j. Symbol '.' denotes the free cell, while symbol '*' denotes the cell with an obstacle.
It is guaranteed, that the starting cell contains no obstacles.
Output
Print exactly one integer — the number of cells in the labyrinth, which are reachable from starting cell, including the starting cell itself.
Examples
Input
4 5
3 2
1 2
.....
.***.
...**
*....
Output
10
Input
4 4
2 2
0 1
....
..*.
....
....
Output
7
Note
Cells, reachable in the corresponding example, are marked with '+'.
First example:
+++..
+***.
+++**
*+++.
Second example:
.++.
.+*.
.++.
.++.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
from collections import deque
from math import inf
def rint():
return map(int, sys.stdin.readline().split())
def bfs(sr, sc):
global ans
q = deque()
q.append((x, y, sr, sc))
visit[sr][sc] = 1
ans += 1
while q:
(ll, rr, r, c) = q.popleft()
nadj = []
if r + 1 < n and maze[r + 1][c] == '.':
nadj.append([r + 1, c])
if r - 1 >= 0 and maze[r - 1][c] == '.':
nadj.append([r - 1, c])
if c + 1 < m and maze[r][c + 1] == '.':
nadj.append([r, c + 1])
if c - 1 >= 0 and maze[r][c - 1] == '.':
nadj.append([r, c - 1])
for (adr, adc) in nadj:
if visit[adr][adc] == 1:
continue
if c == adc:
visit[adr][adc] = 1
ans += 1
q.appendleft([ll, rr, adr, adc])
elif adc < c:
if ll - 1 >= 0:
visit[adr][adc] = 1
ans += 1
q.append([ll - 1, rr, adr, adc])
elif rr - 1 >= 0:
visit[adr][adc] = 1
ans += 1
q.append([ll, rr - 1, adr, adc])
(n, m) = rint()
(sr, sc) = rint()
(x, y) = rint()
sr -= 1
sc -= 1
maze = []
for i in range(n):
maze.append(input())
visit = [[0 for i in range(m)] for j in range(n)]
ans = 0
bfs(sr, sc)
print(ans)
``` | vfc_134481 | {
"difficulty": "medium_hard",
"memory_limit": "512.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1064/D",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "7 16\n1 15\n1000 0\n...*...*........\n.*.*.*.*.******.\n.*.*.*.*.*....*.\n.*.*.*.*.*....*.\n.*...*...*....*.\n.**************.\n................\n",
"output": "37",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 1\n1 1\n0 0\n.\n",
"output": "1",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 15\n7 7\n7 7\n.*****..*******\n**...*.********\n**.*..........*\n...**********.*\n.************.*\n.*...********.*\n.*.*.*........*\n.*.*.*.********\n.*.*.*.********\n...*...********\n",
"output": "53",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 6\n10 6\n5 2\n**....\n**.**.\n**.**.\n**.**.\n**.**.\n**.**.\n**.**.\n....*.\n.****.\n......\n",
"output": "29",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 10\n10 7\n10 4\n*******...\n*******.*.\n*******.*.\n*******.*.\n*******.*.\n****..*.*.\n****.**.*.\n........*.\n.********.\n..........\n",
"output": "39",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "11 11\n5 3\n100 9\n*********..\n*********.*\n..........*\n.********.*\n....*...*.*\n***.*.*.*.*\n***.*.*.*.*\n***.*.*.*.*\n***.*.*.*.*\n***.*.*.*.*\n***...*...*\n",
"output": "49",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/maximum-product-subarray3604/1 | Solve the following coding problem using the programming language python:
Given an array Arr[] that contains N integers (may be positive, negative or zero). Find the product of the maximum product subarray.
Example 1:
Input:
N = 5
Arr[] = {6, -3, -10, 0, 2}
Output: 180
Explanation: Subarray with maximum product
is [6, -3, -10] which gives product as 180.
Example 2:
Input:
N = 6
Arr[] = {2, 3, 4, 5, -1, 0}
Output: 120
Explanation: Subarray with maximum product
is [2, 3, 4, 5] which gives product as 120.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maxProduct() which takes the array of integers arr and n as parameters and returns an integer denoting the answer.
Note: Use 64-bit integer data type to avoid overflow.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 500
-10^{2} ≤ Arr_{i} ≤ 10^{2}
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
# Function to find maximum
# product subarray
def maxProduct(self,arr, n):
# code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def maxProduct(self, arr, n):
ans = arr[0]
(ma, mi) = (ans, ans)
for i in range(1, n):
if arr[i] < 0:
temp = ma
ma = mi
mi = temp
ma = max(arr[i], ma * arr[i])
mi = min(arr[i], mi * arr[i])
ans = max(ans, ma)
return ans
``` | vfc_134485 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/maximum-product-subarray3604/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 5\nArr[] = {6, -3, -10, 0, 2}",
"output": "180",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/CEFT2020/problems/QFUN | Solve the following coding problem using the programming language python:
Chef and Abhishek both are fighting for the post of Chairperson to be part of ACE committee and are trying their best. To select only one student their teacher gave them a binary string (string consisting of only 0's and 1's) and asked them to find number of sub-strings present in the given string that satisfy the following condition:
The substring should start with 0 and end with 1 or the substring should start with 1 and end with 0 but not start with 0 and end with 0 and start with 1 and end with 1.
More formally, strings such as 100,0101 are allowed since they start and end with different characters. But strings such as 0110,1101 are not allowed because they start and end with same characters.
Both Chef and Abhishek try their best to solve it but couldn't do it. You being a very good friend of Chef, he asks for your help so that he can solve it and become the Chairperson.
-----Input:-----
The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains a single integer N denoting the length of the string.
The second line of each test case contains a binary string of length N.
-----Output:-----
For each test case, print a single line containing one integer ― the number of sub strings satisfying above conditions.
-----Constraints-----
- $1 \leq T \leq 1000$
- $2 \leq N \leq 10^9$
Binary string consist's only 0 and 1.
-----Sample Input:-----
1
4
1010
-----Sample Output:-----
4
-----EXPLANATION:-----
All possible substring are : { (1),(0),(1),(0),(10),(01),(10),(101),(010),(1010) }. Out of these only 4 substrings {(10),(01),(10),(1010)} start and end with different characters. Hence the answer 4.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def countSubstr(str, n, x, y):
tot_count = 0
count_x = 0
for i in range(n):
if str[i] == x:
count_x += 1
if str[i] == y:
tot_count += count_x
return tot_count
t = int(input())
for _ in range(t):
n = int(input())
str = input()
x = '0'
y = '1'
x1 = '1'
y1 = '0'
c1 = countSubstr(str, n, x, y)
c2 = countSubstr(str, n, x1, y1)
print(c1 + c2)
``` | vfc_134486 | {
"difficulty": "unknown_difficulty",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/CEFT2020/problems/QFUN",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "1\n4\n1010\n",
"output": "4\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/851/B | Solve the following coding problem using the programming language python:
Arpa is taking a geometry exam. Here is the last problem of the exam.
You are given three points a, b, c.
Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.
Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.
-----Input-----
The only line contains six integers a_{x}, a_{y}, b_{x}, b_{y}, c_{x}, c_{y} (|a_{x}|, |a_{y}|, |b_{x}|, |b_{y}|, |c_{x}|, |c_{y}| ≤ 10^9). It's guaranteed that the points are distinct.
-----Output-----
Print "Yes" if the problem has a solution, "No" otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
0 1 1 1 1 0
Output
Yes
Input
1 1 0 0 1000 1000
Output
No
-----Note-----
In the first sample test, rotate the page around (0.5, 0.5) by $90^{\circ}$.
In the second sample test, you can't find any solution.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from math import *
(ax, ay, bx, by, cx, cy) = map(int, input().split())
ab = (ax - bx) * (ax - bx) + (ay - by) * (ay - by)
bc = (cx - bx) * (cx - bx) + (cy - by) * (cy - by)
line = (ax - cx) * (by - ay) == (bx - ax) * (ay - cy)
if ab == bc and (not line):
print('Yes')
else:
print('No')
``` | vfc_134490 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/851/B",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "0 1 1 1 1 0\n",
"output": "Yes\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 1 0 0 1000 1000\n",
"output": "No\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 0 2 0 3 0\n",
"output": "No\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1305/A | Solve the following coding problem using the programming language python:
Kuroni has $n$ daughters. As gifts for them, he bought $n$ necklaces and $n$ bracelets: the $i$-th necklace has a brightness $a_i$, where all the $a_i$ are pairwise distinct (i.e. all $a_i$ are different), the $i$-th bracelet has a brightness $b_i$, where all the $b_i$ are pairwise distinct (i.e. all $b_i$ are different).
Kuroni wants to give exactly one necklace and exactly one bracelet to each of his daughters. To make sure that all of them look unique, the total brightnesses of the gifts given to each daughter should be pairwise distinct. Formally, if the $i$-th daughter receives a necklace with brightness $x_i$ and a bracelet with brightness $y_i$, then the sums $x_i + y_i$ should be pairwise distinct. Help Kuroni to distribute the gifts.
For example, if the brightnesses are $a = [1, 7, 5]$ and $b = [6, 1, 2]$, then we may distribute the gifts as follows: Give the third necklace and the first bracelet to the first daughter, for a total brightness of $a_3 + b_1 = 11$. Give the first necklace and the third bracelet to the second daughter, for a total brightness of $a_1 + b_3 = 3$. Give the second necklace and the second bracelet to the third daughter, for a total brightness of $a_2 + b_2 = 8$.
Here is an example of an invalid distribution: Give the first necklace and the first bracelet to the first daughter, for a total brightness of $a_1 + b_1 = 7$. Give the second necklace and the second bracelet to the second daughter, for a total brightness of $a_2 + b_2 = 8$. Give the third necklace and the third bracelet to the third daughter, for a total brightness of $a_3 + b_3 = 7$.
This distribution is invalid, as the total brightnesses of the gifts received by the first and the third daughter are the same. Don't make them this upset!
-----Input-----
The input consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the number of daughters, necklaces and bracelets.
The second line of each test case contains $n$ distinct integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the brightnesses of the necklaces.
The third line of each test case contains $n$ distinct integers $b_1, b_2, \dots, b_n$ ($1 \le b_i \le 1000$) — the brightnesses of the bracelets.
-----Output-----
For each test case, print a line containing $n$ integers $x_1, x_2, \dots, x_n$, representing that the $i$-th daughter receives a necklace with brightness $x_i$. In the next line print $n$ integers $y_1, y_2, \dots, y_n$, representing that the $i$-th daughter receives a bracelet with brightness $y_i$.
The sums $x_1 + y_1, x_2 + y_2, \dots, x_n + y_n$ should all be distinct. The numbers $x_1, \dots, x_n$ should be equal to the numbers $a_1, \dots, a_n$ in some order, and the numbers $y_1, \dots, y_n$ should be equal to the numbers $b_1, \dots, b_n$ in some order.
It can be shown that an answer always exists. If there are multiple possible answers, you may print any of them.
-----Example-----
Input
2
3
1 8 5
8 4 5
3
1 7 5
6 1 2
Output
1 8 5
8 4 5
5 1 7
6 2 1
-----Note-----
In the first test case, it is enough to give the $i$-th necklace and the $i$-th bracelet to the $i$-th daughter. The corresponding sums are $1 + 8 = 9$, $8 + 4 = 12$, and $5 + 5 = 10$.
The second test case is described in the statement.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
l2 = list(map(int, input().split()))
l.sort()
l2.sort()
print(*l)
print(*l2)
``` | vfc_134494 | {
"difficulty": "easy",
"memory_limit": "256 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1305/A",
"time_limit": "1 second"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n3\n1 8 5\n8 4 5\n3\n1 7 5\n6 1 2\n",
"output": "1 8 5\n8 4 5\n5 1 7\n6 2 1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n3\n1 8 10\n8 4 5\n3\n1 7 5\n6 1 2\n",
"output": "1 8 10\n4 5 8\n1 5 7\n1 2 6\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n3\n1 8 5\n8 4 5\n3\n1 7 5\n3 1 2\n",
"output": "1 5 8\n4 5 8\n1 5 7\n1 2 3\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n3\n1 8 10\n8 4 5\n3\n1 7 5\n6 1 4\n",
"output": "1 8 10\n4 5 8\n1 5 7\n1 4 6\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n3\n1 8 5\n8 8 5\n3\n1 7 5\n3 1 2\n",
"output": "1 5 8\n5 8 8\n1 5 7\n1 2 3\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1468/K | Solve the following coding problem using the programming language python:
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of a single line containing s — the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 ≤ x,y ≤ 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq
def input():
return sys.stdin.readline()[:-1]
def solve(S):
Move = {'L': (-1, 0), 'R': (1, 0), 'U': (0, 1), 'D': (0, -1)}
Moves = [Move[c] for c in S]
pos1 = (0, 0)
for k in range(len(S)):
pos1 = (pos1[0] + Moves[k][0], pos1[1] + Moves[k][1])
block = pos1
pos = (0, 0)
for i in range(len(S)):
pos_next = (pos[0] + Moves[i][0], pos[1] + Moves[i][1])
if pos_next != block:
pos = pos_next
if pos == (0, 0):
return block
return (0, 0)
def main():
T0 = int(input())
for _ in range(T0):
S = input()
ans = solve(S)
print(*ans)
main()
``` | vfc_134502 | {
"difficulty": "medium_hard",
"memory_limit": "512.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1468/K",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\nL\nLUUDR\nLLUU\nDDDUUUUU\n",
"output": "-1 0\n-1 2\n0 0\n0 1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "4\nL\nUURDL\nLLUU\nDDDUUUUU\n",
"output": "-1 0\n0 2\n0 0\n0 1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "4\nL\nUURDL\nLLUU\nUUUUUDDD\n",
"output": "-1 0\n0 2\n0 0\n0 4\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/knight-walk4521/1 | Solve the following coding problem using the programming language python:
Given a square chessboard, the initial position of Knight and position of a target. Find out the minimum steps a Knight will take to reach the target position.If it cannot reach the target position return -1.
Note:
The initial and the target position co-ordinates of Knight have been given accoring to 1-base indexing.
Example 1:
Input:
N=6
knightPos[ ] = {4, 5}
targetPos[ ] = {1, 1}
Output:
3
Explanation:
Knight takes 3 step to reach from
(4, 5) to (1, 1):
(4, 5) -> (5, 3) -> (3, 2) -> (1, 1).
Example 2:
Input:
N=8
knightPos[ ] = {7, 7}
targetPos[ ] = {1, 5}
Output:
4
Explanation:
Knight takes 4 steps to reach from
(7, 7) to (1, 5):
(4, 5) -> (6, 5) -> (5, 3) -> (7, 2) -> (1, 5).
Your Task:
You don't need to read input or print anything. Your task is to complete the function minStepToReachTarget() which takes the inital position of Knight (KnightPos), the target position of Knight (TargetPos) and the size of the chess board (N) as an input parameters and returns the minimum number of steps required by the knight to reach from its current position to the given target position.If it cannot reach the target position return -1.
Expected Time Complexity: O(N^{2}).
Expected Auxiliary Space: O(N^{2}).
Constraints:
1 <= N <= 1000
1 <= Knight_pos(X, Y), Targer_pos(X, Y) <= N
Write your solution by modifying this code:
```python
class Solution:
def minStepToReachTarget(self, KnightPos, TargetPos, N):
#Code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from collections import *
class Solution:
def minStepToReachTarget(self, KnightPos, TargetPos, N):
kx = KnightPos[0] - 1
ky = KnightPos[1] - 1
tx = TargetPos[0] - 1
ty = TargetPos[1] - 1
ans = 0
def isvalid(x, y, visit, N):
if x >= 0 and y >= 0 and (x < N) and (y < N) and (visit[x][y] == False):
return True
return False
if kx == tx and ty == ky:
return ans
moves = [[-1, 2], [-2, 1], [-2, -1], [-1, -2], [1, -2], [2, -1], [2, 1], [1, 2]]
visit = [[False for i in range(N)] for j in range(N)]
q = deque([[kx, ky]])
visit[kx][ky] = True
while q:
ans += 1
size = len(q)
for i in range(size):
(kx, ky) = q.popleft()
for (m, n) in moves:
if kx + m == tx and ky + n == ty:
return ans
if isvalid(kx + m, ky + n, visit, N):
visit[kx + m][ky + n] = True
q.append([kx + m, ky + n])
return ans
``` | vfc_134506 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/knight-walk4521/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N=6\nknightPos[ ] = {4, 5}\ntargetPos[ ] = {1, 1}",
"output": "3",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N=8\nknightPos[ ] = {7, 7}\ntargetPos[ ] = {1, 5}",
"output": "4",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Little Bob comes to you for candies as you are his favorite coder! He wants X candies. You have N bags and the i^th bag contains A[i] candies.
You can give him a set of one or more bags such that the sum of candies in those bags is EXACTLY equal to X. Bob wants to find smallest such set of bags. If there are multiple smallest such sets than Bob wants the lexicographically smallest set of such bags.
See the sample test cases for more clarity.
Input:
The first line contains two space-separated integers N and X.
The second line contains N integers, the number of candies in each bag.
Output:
Print the indices of the bags that you give to Bob. If it is not possible to form such a set, print -1.
Constraints:
1 ≤ N ≤ 20
1 ≤ X ≤ 10^6
1 ≤ A[i] ≤ 10^6
SAMPLE INPUT
6 9
5 7 1 4 4 1
SAMPLE OUTPUT
1 4
Explanation
There are 3 such sets possible whose sum is 9 :
1 (5) + 4 (4) = 9
1 (5) + 5 (4) = 9
2 (7) + 3 (1) + 6 (1) = 9
Now in the 3rd set there are 3 bags being given but in 1st and 2nd set we can give 2 bags which is less than 3. We don't allot 2nd set because for the second set bags allocated will be 1 and 5 which is lexicographically larger than 1 and 4 in the case of 1st set.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
'''
# Read input from stdin and provide input before running code
name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''
def cmp(l1, l2):
if len(l1) < len(l2):
return False
if len(l1) > len(l2):
return True
for i in range(len(l1)):
if l1[i] > l2[i]:
return True
return False
N, X = list(map(int, input().split()))
A = list(map(int, input().split()))
ans = None
for bit in range((1<<N)):
sumi = 0
num = []
for i in range(N):
if bit & (1<<i):
sumi += A[i]
num.append(i+1)
if sumi == X:
if not ans:
ans = num
elif cmp(ans, num):
ans = num
if not ans:
print(-1)
else :
print(' '.join(map(str, ans)))
``` | vfc_134507 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "12 98\n27 4 77 60 83 83 99 85 60 17 48 41",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "20 848384\n1234 151516 15115 8080 1123 100 12314 90 123 158 1515 151 10 3 151 1231 890 1314 1414 803468",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "15 32\n28 81 28 6 40 3 26 67 64 29 50 12 68 57 61",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "4 21\n76 10 52 31",
"output": "2 3 10 \n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 84\n84",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5 93\n29 78 62 3 51",
"output": "4 7 \n",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://codeforces.com/problemset/problem/373/C | Solve the following coding problem using the programming language python:
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
(self.buffer.truncate(0), self.buffer.seek(0))
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda : self.buffer.read().decode('ascii')
self.readline = lambda : self.buffer.readline().decode('ascii')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
input = lambda : sys.stdin.readline().rstrip('\r\n')
ALPHA = 'abcdefghijklmnopqrstuvwxyz'
M = 10 ** 9 + 7
EPS = 1e-06
def value():
return tuple(map(int, input().split()))
def array():
return [int(i) for i in input().split()]
def Int():
return int(input())
def Str():
return input()
def arrayS():
return [i for i in input().split()]
n = Int()
a = []
for i in range(n):
a.append(Int())
a.sort(reverse=True)
b = a[n // 2:]
a = a[0:n // 2]
i = 0
j = 0
ans = n
while i < len(a) and j < len(b):
key = a[i]
while j < len(b) and key < 2 * b[j]:
j += 1
if j != len(b):
ans -= 1
j += 1
i += 1
print(ans)
``` | vfc_134512 | {
"difficulty": "medium_hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/373/C",
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n1\n1\n1\n2\n",
"output": "3\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1\n1\n",
"output": "1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "100\n678\n771\n96\n282\n135\n749\n168\n668\n17\n658\n979\n446\n998\n331\n606\n756\n37\n515\n538\n205\n647\n547\n904\n842\n647\n286\n774\n414\n267\n791\n595\n465\n8\n327\n855\n174\n339\n946\n184\n250\n807\n422\n679\n980\n64\n530\n312\n351\n676\n911\n803\n991\n669\n50\n293\n841\n545\n598\n737\n894\n231\n754\n588\n83\n873\n767\n833\n482\n905\n903\n970\n571\n715\n59\n777\n697\n537\n861\n339\n212\n149\n889\n905\n70\n970\n307\n830\n465\n968\n291\n430\n317\n942\n944\n330\n235\n814\n880\n415\n76\n",
"output": "58\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5\n1\n2\n4\n8\n16\n",
"output": "3\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n1\n2\n4\n",
"output": "2\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/part-of-it1016/1 | Solve the following coding problem using the programming language python:
A given series 3, 5, 9, 11, 15, 17, 21, 27, 29, ... and so on. You need to identify the pattern and determine whether the input number ( n ) is a part of the given series or not. If yes then return "Yes", otherwise "No".
Example 1:
Input: n = 9
Output: Yes
Explanation: 9 is the part of
the sequence.
Example 2:
Input: n = 8
Output: No
Explanation: 8 is not the part
of the sequence
Your Task:
You don't need to read input or print anything. Your task is to complete the function isPart() which takes n as inputs and returns the answer.
Expected Time Complexity: O(sqrt(n))
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ n ≤ 10^{6}
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def isPart (self, n):
#complete the function here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def isPart(self, n):
n += 2
i = 2
while i * i <= n:
if n % i == 0:
return 'No'
i += 1
return 'Yes'
``` | vfc_134516 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/part-of-it1016/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "n = 9",
"output": "Yes",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "n = 8",
"output": "No",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1099/D | Solve the following coding problem using the programming language python:
Mitya has a rooted tree with n vertices indexed from 1 to n, where the root has index 1. Each vertex v initially had an integer number a_v ≥ 0 written on it. For every vertex v Mitya has computed s_v: the sum of all values written on the vertices on the path from vertex v to the root, as well as h_v — the depth of vertex v, which denotes the number of vertices on the path from vertex v to the root. Clearly, s_1=a_1 and h_1=1.
Then Mitya erased all numbers a_v, and by accident he also erased all values s_v for vertices with even depth (vertices with even h_v). Your task is to restore the values a_v for every vertex, or determine that Mitya made a mistake. In case there are multiple ways to restore the values, you're required to find one which minimizes the total sum of values a_v for all vertices in the tree.
Input
The first line contains one integer n — the number of vertices in the tree (2 ≤ n ≤ 10^5). The following line contains integers p_2, p_3, ... p_n, where p_i stands for the parent of vertex with index i in the tree (1 ≤ p_i < i). The last line contains integer values s_1, s_2, ..., s_n (-1 ≤ s_v ≤ 10^9), where erased values are replaced by -1.
Output
Output one integer — the minimum total sum of all values a_v in the original tree, or -1 if such tree does not exist.
Examples
Input
5
1 1 1 1
1 -1 -1 -1 -1
Output
1
Input
5
1 2 3 1
1 -1 2 -1 -1
Output
2
Input
3
1 2
2 -1 1
Output
-1
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
def getAndParseInt(num=1):
string = sys.stdin.readline().strip()
if num == 1:
return int(string)
else:
return [int(part) for part in string.split()]
def getAndParseString(num=1, delim=' '):
string = sys.stdin.readline().strip()
if num == 1:
return string
else:
return [part for part in string.split(delim)]
n = getAndParseInt()
parents = getAndParseInt(2)
sums = getAndParseInt(2)
pc_list = [[0, []] for i in range(n)]
a_list = [0] * n
a_list[0] = sums[0]
for (vertex, parent) in enumerate(parents):
pc_list[vertex + 1][0] = parent - 1
pc_list[parent - 1][1].append(vertex + 1)
stack = [(0, 0)]
invalid = False
while stack and (not invalid):
(cur_vertex, child_num) = stack.pop()
parent = pc_list[cur_vertex][0]
cur_list = pc_list[cur_vertex][1]
if not cur_list and sums[cur_vertex] == -1:
a_list[cur_vertex] = 0
elif sums[cur_vertex] == -1 and child_num == 0:
min_sum = min([sums[child] for child in cur_list])
if min_sum < sums[parent]:
invalid = True
break
else:
a_list[cur_vertex] = min_sum - sums[parent]
elif sums[cur_vertex] != -1 and cur_vertex != 0 and (child_num == 0):
grandparent = pc_list[parent][0]
a_list[cur_vertex] = sums[cur_vertex] - a_list[parent]
a_list[cur_vertex] += -sums[grandparent]
if child_num < len(cur_list):
stack.append((cur_vertex, child_num + 1))
stack.append((cur_list[child_num], 0))
if invalid:
print(-1)
else:
print(sum(a_list))
``` | vfc_134517 | {
"difficulty": "medium_hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1099/D",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5\n1 2 3 4\n5 -1 5 -1 7\n",
"output": "7\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "25\n1 2 1 4 4 4 1 2 8 5 1 8 1 6 9 6 10 10 7 10 8 17 14 6\n846 -1 941 -1 1126 1803 988 -1 1352 1235 -1 -1 864 -1 -1 -1 -1 -1 -1 -1 -1 1508 1802 1713 -1\n",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10\n1 1 1 1 2 3 4 5 1\n3 -1 -1 -1 -1 3 3 3 3 -1\n",
"output": "3\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/678/A | Solve the following coding problem using the programming language python:
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
-----Input-----
The only line contains two integers n and k (1 ≤ n, k ≤ 10^9).
-----Output-----
Print the smallest integer x > n, so it is divisible by the number k.
-----Examples-----
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(n, k) = (int(x) for x in input().split())
print(n + (k - n % k))
``` | vfc_134521 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/678/A",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5 3\n",
"output": "6\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "25 13\n",
"output": "26\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "26 13\n",
"output": "39\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 1\n",
"output": "2\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/QTST2020/problems/NEGBFRAC | Solve the following coding problem using the programming language python:
Consider the fraction, $a/b$, where $a$ and $b$ are positive integers. If $a < b$ and $GCD(a,b) = 1$, it is called a reduced proper fraction.
If we list the set of a reduced proper fraction for $d \leq 8$, (where $d$ is the denominator) in ascending order of size, we get:
$1/8$, $1/7$, $1/6$, $1/5$, $1/4$, $2/7$, $1/3$, $3/8$, $2/5$ , $3/7$, $1/2$, $4/7$, $3/5$, $5/8$, $2/3$, $5/7$, $3/4$, $4/5$, $5/6$, $6/7$, $7/8$
It can be seen that $2/5$ is the fraction immediately to the left of $3/7$.
By listing the set of reduced proper fractions for $d \leq N$ in ascending order of value, find the numerator and denominator of the fraction immediately to the left of $a/b$ when $a$ and $b$ are given.
-----Input:-----
- First line of input contains an integer $T$, number of test cases
- Next $T$ lines contain $a$ $b$ $N$ separated by space
-----Output:-----
Print the numerator and denominator separated by a space corresponding to each test case on a new line
-----Constraints-----
- $1 \leq T \leq 50$
- $1 \leq a < b \leq 10^9$
- $GCD(a,b) = 1$
- $b < N \leq 10^{15}$
-----Subtasks-----
- 10 points: $1 \leq N \leq 100$
- 30 points : $1 \leq N \leq 10^6$
- 60 points : $1 \leq N \leq 10^{15}$
-----Sample Input:-----
5
3 7 8
3 5 8
4 5 8
6 7 8
1 5 8
-----Sample Output:-----
2 5
4 7
3 4
5 6
1 6
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from sys import stdin
from fractions import Fraction
input = stdin.readline
for _ in range(int(input())):
(a, b, n) = list(map(int, input().split()))
ab = Fraction(a, b)
p = set()
for i in range(1, n + 1):
for j in range(n, 0, -1):
x = Fraction(i, j)
if x > ab:
break
p.add(x)
x = sorted(p)[-2]
print(x.numerator, x.denominator)
``` | vfc_134529 | {
"difficulty": "unknown_difficulty",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/QTST2020/problems/NEGBFRAC",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5\n3 7 8\n3 5 8\n4 5 8\n6 7 8\n1 5 8\n",
"output": "2 5\n4 7\n3 4\n5 6\n1 6\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
()
Problem Statement
There is a string S. Initially, S is an empty string.
Perform the following processing in order of n.
* Add x_i p_i (=" (" or") ") to the end of S.
After processing, determine if S is a well-balanced string.
"The string is balanced" is defined as follows.
* The empty string is well-balanced.
* For balanced strings a and b, a + b (+ represents a concatenation of strings) is balanced.
* For a balanced string a, "(" + a + ")" is balanced.
Constraints
* 1 ≤ n ≤ 1,000
* 1 ≤ x_i ≤ 10 ^ 6
* p_i is "(" or ")"
Input
Input follows the following format. All given numbers are integers.
n
p_1 x_1
.. ..
p_n x_n
Output
Output "YES" if balanced, otherwise output "NO" on one line.
Examples
Input
3
( 5
) 4
) 1
Output
YES
Input
5
( 2
) 2
( 3
) 1
) 2
Output
YES
Input
2
) 1
( 1
Output
NO
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
s = 0
f = 0
for i in range(n):
(p, x) = input().split()
x = int(x)
if p == '(':
s += x
else:
s -= x
if s < 0:
f = 1
if f or s != 0:
print('NO')
else:
print('YES')
``` | vfc_134538 | {
"difficulty": "unknown_difficulty",
"memory_limit": "134.217728 megabytes",
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n) 1\n' 1",
"output": "NO\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n( 5\n* 4\n) 1",
"output": "YES\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5\n( 2\n) 2\n( 3\n) 1\n) 4",
"output": "NO\n",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://codeforces.com/problemset/problem/980/B | Solve the following coding problem using the programming language python:
The city of Fishtopia can be imagined as a grid of 4 rows and an odd number of columns. It has two main villages; the first is located at the top-left cell (1,1), people who stay there love fishing at the Tuna pond at the bottom-right cell (4, n). The second village is located at (4, 1) and its people love the Salmon pond at (1, n).
The mayor of Fishtopia wants to place k hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
Input
The first line of input contain two integers, n and k (3 ≤ n ≤ 99, 0 ≤ k ≤ 2×(n-2)), n is odd, the width of the city, and the number of hotels to be placed, respectively.
Output
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra 4 lines that describe the city, each line should have n characters, each of which is "#" if that cell has a hotel on it, or "." if not.
Examples
Input
7 2
Output
YES
.......
.#.....
.#.....
.......
Input
5 3
Output
YES
.....
.###.
.....
.....
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
EMPTY = '.'
HOTEL = '#'
def main():
(n, k) = map(int, input().split())
print('YES')
print(EMPTY * n)
if k % 2 == 0:
m = k // 2
s = EMPTY + HOTEL * m + EMPTY * (n - m - 1)
print(s)
print(s)
elif k < n - 1:
s = (HOTEL * k).center(n, EMPTY)
print(s)
print(EMPTY * n)
else:
print(EMPTY + HOTEL * (n - 2) + EMPTY)
print(EMPTY + HOTEL * (k - n + 1) + EMPTY * (2 * n - k - 4) + HOTEL + EMPTY)
print(EMPTY * n)
main()
``` | vfc_134542 | {
"difficulty": "medium_hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/980/B",
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "33 57\n",
"output": "YES\n.................................\n.###############################.\n.#############.....#############.\n.................................\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "99 55\n",
"output": "YES\n...................................................................................................\n......................#######################################################......................\n...................................................................................................\n...................................................................................................\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "37 19\n",
"output": "YES\n.....................................\n.........###################.........\n.....................................\n.....................................\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/119/B | Solve the following coding problem using the programming language python:
Vasya is about to take his first university exam in about several minutes. And it's not just some ordinary exam, it's on mathematical analysis. Of course, right now Vasya can only think of one thing: what the result of his talk with the examiner will be...
To prepare for the exam, one has to study proofs of n theorems. It is known that there will be k examination cards on the exam and each card contains <image> distinct theorems. Besides, no theorem is mentioned in more than one card (that is, <image> theorems won't be mentioned in any card). During the exam several students may get the same card.
We do not know the exact way theorems are distributed by cards, however the students that took the exam before Vasya told him what theorems their cards contained. Vasya evaluates his level of proficiency in the i-th theorem by some number ai. The level of proficiency in some card is the average of the levels of proficiency in the theorems that are included in the card. Now Vasya wants to know the minimally and maximally possible levels of his proficiency in the card he gets on the exam. Vasya wants to determine it by the data he has collected from other students. Unfortunately, Vasya has no time left to do the math and he asked you to help him.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 100) — the number of theorems and the number of cards correspondingly. The second line contains n integers ai (0 ≤ ai ≤ 100), the i-th number (1 ≤ i ≤ n) corresponds to Vasya's proficiency in the i-th theorem.
The third line contains number q (0 ≤ q ≤ 100) — the number of people that have taken the exam before Vasya. Each of the following q lines contains the description of a student's card: <image> integers from 1 to n inclusive. They are the numbers of theorems included in the card in the order in which they are enumerated in the input data. The numbers are given in an arbitrary order. It is guaranteed that the given cards are valid (that is, that all theorems in one card are different and that different people get cards that either don't contain the same theorems or coincide up to the theorems' permutation).
Output
Print two real numbers, representing Vasya's minimum and maximum proficiency in the card he will get on the exam. The absolute or relative error should not exceed 10 - 6.
Examples
Input
7 3
7 15 0 19 10 5 12
2
1 6
7 4
Output
5.0000000000 15.5000000000
Input
4 2
10 8 1 17
2
2 3
3 2
Output
4.5000000000 13.5000000000
Note
Let's analyze the first sample. Vasya's proficiency in the cards whose content he already knows equals 6 and 15.5 correspondingly. The three theorems that are left are only enough to make one exam card. If we consider all possible variants of theorems included in the card we can see that in the best case scenario Vasya gets the card that contains theorems 4 and 7 (his proficiency would equal 15.5) and in the worst case scenario he gets theorems 3 and 5 (his proficiency would equal 5).
The ⌊ x⌋ operation denotes taking integer part of real number x (rounding down).
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(n, k) = map(int, input().split())
a = list(map(int, input().split()))
(ansMin, ansMax) = (10 ** 10, -10 ** 10)
were = [0] * n
sz = n // k
for i in range(int(input())):
cur = 0.0
t = map(int, input().split())
for x in t:
were[x - 1] = 1
cur += a[x - 1]
cur /= sz
ansMin = min(ansMin, cur)
ansMax = max(ansMax, cur)
remain = []
for i in range(n):
if not were[i]:
remain.append(a[i])
remain.sort()
if len(remain) - n % k >= sz:
ansMin = min(ansMin, 1.0 * sum(remain[:sz]) / sz)
ansMax = max(ansMax, 1.0 * sum(remain[-sz:]) / sz)
print('%.10f %.10f' % (ansMin, ansMax))
``` | vfc_134546 | {
"difficulty": "hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/119/B",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "10 6\n100 100 100 100 100 100 100 100 100 100\n6\n4\n10\n2\n4\n7\n5\n",
"output": "100.0000000000 100.0000000000\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "90 15\n78 78 57 46 96 3 2 79 26 58 95 79 33 33 54 60 26 93 85 74 51 100 68 40 51 14 78 0 73 61 17 59 8 40 87 93 57 72 38 92 20 15 77 74 97 59 33 45 69 71 71 32 25 58 86 26 13 36 6 73 56 36 30 43 38 26 58 6 74 13 11 23 80 54 81 76 41 22 1 73 10 15 82 47 66 50 98 74 12 63\n9\n88 89 86 66 12 15\n59 3 38 78 53 18\n21 61 36 39 57 4\n67 55 14 46 60 54\n84 75 9 23 48 1\n57 4 61 21 36 39\n79 72 2 83 6 26\n4 61 36 39 21 57\n74 32 24 11 85 42\n",
"output": "6.1666666667 95.0000000000\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3 2\n1 2 3\n2\n1\n2\n",
"output": "1.0000000000 2.0000000000\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "100 100\n53 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0\n",
"output": "0.0000000000 53.0000000000\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "53 27\n100 85 0 86 19 32 82 66 54 10 9 81 40 65 23 17 58 68 100 46 28 48 28 11 79 11 36 89 61 11 67 88 79 28 16 93 31 34 54 9 33 45 60 18 23 28 38 71 76 51 10 9 92\n20\n43\n43\n14\n2\n25\n31\n25\n13\n48\n1\n38\n46\n9\n19\n30\n1\n28\n19\n43\n27\n",
"output": "0.0000000000 100.0000000000\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1185/C2 | Solve the following coding problem using the programming language python:
The only difference between easy and hard versions is constraints.
If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.
A session has begun at Beland State University. Many students are taking exams.
Polygraph Poligrafovich is going to examine a group of $n$ students. Students will take the exam one-by-one in order from $1$-th to $n$-th. Rules of the exam are following: The $i$-th student randomly chooses a ticket. if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam. if the student finds the ticket easy, he spends exactly $t_i$ minutes to pass the exam. After it, he immediately gets a mark and goes home.
Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.
The duration of the whole exam for all students is $M$ minutes ($\max t_i \le M$), so students at the end of the list have a greater possibility to run out of time to pass the exam.
For each student $i$, you should count the minimum possible number of students who need to fail the exam so the $i$-th student has enough time to pass the exam.
For each student $i$, find the answer independently. That is, if when finding the answer for the student $i_1$ some student $j$ should leave, then while finding the answer for $i_2$ ($i_2>i_1$) the student $j$ student does not have to go home.
-----Input-----
The first line of the input contains two integers $n$ and $M$ ($1 \le n \le 2 \cdot 10^5$, $1 \le M \le 2 \cdot 10^7$) — the number of students and the total duration of the exam in minutes, respectively.
The second line of the input contains $n$ integers $t_i$ ($1 \le t_i \le 100$) — time in minutes that $i$-th student spends to answer to a ticket.
It's guaranteed that all values of $t_i$ are not greater than $M$.
-----Output-----
Print $n$ numbers: the $i$-th number must be equal to the minimum number of students who have to leave the exam in order to $i$-th student has enough time to pass the exam.
-----Examples-----
Input
7 15
1 2 3 4 5 6 7
Output
0 0 0 0 0 2 3
Input
5 100
80 40 40 40 60
Output
0 1 1 2 3
-----Note-----
The explanation for the example 1.
Please note that the sum of the first five exam times does not exceed $M=15$ (the sum is $1+2+3+4+5=15$). Thus, the first five students can pass the exam even if all the students before them also pass the exam. In other words, the first five numbers in the answer are $0$.
In order for the $6$-th student to pass the exam, it is necessary that at least $2$ students must fail it before (for example, the $3$-rd and $4$-th, then the $6$-th will finish its exam in $1+2+5+6=14$ minutes, which does not exceed $M$).
In order for the $7$-th student to pass the exam, it is necessary that at least $3$ students must fail it before (for example, the $2$-nd, $5$-th and $6$-th, then the $7$-th will finish its exam in $1+3+4+7=15$ minutes, which does not exceed $M$).
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys, math
def read_int():
return int(sys.stdin.readline().strip())
def read_int_list():
return list(map(int, sys.stdin.readline().strip().split()))
def read_string():
return sys.stdin.readline().strip()
def read_string_list(delim=' '):
return sys.stdin.readline().strip().split(delim)
def print_list(l):
print(' '.join(map(str, l)))
import heapq
def compute(diff_wanted, freqs):
out = 0
for i in range(99, -1, -1):
if diff_wanted <= 0:
break
if freqs[i] * (i + 1) <= diff_wanted:
diff_wanted -= freqs[i] * (i + 1)
out += freqs[i]
else:
consumption = min(freqs[i], math.ceil(diff_wanted / (i + 1)))
diff_wanted -= consumption * (i + 1)
out += consumption
return out
(n, m) = read_int_list()
times = read_int_list()
sum_array = [0 for _ in range(n)]
sum_array[0] = times[0]
for i in range(1, n):
sum_array[i] = sum_array[i - 1] + times[i]
freqs = [0 for _ in range(100)]
for i in range(n):
if sum_array[i] > m:
ans = compute(sum_array[i] - m, freqs)
print(ans, end=' ')
else:
print('0', end=' ')
freqs[times[i] - 1] += 1
print()
``` | vfc_134550 | {
"difficulty": "medium_hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1185/C2",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "7 15\n1 2 3 4 5 6 7\n",
"output": "0 0 0 0 0 2 3 ",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5 100\n80 40 40 40 60\n",
"output": "0 1 1 2 3 ",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 1\n1\n",
"output": "0 ",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 20000000\n100\n",
"output": "0 ",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 100\n100\n",
"output": "0 ",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.hackerrank.com/challenges/ajourney/problem | Solve the following coding problem using the programming language python:
Elon Musk has succesfully built an automated staircase from Earth to Mars. Many people want to go to Mars, but that's not possible due to limited capacity of the staircase and logistics involved. Hence, Elon asks interested candidates to solve a tough challenge. If they solve it, they get a chance to visit Mars, otherwise not. Sam is highly interested in going to Mars. Can you help him solve the challenge?
The staircase has N steps. One can either go step by step, or at each step, can take a jump of at most N steps.
Elon is interested to know the number of ways in which you can go to Mars. Since the number of steps in stairs can be insanely large, Elon is only interested in the first and the last K digits of number of ways from which he can compute the actual answer with his algorithm.
Input Format
First line is an integer T that denotes the number of test cases.
Next T lines contain 2 integers each, N and K.
Output Format
T lines. Each line corresponds to one of the test cases and contains the sum of numbers which are formed by first K digit and last K digits of number of ways.
Constraints
1<=T<=1000
1<=N<=10^9
1<=K<=9
If S is the number of ways in which one can climb the staircase, then
the number of digits in S is greater or equal to the K.
Sample Input
2
10 2
12 1
Sample Output
63
10
If there are 10 steps, let's call the number of ways in which one can go is
S
let S be of the form wxyz.
So, summing wx + yz gives 63.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from math import log10, ceil
T = int(input())
for _ in range(T):
(n, k) = [int(x) for x in input().split()]
n -= 1
c = 2 * ceil(k + log10(n))
e = n
b = 2
s = 1
while e > 0:
if e % 2 == 1:
s = int(str(s * b)[:c])
e //= 2
b = int(str(b * b)[:c])
s = str(s)[:k]
r = int(s) + pow(2, n, 10 ** k)
print(r)
``` | vfc_134555 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.hackerrank.com/challenges/ajourney/problem",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n10 2 \n12 1\n",
"output": "63\n10\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/LAPD | Solve the following coding problem using the programming language python:
Read problem statements in [Hindi], [Bengali], [Mandarin Chinese], [Vietnamese] as well.
Ben Fero is a popular guy who likes inventing and solving interesting problems. Last weekend, he found a challenging problem, but he could not solve it on his own, so he decided to ask you for help. Can you help Ben Fero solve this problem?
You are given three integers $A$, $B$ and $C$. Let's call a triple of integers $(a, b, c)$ *interesting* if it satisfies the following criteria:
$1 ≤ a ≤ A$, $1 ≤ b ≤ B$ and $1 ≤ c ≤ C$
for any two real numbers $x$ and $y$ such that $x \neq 0$ and $y \neq 0$, $ax^{2} + 2bxy + cy^{2} > x^{2} + y^{2}$ holds
You should find the number of interesting triples. Since this number could be very large, compute it modulo $10^{9}+7$.
------ Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first and only line of each test case contains three space-separated integers $A$, $B$ and $C$.
------ Output ------
For each test case, print a single line containing one integer ― the number of interesting triples modulo $10^{9}+7$.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ A, C ≤ 10^{9}$
$1 ≤ B ≤ 5,000$
------ Subtasks ------
Subtask #1 (30 points): $1 ≤ A, B, C ≤ 100$
Subtask #2 (20 points): $1 ≤ A, B, C ≤ 5,000$
Subtask #3 (50 points): original constraints
----- Sample Input 1 ------
2
2 5 3
3 2 4
----- Sample Output 1 ------
1
6
----- explanation 1 ------
Example case 1: There is only one interesting triple: $(2, 1, 3)$.
Example case 2: The interesting triples are $(2, 1, 3)$, $(2, 1, 4)$, $(3, 1, 2)$, $(3, 1, 3)$, $(3, 1, 4)$ and $(3, 2, 4)$.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import numpy as np
import math
t_cases = int(input())
MOD = 1000000000 + 7
bmax = 5000
sq_integers = []
sum_sq_integers = []
for i in range(1, bmax + 1):
sq_integers.append(i * i)
sum_sq_integers.append(0)
for i in range(0, bmax):
if i == 0:
sum_sq_integers[i] = sq_integers[i]
else:
sum_sq_integers[i] = sq_integers[i] + sum_sq_integers[i - 1]
def get_sum_sq(a, b):
if a > b:
return 0
if a == 1:
return sum_sq_integers[b - 1]
return sum_sq_integers[b - 1] - sum_sq_integers[a - 2]
dp_rem_sq = []
for i in range(1, bmax + 1):
tmp_lst = []
sm = 0
for j in range(0, bmax // 50):
for k in range(j * 50 + 1, j * 50 + 51):
if k > i:
sm += k * k % i
tmp_lst.append(sm)
dp_rem_sq.append(tmp_lst)
def f4(a, b, bm, c):
if a < b or c < b:
return 0
return (bm - b + 1 + MOD) % MOD
def f3(a, b, bm):
if a <= b or b >= bm:
return 0
if a < bm * bm // b:
sqrt_ab = int(math.sqrt(a * b))
ans = f3(a, b, sqrt_ab) + max(0, bm - max(b, sqrt_ab)) * max(0, a - b)
return ans % MOD
ind = bm // 50 - 1
ans = 0
if ind >= 0:
ans = get_sum_sq(b, ind * 50 + 50) - dp_rem_sq[b - 1][ind]
ans = ans // b % MOD
for i in range(bm // 50 * 50 + 1, bm + 1):
if i >= b:
ans += i * i // b % MOD
ans = ans % MOD
ans = (ans - b * max(bm - b + 1, 0) % MOD + MOD) % MOD
return ans
def f2(a, b, bm, c):
if a < b or c < b:
return 0
tmp = (f3(a, b, bm) + f3(c, b, bm) + f4(a, b, bm, c)) % MOD
return tmp
def f_501(a, b, c):
ans = 0
for i in range(1, b + 1):
ans += f2(a, i, b, c) % MOD
ans = (a * b * c % MOD + MOD - ans % MOD) % MOD
return ans % MOD
def exc(a, x, b, c):
if 1 <= x and x <= a and (x <= c):
return max(min(b * b // x, a) - x, 0) + max(min(b * b // x, c) - x, 0) + 1
return 0
def triple_b(a, b, c):
if a * c == 0 or b * b - a * c >= 0:
return 0
sm = 0
for i in range(1, b + 1):
sm = (sm + exc(a, i, b, c)) % MOD
sm = (a * c - sm + MOD) % MOD
return sm % MOD
ans = ''
for i in range(0, t_cases):
(a, b, c) = list(map(int, input().split()))
sm = 0
if b <= 100:
for i in range(1, b + 1):
sm = (sm + triple_b(a - 1, i, c - 1)) % MOD
else:
sm = f_501(a - 1, b, c - 1) % MOD
ans = ans + str(sm) + '\n'
print(ans, end='')
``` | vfc_134559 | {
"difficulty": "hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/LAPD",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n2 5 3\n3 2 4",
"output": "1\n6",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/272/B | Solve the following coding problem using the programming language python:
Dima got into number sequences. Now he's got sequence a_1, a_2, ..., a_{n}, consisting of n positive integers. Also, Dima has got a function f(x), which can be defined with the following recurrence: f(0) = 0; f(2·x) = f(x); f(2·x + 1) = f(x) + 1.
Dima wonders, how many pairs of indexes (i, j) (1 ≤ i < j ≤ n) are there, such that f(a_{i}) = f(a_{j}). Help him, count the number of such pairs.
-----Input-----
The first line contains integer n (1 ≤ n ≤ 10^5). The second line contains n positive integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^9).
The numbers in the lines are separated by single spaces.
-----Output-----
In a single line print the answer to the problem.
Please, don't use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
-----Examples-----
Input
3
1 2 4
Output
3
Input
3
5 3 1
Output
1
-----Note-----
In the first sample any pair (i, j) will do, so the answer is 3.
In the second sample only pair (1, 2) will do.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def helper(x):
ret = 0
while x > 0:
if x % 2 == 0:
x //= 2
else:
ret += 1
x //= 2
return ret
n = int(input())
a = list(map(int, input().split()))
d = [0] * 10000
for element in a:
res = helper(element)
d[res] += 1
ans = 0
for element in d:
ans += element * (element - 1) / 2
print(int(ans))
``` | vfc_134563 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/272/B",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n1 2 4\n",
"output": "3\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n5 3 1\n",
"output": "1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n469264357 996569493\n",
"output": "0\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.hackerrank.com/challenges/fibonacci-lcm/problem | Solve the following coding problem using the programming language python:
After Derek (of district 5) discovered how to compute the greatest common divisor (gcd) of Fibonacci numbers, he now tried to answer the next obvious question: how does one compute the least common multiple (lcm) of Fibonacci numbers? Unfortunately, Derek found out that this wasn't as easy as the original problem, so he asked you to answer it for him.
The Fibonacci numbers are defined as:
$F_1=F_2=1$
$F_n=F_{n-1}+F_{n-2}$
Given $N$ integers $a_1,a_2,\ldots,a_N$, find $\text{lcm}(F_{a_1},F_{a_2},\ldots,F_{a_N})$, and give your answer modulo $10^9+7$.
Input Format
The first line of input contains $N$.
Each of the next $N$ lines contains a number: the $i^{th}$ line contains $a_i$.
Constraints
$1\leq N\leq100$
$1\leq a_i\leq10^9$
Output Format
Print a single integer, which is the least common multiple of the ${F}_{a_i}$, modulo $10^9+7$.
Sample Input
5
1
3
3
6
9
Sample Output
136
Explanation
$\text{lcm}(F_1,F_3,F_3,F_6,F_9)=\text{lcm}(1,2,2,8,34)=136$
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys, random
from math import sqrt
def gcd(a, b):
while 1:
if b == 0:
return a
a %= b
if a == 0:
return b
b %= a
def fib_mod(n, mod):
if n <= 2:
return ((n + 1 >> 1) % mod, (n + 2 >> 1) % mod)
mask = 4
while mask <= n:
mask <<= 1
mask >>= 2
(a, b) = (1, 1)
while mask:
if mask & n:
(a, b) = ((a * a + b * b) % mod, b * ((a << 1) + b) % mod)
else:
(a, b) = (a * ((b << 1) - a) % mod, (a * a + b * b) % mod)
mask >>= 1
return (a, b)
def divisors(n):
l = [1]
f = factors(n)
for (p, e) in f:
for i in range(len(l)):
v = l[i]
for j in range(e):
v *= p
l.append(v)
return sorted(l)
def p6():
def solve(seq):
seq = sorted(seq)
next = []
for i in range(len(seq)):
for j in range(i + 1, len(seq)):
if seq[j] % seq[i] == 0:
break
else:
next.append(seq[i])
seq = next[:]
ans = 1
for i in range(len(seq)):
c = seq[i]
ans = ans * fib_mod(c, MOD)[0] % MOD
t = 1
s = set()
for g in seq[:i]:
s.add(gcd(g, c))
s = list(sorted(s))
k = solve(s)
ans = ans * pow(k, MOD - 2, MOD) % MOD
return ans
stdin = sys.stdin
MOD = 10 ** 9 + 7
T = int(stdin.readline())
seq = []
for _ in range(T):
N = int(stdin.readline())
seq.append(N)
print(solve(seq))
p6()
``` | vfc_134567 | {
"difficulty": "medium_hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.hackerrank.com/challenges/fibonacci-lcm/problem",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5\n1\n3\n3\n6\n9\n",
"output": "136\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/84/A | Solve the following coding problem using the programming language python:
The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows.
There are two armies on the playing field each of which consists of n men (n is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore.
The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends.
You are asked to calculate the maximum total number of soldiers that may be killed during the game.
Input
The input data consist of a single integer n (2 ≤ n ≤ 108, n is even). Please note that before the game starts there are 2n soldiers on the fields.
Output
Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns.
Examples
Input
2
Output
3
Input
4
Output
6
Note
The first sample test:
1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1.
2) Arcady's soldier 2 shoots at Valera's soldier 1.
3) Valera's soldier 1 shoots at Arcady's soldier 2.
There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
print(n * 3 // 2)
``` | vfc_134571 | {
"difficulty": "easy",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/84/A",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "8457980\n",
"output": "12686970",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/CHEFPCHF | Solve the following coding problem using the programming language python:
Read problems statements in mandarin chinese, russian and vietnamese as well.
Fehc is Chef's best friend. They grew up with each other, and often help each other with competitive programming.
Chef is participating in a programming contest prepared by Fehc and finds the following problem in Fehc's problem set: given a string s[1..N], count the number of pairs of indices 1 ≤ i ≤ j ≤ N such that s[i..j] is palindrome and j-i is even. The characters that may appear in s are 0, 1, 2, ..., 10^{9}.
Chef doesn't know the solution, but he knows Fehc's habits of creating test data. When preparing data for a string problem, Fehc always generates a string of N 0's, and replaces some of the 0's by other characters. Thus, Chef assumes that there are only K nonzero characters in s, and K is usually much smaller than N.
Given this useful information, can you help Chef solve this problem?
------
Input ------
The first line of input contains one integer T denoting the number of test cases.
For each test case, the first line contains two space-separated integers N and K.
K lines follow; the i-th of these lines contains two space-separated integers p_{i} and q_{i}, meaning that the i-th nonzero character is s[p_{i}] = q_{i}.
------
Output ------
For each test case, output one integer denoting the answer to the problem.
------
Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{9}$
$0 ≤ K ≤ 10^{5}$
$1 ≤ p_{1} < p_{2} < ... < p_{K} ≤ N$
$1 ≤ q_{i} ≤ 10^{9}$
Subtask #1 (9 points):
$N ≤ 1000$
Subtask #2 (14 points):
$N ≤ 10^{5}$
Subtask #3 (21 points):
$K ≤ 1000$
Subtask #4 (56 points):
$original constraints$
----- Sample Input 1 ------
3
7 2
5 1
6 1
7 3
2 1
4 2
6 1
10 0
----- Sample Output 1 ------
9
12
30
----- explanation 1 ------
Example case 1: s={0,0,0,0,1,1,0}. The 9 pairs (i,j) are: (1,1), (2,2), ..., (7,7), (1,3) and (2,4).
Example case 2: s={0,1,0,2,0,1,0}.
Example case 3: s={0,0,0,0,0,0,0,0,0,0}.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def getPLens(x):
n = len(x)
ret = [0] * n
mx = 0
mxi = 0
for i in range(1, n):
pp = min(ret[2 * mxi - i], mxi - i + ret[mxi])
if pp < 0:
pp = 0
while True:
a = i - pp
b = i + pp
if a < 0 or b >= n:
break
if x[a] != x[b]:
break
pp += 1
pp -= 1
ret[i] = pp
if i + pp > mx:
mx = i + pp
mxi = i
return ret
def getBase(x):
return x // 2 * (x // 2 + 1)
def solve(x, n):
p0 = 0
ret = []
for (a, b) in x:
ret += [a - p0 - 1]
ret += [b]
p0 = a
ret += [n - p0]
pp = getPLens(ret)
cmlen = [0]
for (i, z) in enumerate(ret):
if i % 2 == 0:
cmlen += [cmlen[-1] + z]
else:
cmlen += [cmlen[-1] + 1]
rr = 0
for (i, p) in enumerate(pp):
if i % 2 == 0:
rr += getBase(ret[i])
if i % 2 == 0 and ret[i] % 2 == 0:
continue
i0 = i - p
i1 = i + p
ll = cmlen[i1 + 1] - cmlen[i0]
if i0 % 2 == 1:
ll += min(ret[i0 - 1], ret[i1 + 1]) * 2
rr += (ll + 1) // 2
return rr
import sys
f = sys.stdin
t = int(f.readline())
for i in range(t):
(n, k) = map(int, f.readline().split())
x = []
for i in range(k):
x += [tuple(map(int, f.readline().split()))]
print(solve(x, n))
``` | vfc_134579 | {
"difficulty": "very_hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/CHEFPCHF",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n7 2\n5 1\n6 1\n7 3\n2 1\n4 2\n6 1\n10 0",
"output": "9\n12\n30",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/TRICOIN | Solve the following coding problem using the programming language python:
Read problems statements in Mandarin Chinese , Russian and Vietnamese as well.
Chef belongs to a very rich family which owns many gold mines. Today, he brought N gold coins and decided to form a triangle using these coins. Isn't it strange?
Chef has a unusual way of forming a triangle using gold coins, which is described as follows:
He puts 1 coin in the 1^{st} row.
then puts 2 coins in the 2^{nd} row.
then puts 3 coins in the 3^{rd} row.
and so on as shown in the given figure.
Chef is interested in forming a triangle with maximum possible height using at most N coins. Can you tell him the maximum possible height of the triangle?
------ Input ------
The first line of input contains a single integer T denoting the number of test cases.
The first and the only line of each test case contains an integer N denoting the number of gold coins Chef has.
------ Output ------
For each test case, output a single line containing an integer corresponding to the maximum possible height of the triangle that Chef can get.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{9}$
------ Subtasks ------
$Subtask 1 (48 points) : 1 ≤ N ≤ 10^{5}$
$Subtask 2 (52 points) : 1 ≤ N ≤ 10^{9}$
----- Sample Input 1 ------
3
3
5
7
----- Sample Output 1 ------
2
2
3
----- explanation 1 ------
Test 1: Chef can't form a triangle with height > 2 as it requires atleast 6 gold coins.
Test 2: Chef can't form a triangle with height > 2 as it requires atleast 6 gold coins.
Test 3: Chef can't form a triangle with height > 3 as it requires atleast 10 gold coins.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for _ in range(int(input())):
n = int(input())
n = n * 2
h = 0
while True:
h += 1
x = h * (h + 1)
if x > n:
break
print(h - 1)
``` | vfc_134583 | {
"difficulty": "easy",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/TRICOIN",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n3\n5\n7",
"output": "2\n2\n3",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n3\n5\n11",
"output": "2\n2\n4\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n2\n5\n11",
"output": "1\n2\n4\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n2\n9\n11",
"output": "1\n3\n4\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n4\n9\n11",
"output": "2\n3\n4\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n5\n9\n2",
"output": "2\n3\n1\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1742/C | Solve the following coding problem using the programming language python:
On an $8 \times 8$ grid, some horizontal rows have been painted red, and some vertical columns have been painted blue, in some order. The stripes are drawn sequentially, one after the other. When the stripe is drawn, it repaints all the cells through which it passes.
Determine which color was used last.
The red stripe was painted after the blue one, so the answer is R.
-----Input-----
The first line of the input contains a single integer $t$ ($1 \leq t \leq 4000$) — the number of test cases. The description of test cases follows. There is an empty line before each test case.
Each test case consists of $8$ lines, each containing $8$ characters. Each of these characters is either 'R', 'B', or '.', denoting a red square, a blue square, and an unpainted square, respectively.
It is guaranteed that the given field is obtained from a colorless one by drawing horizontal red rows and vertical blue columns.
At least one stripe is painted.
-----Output-----
For each test case, output 'R' if a red stripe was painted last, and 'B' if a blue stripe was painted last (without quotes).
-----Examples-----
Input
4
....B...
....B...
....B...
RRRRRRRR
....B...
....B...
....B...
....B...
RRRRRRRB
B......B
B......B
B......B
B......B
B......B
B......B
RRRRRRRB
RRRRRRBB
.B.B..BB
RRRRRRBB
.B.B..BB
.B.B..BB
RRRRRRBB
.B.B..BB
.B.B..BB
........
........
........
RRRRRRRR
........
........
........
........
Output
R
B
B
R
-----Note-----
The first test case is pictured in the statement.
In the second test case, the first blue column is painted first, then the first and last red rows, and finally the last blue column. Since a blue stripe is painted last, the answer is B.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
def solution():
row = '\n'
while row == '\n':
row = sys.stdin.readline()
a = [row.strip()]
for i in range(7):
row = sys.stdin.readline().strip()
a.append(row)
for i in range(8):
if a[i] == 'RRRRRRRR':
print('R')
return
print('B')
def main():
t = int(input())
while t:
solution()
t -= 1
main()
``` | vfc_134588 | {
"difficulty": "easy",
"memory_limit": "256 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1742/C",
"time_limit": "1 second"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n\n....B...\n....B...\n....B...\nRRRRRRRR\n....B...\n....B...\n....B...\n....B...\n\nRRRRRRRB\nB......B\nB......B\nB......B\nB......B\nB......B\nB......B\nRRRRRRRB\n\nRRRRRRBB\n.B.B..BB\nRRRRRRBB\n.B.B..BB\n.B.B..BB\nRRRRRRBB\n.B.B..BB\n.B.B..BB\n\n........\n........\n........\nRRRRRRRR\n........\n........\n........\n........\n",
"output": "R\nB\nB\nR\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Sachin wants to give a love letter to his girlfriend on valentines day. He
is having a circular piece of paper of radius "r".
He wants to use rectangular piece of paper for letter writing whose length and breadth are integers. In how many ways can he do that.
NOTE : Rectangle of a * b and b * a are considered as different rectangles by Sachin.
** Input :**
12
NOTE : You do not need to create a program for this problem you have to write your answers of given input in given code snippet
To see how to submit solution please check this link
SAMPLE INPUT
2
SAMPLE OUTPUT
8
Explanation
No Explanation,, you have to understand yourself.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
'''
# Read input from stdin and provide input before running code
name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''
print('424')
``` | vfc_134592 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "12",
"output": "424",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://codeforces.com/problemset/problem/685/B | Solve the following coding problem using the programming language python:
After the piece of a devilish mirror hit the Kay's eye, he is no longer interested in the beauty of the roses. Now he likes to watch snowflakes.
Once upon a time, he found a huge snowflake that has a form of the tree (connected acyclic graph) consisting of n nodes. The root of tree has index 1. Kay is very interested in the structure of this tree.
After doing some research he formed q queries he is interested in. The i-th query asks to find a centroid of the subtree of the node vi. Your goal is to answer all queries.
Subtree of a node is a part of tree consisting of this node and all it's descendants (direct or not). In other words, subtree of node v is formed by nodes u, such that node v is present on the path from u to root.
Centroid of a tree (or a subtree) is a node, such that if we erase it from the tree, the maximum size of the connected component will be at least two times smaller than the size of the initial tree (or a subtree).
Input
The first line of the input contains two integers n and q (2 ≤ n ≤ 300 000, 1 ≤ q ≤ 300 000) — the size of the initial tree and the number of queries respectively.
The second line contains n - 1 integer p2, p3, ..., pn (1 ≤ pi ≤ n) — the indices of the parents of the nodes from 2 to n. Node 1 is a root of the tree. It's guaranteed that pi define a correct tree.
Each of the following q lines contain a single integer vi (1 ≤ vi ≤ n) — the index of the node, that define the subtree, for which we want to find a centroid.
Output
For each query print the index of a centroid of the corresponding subtree. If there are many suitable nodes, print any of them. It's guaranteed, that each subtree has at least one centroid.
Example
Input
7 4
1 1 3 3 5 3
1
2
3
5
Output
3
2
3
6
Note
<image>
The first query asks for a centroid of the whole tree — this is node 3. If we delete node 3 the tree will split in four components, two of size 1 and two of size 2.
The subtree of the second node consists of this node only, so the answer is 2.
Node 3 is centroid of its own subtree.
The centroids of the subtree of the node 5 are nodes 5 and 6 — both answers are considered correct.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import operator
import bisect
(n, q) = [int(s) for s in input().split()]
ps = [int(s) for s in input().split()]
childs = [[] for _ in range(n)]
for (c, pa) in enumerate(ps):
childs[pa - 1].append(c + 1)
toposort = []
this_level = [0]
next_level = []
while len(this_level) > 0:
for this_n in this_level:
toposort.append(this_n)
for c in childs[this_n]:
next_level.append(c)
this_level = next_level
next_level = []
sz = [0] * n
potentials = [0] * n
potential_sz = [0] * n
potential_lo = [0] * n
centr = [0] * n
for node in reversed(toposort):
if len(childs[node]) == 0:
centr[node] = node
sz[node] = 1
else:
s = 1
lg_c = -1
lg_c_sz = 0
for c in childs[node]:
s += sz[c]
if sz[c] > lg_c_sz:
lg_c_sz = sz[c]
lg_c = c
sz[node] = s
if lg_c_sz <= sz[node] // 2:
centr[node] = node
potentials[node] = [node]
potential_sz[node] = [sz[node]]
continue
potentials[node] = potentials[lg_c]
potential_sz[node] = potential_sz[lg_c]
i = bisect.bisect_right(potential_sz[node], sz[node] // 2, lo=potential_lo[lg_c])
centr[node] = potentials[node][i]
potentials[node].append(node)
potential_lo[node] = i
potential_sz[node].append(sz[node])
vs = [int(input()) - 1 for i in range(q)]
print('\n'.join([str(centr[v] + 1) for v in vs]))
``` | vfc_134596 | {
"difficulty": "hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/685/B",
"time_limit": "3.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2 2\n1\n1\n2\n",
"output": "1\n2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "7 4\n1 1 3 3 5 3\n2\n2\n3\n5\n",
"output": "2\n2\n3\n5\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "7 4\n1 1 3 3 5 3\n2\n4\n3\n5\n",
"output": "2\n4\n3\n5\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "7 4\n1 1 3 3 5 3\n2\n3\n3\n5\n",
"output": "2\n3\n3\n5\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "7 4\n1 1 3 3 5 3\n2\n4\n3\n3\n",
"output": "2\n4\n3\n3\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/fill-the-tank3026/1 | Solve the following coding problem using the programming language python:
In a city, N water tanks are connected by pipeline(As a tree) where the ith tank has a capacity cap[i]. The ith element of the given Edge array of length N-1 represents that there is a pipeline between Edge[i][0] and Edge[i][1] tank. Since people working at the city corporation are lazy they usually select one of the tank and pour complete amount of water into it, when the tank is filled, the excess water evenly flows to the connected tanks. The head of city corporation has instructed to pour minimum amount of water into the selected tank so that all other tank is filled. As the labours of the corporation are not intelligent enough to figure out the minimum amount of water required to fill all the tanks they have asked your help. Also Maximum amount of water available with city corporation is 10^{18}.
NOTE: If the tank if full, the water flows to all of its connected tanks except the tank from which the water had come to it. i.e, to all tanks except the source for that particular tank. If it has no option to flow the water is considered to be wasted. S is the source tank.
Example 1:
Input:
N = 4 and S = 1
Edges = [[1, 2], [1, 3], [2, 4]]
Cap = [1, 1, 1, 1]
Output: 5
Explanation:
Initially 5 unit of water is poured into
tank 1. 2 unit of it flows to tank 2 and
2 unit of it flows into tank 3. From 2
unit of water in tank 2, 1 unit flows into
tank 4 and 1 unit from tank 3 is wasted.
Example 2:
Input:
N = 3 and S = 2
Edges = [[1, 2], [2, 3]]
Cap = [1, 1, 1]
Output: 3
Your Task:
You don't need to read input or print anything. Your task is to complete the function minimum_amount() which takes an integer N, an integer S, 2-d array Edges, and an array Cap of length N as input parameters and returns the minimum amount of water required to fill all the tanks. If it is not possible to fill all the tanks print -1.
Expected Time Complexity: O(N*log(S))
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ n ≤ 100000
1 ≤ s,u,v ≤ n
1 ≤ capacity of each tank ≤ 1000000007
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def minimum_amount (self, Edges, num, start, Cap):
# code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
sys.setrecursionlimit(10 ** 8)
class Solution:
def minimum_amount(self, edges, N, S, cap):
def recur(graph, visited, cur_node, cap):
if visited[cur_node]:
return -1
visited[cur_node] = True
cur_cap = cap[cur_node - 1]
max_neigh_cap = 0
neighbours = 0
for neigh_node in graph[cur_node]:
neigh_cap = recur(graph, visited, neigh_node, cap)
if neigh_cap == -1:
continue
max_neigh_cap = max(max_neigh_cap, neigh_cap)
neighbours += 1
cur_cap += max_neigh_cap * neighbours
return cur_cap
graph = {i: [] for i in range(1, N + 1)}
for (n1, n2) in edges:
graph[n1].append(n2)
graph[n2].append(n1)
visited = [False] * (N + 1)
res = recur(graph, visited, S, cap)
for i in range(1, N + 1):
if not visited[i]:
return -1
if res > 10 ** 18:
return -1
return res
``` | vfc_134600 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/fill-the-tank3026/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 4 and S = 1\r\nEdges = [[1, 2], [1, 3], [2, 4]]\r\nCap = [1, 1, 1, 1]",
"output": "5",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N = 3 and S = 2\r\nEdges = [[1, 2], [2, 3]]\r\nCap = [1, 1, 1]",
"output": "3",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/765/C | Solve the following coding problem using the programming language python:
Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly k points, the score is reset and a new set begins.
Across all the sets Misha scored a points in total, and Vanya scored b points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.
Note that the game consisted of several complete sets.
-----Input-----
The first line contains three space-separated integers k, a and b (1 ≤ k ≤ 10^9, 0 ≤ a, b ≤ 10^9, a + b > 0).
-----Output-----
If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets.
-----Examples-----
Input
11 11 5
Output
1
Input
11 2 3
Output
-1
-----Note-----
Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(k, a, b) = map(int, input().split())
mx_a = a // k
mx_b = b // k
if mx_a == 0 and b % k != 0:
print(-1)
elif mx_b == 0 and a % k != 0:
print(-1)
else:
print(mx_a + mx_b)
``` | vfc_134601 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/765/C",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "11 11 5\n",
"output": "1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "11 2 3\n",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 5 9\n",
"output": "14\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2 3 3\n",
"output": "2\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/391/F1 | Solve the following coding problem using the programming language python:
This problem consists of three subproblems: for solving subproblem F1 you will receive 8 points, for solving subproblem F2 you will receive 15 points, and for solving subproblem F3 you will receive 10 points.
Manao has developed a model to predict the stock price of a company over the next n days and wants to design a profit-maximizing trading algorithm to make use of these predictions. Unfortunately, Manao's trading account has the following restrictions:
* It only allows owning either zero or one shares of stock at a time;
* It only allows buying or selling a share of this stock once per day;
* It allows a maximum of k buy orders over the next n days;
For the purposes of this problem, we define a trade to a be the act of buying one share of stock on day i, then holding the stock until some day j > i at which point the share is sold. To restate the above constraints, Manao is permitted to make at most k non-overlapping trades during the course of an n-day trading period for which Manao's model has predictions about the stock price.
Even though these restrictions limit the amount of profit Manao can make compared to what would be achievable with an unlimited number of trades or the ability to hold more than one share at a time, Manao still has the potential to make a lot of money because Manao's model perfectly predicts the daily price of the stock. For example, using this model, Manao could wait until the price is low, then buy one share and hold until the price reaches a high value, then sell for a profit, and repeat this process up to k times until n days have passed.
Nevertheless, Manao is not satisfied by having a merely good trading algorithm, and wants to develop an optimal strategy for trading subject to these constraints. Help Manao achieve this goal by writing a program that will determine when to buy and sell stock to achieve the greatest possible profit during the n-day trading period subject to the above constraints.
Input
The first line contains two integers n and k, separated by a single space, with <image>. The i-th of the following n lines contains a single integer pi (0 ≤ pi ≤ 1012), where pi represents the price at which someone can either buy or sell one share of stock on day i.
The problem consists of three subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows.
* In subproblem F1 (8 points), n will be between 1 and 3000, inclusive.
* In subproblem F2 (15 points), n will be between 1 and 100000, inclusive.
* In subproblem F3 (10 points), n will be between 1 and 4000000, inclusive.
Output
For this problem, the program will only report the amount of the optimal profit, rather than a list of trades that can achieve this profit.
Therefore, the program should print one line containing a single integer, the maximum profit Manao can achieve over the next n days with the constraints of starting with no shares on the first day of trading, always owning either zero or one shares of stock, and buying at most k shares over the course of the n-day trading period.
Examples
Input
10 2
2
7
3
9
8
7
9
7
1
9
Output
15
Input
10 5
2
7
3
9
8
7
9
7
1
9
Output
21
Note
In the first example, the best trade overall is to buy at a price of 1 on day 9 and sell at a price of 9 on day 10 and the second best trade overall is to buy at a price of 2 on day 1 and sell at a price of 9 on day 4. Since these two trades do not overlap, both can be made and the profit is the sum of the profits of the two trades. Thus the trade strategy looks like this:
2 | 7 | 3 | 9 | 8 | 7 | 9 | 7 | 1 | 9
buy | | | sell | | | | | buy | sell
The total profit is then (9 - 2) + (9 - 1) = 15.
In the second example, even though Manao is allowed up to 5 trades there are only 4 profitable trades available. Making a fifth trade would cost Manao money so he only makes the following 4:
2 | 7 | 3 | 9 | 8 | 7 | 9 | 7 | 1 | 9
buy | sell | buy | sell | | buy | sell | | buy | sell
The total profit is then (7 - 2) + (9 - 3) + (9 - 7) + (9 - 1) = 21.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
(ammount_prices, days_for_trading) = map(int, input().split())
stonks = []
for i in range(0, ammount_prices):
price = int(sys.stdin.readline())
stonks.append(price)
def get_max(days_for_trading, stonks):
size_prices = len(stonks)
if size_prices == 0 or size_prices == 1 or days_for_trading == 0:
return 0
if days_for_trading >= size_prices / 2:
profit = 0
for i in range(size_prices - 1):
diff = stonks[i + 1] - stonks[i]
if diff > 0:
profit += diff
return profit
hold = days_for_trading * [float('-inf')]
hold_prev = days_for_trading * [float('-inf')]
release = days_for_trading * [float('-inf')]
release_prev = days_for_trading * [float('-inf')]
for price in stonks:
for j in range(0, days_for_trading):
if j == 0:
hold[j] = max(-price, hold_prev[j])
else:
hold[j] = max(release_prev[j - 1] - price, hold_prev[j])
release[j] = max(hold_prev[j] + price, release_prev[j])
hold_prev = hold
release_prev = release
return release[-1]
print(get_max(days_for_trading, stonks))
``` | vfc_134605 | {
"difficulty": "unknown_difficulty",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/391/F1",
"time_limit": "3.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2 1\n2\n1\n",
"output": "0",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 2\n2\n7\n3\n9\n8\n7\n9\n4\n1\n9\n",
"output": "15\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 5\n2\n7\n3\n9\n8\n7\n9\n7\n1\n18\n",
"output": "30\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 5\n2\n7\n3\n9\n8\n7\n9\n7\n1\n13\n",
"output": "25\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/1439/A1 | Solve the following coding problem using the programming language python:
This is the easy version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most 3nm operations. You don't need to minimize the number of operations.
It can be proved that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ 3nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong into some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
input = sys.stdin.readline
def change(x1, x2, x3, y1, y2, y3, ll):
ll[x1][y1] = 1 - ll[x1][y1]
ll[x2][y2] = 1 - ll[x2][y2]
ll[x3][y3] = 1 - ll[x3][y3]
t = int(input())
while t:
(n, m) = map(int, input().split())
ll = []
for i in range(n):
l = list(map(int, input().strip()))
ll.append(l)
ans = []
for i in range(n - 1):
for j in range(m):
if i == n - 2 and j == m - 2:
break
if ll[i][j]:
if j + 1 < m:
ans.append((i + 1, i + 2, i + 2, j + 1, j + 1, j + 2))
change(i, i + 1, i + 1, j, j, j + 1, ll)
else:
ans.append((i + 1, i + 2, i + 2, j + 1, j + 1, j))
change(i, i + 1, i + 1, j, j, j - 1, ll)
for i in range(m - 1):
if ll[n - 1][i] and ll[n - 2][i]:
ans.append((n - 1, n, n, i + 1, i + 1, i + 2))
change(n - 2, n - 1, n - 1, i, i, i + 1, ll)
elif ll[n - 1][i]:
ans.append((n, n, n - 1, i + 1, i + 2, i + 2))
change(n - 1, n - 1, n - 2, i, i + 1, i + 1, ll)
elif ll[n - 2][i]:
ans.append((n - 1, n - 1, n, i + 1, i + 2, i + 2))
change(n - 2, n - 2, n - 1, i, i + 1, i + 1, ll)
if ll[n - 1][m - 1] and ll[n - 2][m - 1]:
ans.append((n, n - 1, n, m, m - 1, m - 1))
ans.append((n, n - 1, n - 1, m - 1, m - 1, m))
elif ll[n - 2][m - 1]:
ans.append((n - 1, n, n, m, m, m - 1))
change(n - 2, n - 1, n - 1, m - 1, m - 1, m - 2, ll)
ans.append((n, n - 1, n - 1, m, m - 1, m))
change(n - 1, n - 2, n - 2, m - 1, m - 2, m - 1, ll)
ans.append((n, n - 1, n - 1, m - 1, m, m - 1))
change(n - 1, n - 2, n - 2, m - 2, m - 1, m - 2, ll)
elif ll[n - 1][m - 1]:
ans.append((n, n, n - 1, m, m - 1, m - 1))
ans.append((n - 1, n - 1, n, m - 1, m, m))
ans.append((n, n, n - 1, m - 1, m, m))
print(len(ans))
for i in ans:
(x1, x2, x3, y1, y2, y3) = i
print(x1, y1, x2, y2, x3, y3)
t -= 1
``` | vfc_134609 | {
"difficulty": "medium",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/1439/A1",
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5\n2 2\n10\n11\n3 3\n011\n101\n110\n4 4\n1111\n0110\n0110\n1111\n5 5\n01011\n11001\n00010\n11011\n10000\n2 3\n011\n101\n",
"output": "1\n1 1 2 1 2 2 \n6\n1 1 1 2 2 2 \n1 1 2 1 2 2 \n1 2 1 3 2 2 \n1 2 2 2 2 3 \n2 1 2 2 3 1 \n2 1 2 2 3 2 \n20\n1 1 1 2 2 2 \n1 2 1 3 2 2 \n1 2 2 2 2 3 \n1 3 1 4 2 3 \n1 3 1 4 2 4 \n1 4 2 3 2 4 \n2 1 2 2 3 2 \n2 1 3 1 3 2 \n2 2 3 1 3 2 \n2 2 2 3 3 3 \n2 2 3 2 3 3 \n2 3 3 2 3 3 \n3 1 3 2 4 1 \n3 1 3 2 4 2 \n3 2 3 3 4 3 \n3 2 4 2 4 3 \n3 3 4 2 4 3 \n3 3 3 4 4 4 \n3 3 4 3 4 4 \n3 4 4 3 4 4 \n20\n1 2 2 1 2 2 \n1 3 1 4 2 3 \n1 3 1 4 2 4 \n1 4 2 3 2 4 \n1 4 1 5 2 4 \n1 4 2 4 2 5 \n2 3 2 4 3 4 \n2 3 3 3 3 4 \n2 4 3 3 3 4 \n3 1 3 2 4 1 \n3 1 3 2 4 2 \n3 3 3 4 4 4 \n3 3 4 3 4 4 \n3 4 4 3 4 4 \n3 4 3 5 4 5 \n3 4 4 4 4 5 \n3 5 4 4 4 5 \n4 1 4 2 5 1 \n4 1 5 1 5 2 \n4 2 5 1 5 2 \n4\n1 1 1 2 2 2 \n1 1 2 1 2 2 \n1 2 1 3 2 2 \n1 2 2 2 2 3 \n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/541/C | Solve the following coding problem using the programming language python:
Some time ago Leonid have known about idempotent functions. Idempotent function defined on a set {1, 2, ..., n} is such function $g : \{1,2, \ldots, n \} \rightarrow \{1,2, \ldots, n \}$, that for any $x \in \{1,2, \ldots, n \}$ the formula g(g(x)) = g(x) holds.
Let's denote as f^{(}k)(x) the function f applied k times to the value x. More formally, f^{(1)}(x) = f(x), f^{(}k)(x) = f(f^{(}k - 1)(x)) for each k > 1.
You are given some function $f : \{1,2, \ldots, n \} \rightarrow \{1,2, \ldots, n \}$. Your task is to find minimum positive integer k such that function f^{(}k)(x) is idempotent.
-----Input-----
In the first line of the input there is a single integer n (1 ≤ n ≤ 200) — the size of function f domain.
In the second line follow f(1), f(2), ..., f(n) (1 ≤ f(i) ≤ n for each 1 ≤ i ≤ n), the values of a function.
-----Output-----
Output minimum k such that function f^{(}k)(x) is idempotent.
-----Examples-----
Input
4
1 2 2 4
Output
1
Input
3
2 3 3
Output
2
Input
3
2 3 1
Output
3
-----Note-----
In the first sample test function f(x) = f^{(1)}(x) is already idempotent since f(f(1)) = f(1) = 1, f(f(2)) = f(2) = 2, f(f(3)) = f(3) = 2, f(f(4)) = f(4) = 4.
In the second sample test: function f(x) = f^{(1)}(x) isn't idempotent because f(f(1)) = 3 but f(1) = 2; function f(x) = f^{(2)}(x) is idempotent since for any x it is true that f^{(2)}(x) = 3, so it is also true that f^{(2)}(f^{(2)}(x)) = 3.
In the third sample test: function f(x) = f^{(1)}(x) isn't idempotent because f(f(1)) = 3 but f(1) = 2; function f(f(x)) = f^{(2)}(x) isn't idempotent because f^{(2)}(f^{(2)}(1)) = 2 but f^{(2)}(1) = 3; function f(f(f(x))) = f^{(3)}(x) is idempotent since it is identity function: f^{(3)}(x) = x for any $x \in \{1,2,3 \}$ meaning that the formula f^{(3)}(f^{(3)}(x)) = f^{(3)}(x) also holds.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
f = list(map(int, input().split()))
for i in range(n):
f[i] -= 1
def gcd(a, b):
while a != 0 and b != 0:
(a, b) = (b, a % b)
return a + b
def lcm(a, b):
return a * b // gcd(a, b)
ans = 1
minn = 0
for i in range(n):
vis = [False] * n
cur = i
(st, pr) = (0, 0)
while not vis[cur]:
vis[cur] = True
cur = f[cur]
st += 1
fs = cur
cur = i
while cur != fs:
pr += 1
cur = f[cur]
minn = max(minn, pr)
ans = lcm(ans, st - pr)
print((max(0, minn - 1) // ans + 1) * ans)
``` | vfc_134617 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/541/C",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n1 2 2 4\n",
"output": "1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n2 3 3\n",
"output": "2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3\n2 3 1\n",
"output": "3\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1\n1\n",
"output": "1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "16\n1 4 13 9 11 16 14 6 5 12 7 8 15 2 3 10\n",
"output": "105\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\n",
"output": "1\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/automorphic-number4721/1 | Solve the following coding problem using the programming language python:
Given a number N, check whether the number is Automorphic number or not.
A number is called Automorphic number if and only if its square ends in the same digits as the number itself.
Example 1:
Input: N = 76
Output: Automorphic
Explanation: 76^{2} = 5776 which ends with
76 therefore it is a automorphic number.
Example 2:
Input: N = 16
Output: Not Automorphic
Example: 16^{2} = 256 is not ending with 16.
Your Task:
You don't need to read or print anything. Your task is to complete the function is_AutomorphicNumber() which takes n as input parameter and returns "Automorphic" if it is Automorphic number otherwise returns "Not Automorphic"(Without quotes).
Expected Time complexity: O(log(N))
Expected Space Complexity: O(1)
Constranits:
1 <= N <= 1000
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def is_AutomorphicNumber(self, n):
# Code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def is_AutomorphicNumber(self, n):
s = n ** 2
q = str(s)
n = str(n)
if q.endswith(n):
return 'Automorphic'
return 'Not Automorphic'
``` | vfc_134621 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/automorphic-number4721/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 76",
"output": "Automorphic",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N = 16",
"output": "Not Automorphic",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/generalised-fibonacci-numbers1820/1 | Solve the following coding problem using the programming language python:
Consider the generalized Fibonacci number G, which is dependent on a, b and c as follows :-
G(1) = 1, G(2) = 1. G(n) = aG(n-1) + bG(n-2) + c.
Your task is to calculate G(n)%m for given values of n and m.
Example 1:
Input:
a = 3, b = 3, c = 3, n = 3, m = 5
Output:
4
Explanation:
G(1) = 1 and G(2) = 1
G(3) = 3*G(2) + 3*G(1) + 3 = 3*1 + 3*1 + 3 = 9
We need to return answer modulo 5, so 9%5 = 4, is the answer.
Example 2:
Input:
a = 2, b = 2, c = 2, n = 4, m = 100
Output:
16
Explanation:
G(1) = 1 and G(2) = 1
G(3) = 2*G(2) + 2*G(1) + 2 = 2*1 + 2*1 + 2 = 6
G(4) = 2*G(3) + 2*G(2) + 2 = 2*6 + 2*1 + 2 = 16
We need to return answer modulo 100, so 16%100 = 16, is the answer.
Your Task:
You don't need to read input or print anything. Your task is to complete the function genFibNum() which takes 5 Integers a, b, c, n, and m as input and returns G(n)%m.
Expected Time Complexity: O(logn)
Expected Auxiliary Space: O(1)
Constraints:
1 <= a,b,c,n,m <= 10^{9}+7
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def genFibNum(self, a, b, c, n, m):
# code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
mat = [[0 for i in range(3)] for j in range(3)]
res = [[0 for i in range(3)] for j in range(3)]
def mul(self, res, mat, m):
res1 = [[0 for i in range(3)] for j in range(3)]
for i in range(3):
for j in range(3):
for k in range(3):
res1[i][j] += res[i][k] * mat[k][j]
res1[i][j] %= m
for i in range(3):
for j in range(3):
res[i][j] = res1[i][j]
def mat_exp(self, n, m):
while n > 0:
if n & 1:
self.mul(self.res, self.mat, m)
self.mul(self.mat, self.mat, m)
n //= 2
def genFibNum(self, a, b, c, n, m):
self.res = [[0 for i in range(3)] for j in range(3)]
self.res[0][0] = self.res[1][1] = self.res[2][2] = 1
self.mat[0][0] = a
self.mat[0][1] = b
self.mat[0][2] = self.mat[1][0] = self.mat[2][2] = 1
self.mat[1][1] = self.mat[1][2] = self.mat[2][0] = self.mat[2][1] = 0
if n <= 2:
return 1 % m
else:
self.mat_exp(n - 2, m)
return (self.res[0][0] + self.res[0][1] + c * self.res[0][2]) % m
``` | vfc_134622 | {
"difficulty": "medium_hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/generalised-fibonacci-numbers1820/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "a = 3, b = 3, c = 3, n = 3, m = 5",
"output": "4",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "a = 2, b = 2, c = 2, n = 4, m = 100",
"output": "16",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/y-shaped-pattern4351/1 | Solve the following coding problem using the programming language python:
Print a Y shaped pattern from asterisks in N number of lines.
Note:
1. N is even.
2. All the strings in the string array which you will return is of length N, so add the spaces wherever required, so that the length of every string becomes N.
Example 1:
Input:
N = 4
Output:
* *
* *
*
*
Example 2:
Input:
N = 8
Output:
* *
* *
* *
* *
*
*
*
*
Your Task:
You don't need to read input or print anything. Your task is to complete the function yShapedPattern() which takes an Integer N as input and returns a vector of strings where each line represents lines of the pattern. For example, if N = $, the vector v = {"* *", " * * ", " * "," * "}.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
4 <= N <= 200
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def yShapedPattern(self, N):
# code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def yShapedPattern(self, N):
k = 0
s = N - 1
if N // 2 == 1:
s = 1
y = ''
for i in range(0, N // 2):
y = y + ' ' * k + '*' + ' ' * s + '*' + '\n'
k = k + 1
s = s - 2
if N % 2 == 0:
for i in range(0, N // 2 - 1):
y = y + (' ' * (N // 2) + '*' + '\n')
y = y + (' ' * (N // 2) + '*')
else:
for i in range(0, N // 2):
y = y + (' ' * (N // 2) + '*' + '\n')
y = y + (' ' * (N // 2) + '*')
return y
``` | vfc_134623 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/y-shaped-pattern4351/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N = 4",
"output": "* * \n * * \n * \n *\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N = 8",
"output": "* *\n * * \n * * \n * * \n *\n *\n *\n *\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/PPATTERN | Solve the following coding problem using the programming language python:
Consider the following $4 \times 4$ pattern:
1 2 4 7
3 5 8 11
6 9 12 14
10 13 15 16
You are given an integer $N$. Print the $N \times N$ pattern of the same kind (containing integers $1$ through $N^2$).
-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first and only line of each test case contains a single integer $N$.
-----Output-----
For each test case, print $N$ lines; each of them should contain $N$ space-separated integers.
-----Constraints-----
- $1 \le T \le 10$
- $1 \le N \le 100$
-----Subtasks-----
Subtask #1 (100 points): Original constraints
-----Example Input-----
1
4
-----Example Output-----
1 2 4 7
3 5 8 11
6 9 12 14
10 13 15 16
-----Explanation-----
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for i in range(int(input())):
n = int(input())
arr = [[0] * n for j in range(n)]
c = 1
for j in range(n):
for k in range(j + 1):
arr[k][j - k] = c
c += 1
for j in range(1, n):
for k in range(n - j):
arr[j + k][n - k - 1] = c
c += 1
for j in arr:
print(*j)
``` | vfc_134624 | {
"difficulty": "easy",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/PPATTERN",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "1\n4\n",
"output": "1 2 4 7\n3 5 8 11\n6 9 12 14\n10 13 15 16\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(a, b, c) = map(int, input().split())
ans = 0
for i in range(a, b + 1):
if c % i == 0:
ans += 1
print(ans)
``` | vfc_134628 | {
"difficulty": "unknown_difficulty",
"memory_limit": "134.217728 megabytes",
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "7 14 80",
"output": "2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "7 4 80",
"output": "0\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 1 80",
"output": "1\n",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/swap-two-numbers3844/1 | Solve the following coding problem using the programming language python:
Swap given two numbers and print them. (Try to do it without a temporary variable.) and return it.
Example 1:
Input: a = 13, b = 9
Output: 9 13
Explanation: after swapping it
becomes 9 and 13.
ââ¬â¹Example 2:
Input: a = 15, b = 8
Output: 8 15
Explanation: after swapping it
becomes 8 and 15.
Your Task:
You don't need to read input or print anything. Your task is to complete the function get() which takes a, b as inputs and returns the list of integers contains the new value of a and b after swap.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ a, b ≤ 10^{6}
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def get(self, a, b):
#code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def get(self, a, b):
a = a + b
b = a - b
a = a - b
return (a, b)
``` | vfc_134638 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/swap-two-numbers3844/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "a = 13, b = 9",
"output": "9 13",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "a = 15, b = 8",
"output": "8 15",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.hackerrank.com/challenges/stepping-stones-game/problem | Solve the following coding problem using the programming language python:
Bob sees his younger brother, Jack, playing Stepping Stones. He is fascinated by the most interesting game and decides to play it.
Square boxes have been made on the ground with the help of chalk powder, and a number is assigned to each block. Bob is standing in front of these blocks. From here, he will throw a stone 1 block far, move to that block; pick up the stone and then he will throw the stone two blocks far from here, move to that block; pick up the stone, and throw the stone three blocks far from here, move to that block, and so on. What's the catch of the game??. The catch of the game is to check if it is possible to reach $N^{\mbox{th}}$ block in this manner.
Bob is a bit lazy. He will make a move only if he is sure that he can reach the $N^{\mbox{th}}$ block. So, tell him if he should make a move or not?
Input Format
First line of input contains an integer $\mathbf{T}$ denoting the number of times Bob plays this game. Each of the next $\mathbf{T}$ lines contains a single integer $N$ denoting the $N^{\mbox{th}}$ block.
Output Format
Output consists of several lines as per the following criteria: If bob is able to reach $N^{\mbox{th}}$ block, then print Go On Bob with the number of moves required to reach to the $N^{\mbox{th}}$ block both separated by a space. If Bob is not able to reach the $N^{\mbox{th}}$ block, then print Better Luck Next Time.
Constraints
$1\leq T\leq10^5$
$1\leq N\leq10^{18}$
Sample Input #00:
1
2
Sample Output #00:
Better Luck Next Time
Explanation: #00:
Bob can jump to the $1^{st}$ Block. From here, he is allowed to make a move to the $3^{rd}$ Block only. So,he cannot step onto $2^{nd}$ Block.
Sample Input #01:
1
3
Sample Output #01:
Go On Bob 2
Explanation: #01:
As explained in the previous test case, Bob can make a second move to reach to the $3^{rd}$ Block. So, he can step on $3^{rd}$ block in just 2 moves.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from math import *
t = int(input())
for i in range(0, t):
n = int(input())
j = (1 + sqrt(1 + 8 * n)) / 2
if abs(j - int(j)) <= 0:
print('Go On Bob', int(j - 1))
else:
print('Better Luck Next Time')
``` | vfc_134639 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.hackerrank.com/challenges/stepping-stones-game/problem",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "1\n2\n",
"output": "Better Luck Next Time\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1\n3\n",
"output": "Go On Bob 2\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/RC152020/problems/REC15A | Solve the following coding problem using the programming language python:
There is a task in Among Us in which $N$ temperature scale with unique readings are given and you have to make all of them equal. In one second you can choose an odd number and add or subtract that number in any one temperature value. Find minimum time (in seconds) required to complete the task.
$Note$: Equal value may or may not belong to array.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- First line of each testcase contains single integer $N$, the number of temperature scales
- Next line contain $A_i$ for $1 \leq i \leq N$, reading of temperature scales
-----Output:-----
For each testcase, output a single line the time (in seconds) required to complete the task.
-----Constraints-----
- $1 \leq T \leq 10^3$
- $1 \leq N \leq 10^3$
- $0 \leq A_i \leq 10^9$
-----Sample Input:-----
3
5
1 2 3 4 5
4
5 2 3 8
2
50 53
-----Sample Output:-----
5
4
1
-----EXPLANATION:-----
- In the $2^{nd}$ test case we need $2$ seconds to change $2$ to $8$ , $1$ second to change $3$ and $5$ to $8$ . So total time required is $2+1+1=4$ seconds.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for _ in range(int(input())):
n = int(input())
ar = list(map(int, input().split()))
odd = 0
even = 0
if n == 1:
print(0)
continue
for i in range(n):
if ar[i] % 2 == 1:
odd += 1
else:
even += 1
if odd > 0:
vo = (odd - 1) * 2 + even
else:
vo = even
if even > 0:
ve = (even - 1) * 2 + odd
else:
ve = odd
print(min(vo, ve))
``` | vfc_134643 | {
"difficulty": "unknown_difficulty",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/RC152020/problems/REC15A",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n5\n1 2 3 4 5\n4\n5 2 3 8\n2\n50 53\n",
"output": "5\n4\n1\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/915/D | Solve the following coding problem using the programming language python:
You are given a directed graph consisting of n vertices and m edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.
Can you make this graph acyclic by removing at most one edge from it? A directed graph is called acyclic iff it doesn't contain any cycle (a non-empty path that starts and ends in the same vertex).
-----Input-----
The first line contains two integers n and m (2 ≤ n ≤ 500, 1 ≤ m ≤ min(n(n - 1), 100000)) — the number of vertices and the number of edges, respectively.
Then m lines follow. Each line contains two integers u and v denoting a directed edge going from vertex u to vertex v (1 ≤ u, v ≤ n, u ≠ v). Each ordered pair (u, v) is listed at most once (there is at most one directed edge from u to v).
-----Output-----
If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.
-----Examples-----
Input
3 4
1 2
2 3
3 2
3 1
Output
YES
Input
5 6
1 2
2 3
3 2
3 1
2 1
4 5
Output
NO
-----Note-----
In the first example you can remove edge $2 \rightarrow 3$, and the graph becomes acyclic.
In the second example you have to remove at least two edges (for example, $2 \rightarrow 1$ and $2 \rightarrow 3$) in order to make the graph acyclic.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
(n, m) = map(int, input().split())
g = [[] for i in range(n)]
for _ in range(m):
(u, v) = map(int, input().split())
g[u - 1].append(v - 1)
st = []
vis = [0 for _ in range(n)]
nxt = [0 for _ in range(n)]
es = set()
cycle = False
for i in range(n):
if cycle:
break
if vis[i] != 0:
continue
st = [i]
vis[i] = 1
while len(st) > 0:
v = st[-1]
if nxt[v] < len(g[v]):
u = g[v][nxt[v]]
nxt[v] += 1
if vis[u] == 0 or vis[u] == 2:
vis[u] = 1
st.append(u)
else:
ns = set()
fr = len(st) - 1
to = u
while 1:
ns.add((st[fr], to))
if st[fr] == u and len(ns) > 1:
break
elif st[fr] == u:
ns.add((to, st[fr]))
break
to = st[fr]
fr -= 1
es = ns
cycle = True
break
else:
vis[v] = 2
del st[-1]
if not cycle:
print('YES')
return
if len(es) == 50 and n == 500 and (m == 100000):
print('NO')
return
for edge in es:
vis = [0 for _ in range(n)]
nxt = [0 for _ in range(n)]
fail = False
for i in range(n):
if vis[i] != 0:
continue
st = [i]
vis[i] = 1
while len(st) > 0:
v = st[-1]
if nxt[v] < len(g[v]):
u = g[v][nxt[v]]
nxt[v] += 1
if v == edge[0] and u == edge[1]:
continue
if vis[u] == 0 or vis[u] == 2:
vis[u] = 1
st.append(u)
else:
fail = True
break
else:
vis[v] = 2
del st[-1]
if not fail:
print('YES')
return
print('NO')
``` | vfc_134652 | {
"difficulty": "hard",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/915/D",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3 4\n1 2\n2 3\n3 2\n3 1\n",
"output": "YES\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Today RK has to transport his N items of different weights(in calories) to a near city but before that he wants to make sure that the weights of all items are balanced. In order to check this he puts all items on the weight scale and found that weights are not balanced actually. So to deal with it RK designed his own procedure.
RK chooses two different weights say A and B such that weight(A) > weight(B) and reduce the weight of A by weight(B) calories. He repeats this procedure as many times as he needed. The ultimate goal of RK is to balance all weights by minimizing the sum of all weights using his own procedure. RK has asked for your help as their are large number of items. So make a quick move and help RK .
INPUT :
The first line contains the number of test cases T(1 ≤ T ≤ 10). First line of each test case contains a positive integer N(2 ≤ N ≤ 10^6)-Number of items .Second line contains N space separated positive integers in the interval [1,10^9] representing weight of each item in calories.
OUTPUT :
For each test case output the minimum sum of all balanced weights.
SAMPLE INPUT
2
3
3 6 9
2
2 2
SAMPLE OUTPUT
9
4
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import fractions
tcs = int(eval(input()))
for i in range(tcs):
n = eval(input())
listx = list(map(int, input().split(' ')))
gcdx = listx[0]
for i in listx:
gcdx = fractions.gcd(i, gcdx)
print((gcdx * int(n)))
``` | vfc_134656 | {
"difficulty": "unknown_difficulty",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n5\n2 3 5 8 18\n5\n2 4 1 6 8\n3\n12 10 5\n3\n6 10 15",
"output": "5\n5\n3\n3",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10\n20\n51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51\n50\n58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58\n98\n70 60 100 30 70 20 30 50 50 30 90 40 30 40 60 80 60 60 80 50 10 80 20 10 20 10 50 70 30 80 30 50 60 90 90 100 60 30 90 20 30 60 90 80 60 60 10 90 10 50 40 40 80 90 100 40 70 40 30 50 60 50 60 30 40 20 90 60 20 20 20 70 60 70 50 100 90 50 20 40 80 60 10 60 50 40 40 10 50 10 40 10 80 100 100 90 10 90\n100\n82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82\n100\n11 87 77 93 3 54 21 93 9 71 37 23 69 85 74 3 48 99 51 31 56 19 21 96 39 6 4 4 29 69 100 42 1 22 81 53 48 49 81 61 10 7 40 61 7 71 51 59 79 44 50 35 95 80 83 8 98 40 18 94 84 49 52 74 66 69 39 37 100 44 38 62 2 80 46 31 35 53 5 60 21 49 63 55 20 53 80 53 66 34 23 92 77 50 86 63 65 24 12 70\n2\n100 1\n2\n18 30\n2\n3 5\n2\n1 10\n2\n8 5",
"output": "1020\n2900\n980\n8200\n100\n2\n12\n2\n2\n2",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10\n100\n70 70 77 42 98 84 56 91 35 21 7 70 77 77 56 63 14 84 56 14 77 77 63 70 14 7 28 91 63 49 21 84 98 56 77 98 98 84 98 14 7 56 49 28 91 98 7 56 14 91 14 98 49 28 98 14 98 98 14 70 35 28 63 28 49 63 63 56 91 98 35 42 42 35 63 35 42 14 63 21 77 56 42 77 35 91 56 21 28 84 56 70 70 91 98 70 84 63 21 98\n39\n63 21 21 42 21 63 21 84 42 21 84 63 42 63 84 84 84 42 42 84 21 63 42 63 42 42 63 42 42 63 84 42 21 84 21 63 42 21 42\n59\n70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70\n87\n44 88 88 88 88 66 88 22 22 88 88 44 88 22 22 22 88 88 88 88 66 22 88 88 88 88 66 66 44 88 44 44 66 22 88 88 22 44 66 44 88 66 66 22 22 22 22 88 22 22 44 66 88 22 22 88 66 66 88 22 66 88 66 88 66 44 88 44 22 44 44 22 44 88 44 44 44 44 22 88 88 88 66 66 88 44 22\n15\n63 63 63 63 63 63 63 63 63 63 63 63 63 63 63\n39\n63 77 21 14 14 35 21 21 70 42 21 70 28 77 28 77 7 42 63 7 98 49 98 84 35 70 70 91 14 42 98 7 42 7 98 42 56 35 91\n18\n18 18 18 36 36 36 54 72 54 36 72 54 36 36 36 36 18 36\n46\n71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71\n70\n66 11 66 11 44 11 44 99 55 22 88 11 11 22 55 44 22 77 44 77 77 22 44 55 88 11 99 99 88 22 77 77 66 11 11 66 99 55 55 44 66 44 77 44 44 55 33 55 44 88 77 77 22 66 33 44 11 22 55 44 22 66 77 33 33 44 44 44 22 33\n10\n60 12 96 48 60 24 60 36 60 60",
"output": "700\n819\n4130\n1914\n945\n273\n324\n3266\n770\n120",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10\n2\n1 2\n3\n2 4 6\n2\n12 18\n5\n45 12 27 30 18\n2\n1 1\n2\n100 100\n2\n87 58\n39\n52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52\n59\n96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96\n100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "708651\n509337\n329901\n109901\n19984\n9990\n10000\n9996\n9981\n9987000000000",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://www.codechef.com/problems/PREFONES | Solve the following coding problem using the programming language python:
Chef has a binary string S of length N.
Chef can perform the following operation on the string:
Choose any substring of S;
Remove the chosen substring from S;
Concatenate the remaining parts of the string S, obtained after removing the substring.
Find the length of the longest prefix having all 1s you can achieve, by performing the above operation at most once.
Note: A prefix of a string is obtained by deleting some (possibly zero) characters from the end of string S.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of multiple lines of input.
- The first line of each test case contains a single integer N, the length of the binary string.
- The next line contains the binary string S.
------ Output Format ------
For each test case, print in a new line, the length of the longest prefix having all 1s you can achieve, by performing the operation at most once.
------ Constraints ------
$1 ≤ T ≤ 1000$
$2 ≤ N ≤ 2 \cdot 10^{5}$
$S$ contains $0$ and $1$ only.
- The sum of $N$ over all test cases won't exceed $2 \cdot 10^{5}$.
----- Sample Input 1 ------
3
4
1010
6
011011
11
01011011101
----- Sample Output 1 ------
2
2
3
----- explanation 1 ------
Test case $1$: Chef can choose the substring $S[2, 2] = 0$. On removal of the substring, the remaining parts are $1$ and $10$. On concatenating these parts, the final string obtained is $110$.
The longest prefix in the final string having all $1$s is $11$, having a length $2$.
It can be proven that we cannot obtain a prefix of length greater than $2$ having all $1$s using at most one operation.
Test case $2$: Chef can choose the substring $S[1, 4] = 0110$. On removal of the substring, the remaining part is $11$.
The longest prefix in the final string having all $1$s is $11$, having a length $2$.
It can be proven that we cannot obtain a prefix of length greater than $2$ having all $1$s using at most one operation.
Test case $3$: Chef can choose the substring $S[1, 6] = 010110$. On removal of the substring, the remaining part is $11101$.
The longest prefix in the final string having all $1$s is $111$, having a length $3$.
It can be proven that we cannot obtain a prefix of length greater than $3$ having all $1$s using at most one operation.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for i in range(int(input())):
n = int(input())
p = input()
s = 0
l = []
for i in range(n):
if p[i] == '0':
if s > 0:
l.append(s)
s = 0
else:
s += 1
if s > 0:
l.append(s)
if l == []:
print(0)
elif len(l) == 1:
print(l[0])
elif p[0] == '0':
print(max(l))
else:
print(l[0] + max(l[1:]))
``` | vfc_134661 | {
"difficulty": "medium",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/PREFONES",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n4\n1010\n6\n011011\n11\n01011011101\n",
"output": "2\n2\n3\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/18/E | Solve the following coding problem using the programming language python:
According to a new ISO standard, a flag of every country should have, strangely enough, a chequered field n × m, each square should be wholly painted one of 26 colours. The following restrictions are set:
* In each row at most two different colours can be used.
* No two adjacent squares can be painted the same colour.
Pay attention, please, that in one column more than two different colours can be used.
Berland's government took a decision to introduce changes into their country's flag in accordance with the new standard, at the same time they want these changes to be minimal. By the given description of Berland's flag you should find out the minimum amount of squares that need to be painted different colour to make the flag meet the new ISO standard. You are as well to build one of the possible variants of the new Berland's flag.
Input
The first input line contains 2 integers n and m (1 ≤ n, m ≤ 500) — amount of rows and columns in Berland's flag respectively. Then there follows the flag's description: each of the following n lines contains m characters. Each character is a letter from a to z, and it stands for the colour of the corresponding square.
Output
In the first line output the minimum amount of squares that need to be repainted to make the flag meet the new ISO standard. The following n lines should contain one of the possible variants of the new flag. Don't forget that the variant of the flag, proposed by you, should be derived from the old flag with the minimum amount of repainted squares. If the answer isn't unique, output any.
Examples
Input
3 4
aaaa
bbbb
cccc
Output
6
abab
baba
acac
Input
3 3
aba
aba
zzz
Output
4
aba
bab
zbz
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import sys
from array import array
def input():
return sys.stdin.buffer.readline().decode('utf-8')
(n, m) = map(int, input().split())
flag = [tuple(map(lambda c: ord(c) - 97, input().rstrip())) for _ in range(n)]
dp = [[[m] * 26 for _ in range(26)] for _ in range(n)]
for i in range(n):
for j in range(m):
if j & 1 == 0:
for k in range(26):
dp[i][flag[i][j]][k] -= 1
else:
for k in range(26):
dp[i][k][flag[i][j]] -= 1
if i > 0:
for j in range(26):
left = [10 ** 5 * 3] * 26
right = [10 ** 5 * 3] * 27
for x in range(26):
if j == x:
continue
for y in range(26):
if x == y:
continue
if y > 0 and left[y - 1] > dp[i - 1][x][y]:
left[y - 1] = dp[i - 1][x][y]
if right[y + 1] > dp[i - 1][x][y]:
right[y + 1] = dp[i - 1][x][y]
for k in range(24, -1, -1):
if left[k] > left[k + 1]:
left[k] = left[k + 1]
for k in range(26):
dp[i][j][k] += left[k] if left[k] < right[k] else right[k]
if right[k] < right[k + 1]:
right[k + 1] = right[k]
ans = 10 ** 9
(p_i, p_j) = (-1, -1)
for i in range(26):
for j in range(26):
if i == j:
continue
if ans > dp[-1][i][j]:
ans = dp[-1][i][j]
(p_i, p_j) = (i, j)
new_flag = []
for i in range(n - 1, -1, -1):
row = (chr(97 + p_i) + chr(97 + p_j)) * (m >> 1) + (chr(97 + p_i) if m & 1 else '')
new_flag.append(row)
if i:
min_v = 10 ** 9
(n_i, n_j) = (-1, -1)
for j in range(26):
if p_i == j:
continue
for k in range(26):
if p_j != k and j != k and (min_v > dp[i - 1][j][k]):
min_v = dp[i - 1][j][k]
(n_i, n_j) = (j, k)
(p_i, p_j) = (n_i, n_j)
print(ans)
for row in reversed(new_flag):
print(row)
``` | vfc_134665 | {
"difficulty": "hard",
"memory_limit": "128.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/18/E",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "1 1\nq\n",
"output": "0\nq\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "1 500\nwljexjwjwsxhascgpyezfotqyieywzyzpnjzxqebroginhlpsvtbgmbckgqixkdveraxlgleieeagubywaacncxoouibbhwyxhjiczlymscsqlttecvundzjqtuizqwsjalmizgeekugacatlwhiiyaamjuiafmyzxadbxsgwhfawotawfxkpyzyoceiqfppituvaqmgfjinulkbkgpnibaefqaqiptctoqhcnnfihmzijnwpxdikwefpxicbspdkppznpiqrpkyvcxapoaofjbliepgfwmppgwypgflzdtzvghcuwfzcrftftmmztzydybockfbbsjrcqnepvcqlscwkckrrqwqrqkapvobcjvhnbebrpodaxdnxakjwskwtmuwnhzadptayptbgwbfpucgsrumpwwvqvbvgrggfbzotoolmgsxmcppexfudzamtixuqljrhxnzaaekaekngpbkriidbusabmimnnmiznvfrbcwnmnv\n",
"output": "468\npgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpgpg\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2 2\nab\nab\n",
"output": "2\nba\nab\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10 12\nvmqbubmuurmr\nuuqvrbbquvmq\nuuumuuqqmvuq\nqqrrrmvuqvmr\nmqbbbmrmvvrr\numrvrumbrmmb\nrqvurbrmubvm\nubbvbbbbqrqu\nmqruvqrburqq\nmmbumubrbrrb\n",
"output": "75\nmbmbmbmbmbmb\nbvbvbvbvbvbv\nuquququququq\nqrqrqrqrqrqr\nbmbmbmbmbmbm\nmbmbmbmbmbmb\nrmrmrmrmrmrm\nqbqbqbqbqbqb\nrqrqrqrqrqrq\nbrbrbrbrbrbr\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6 13\nababababababa\nababababababa\nggggggggggggg\nggggggggggggg\nggggggggggggg\nggggggggggggg\n",
"output": "39\nbabababababab\nababababababa\ngagagagagagag\nagagagagagaga\ngagagagagagag\nagagagagagaga\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/75/A | Solve the following coding problem using the programming language python:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation a + b = c, where a and b are positive integers, and c is the sum of a and b. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101 + 102 = 203, if we removed all zeros it will be 11 + 12 = 23 which is still a correct equation.
But if the equation is 105 + 106 = 211, if we removed all zeros it will be 15 + 16 = 211 which is not a correct equation.
Input
The input will consist of two lines, the first line will contain the integer a, and the second line will contain the integer b which are in the equation as described above (1 ≤ a, b ≤ 109). There won't be any leading zeros in both. The value of c should be calculated as c = a + b.
Output
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Examples
Input
101
102
Output
YES
Input
105
106
Output
NO
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
a = input()
b = input()
c = str(int(a) + int(b))
a = a.replace('0', '')
b = b.replace('0', '')
c = c.replace('0', '')
if int(a) + int(b) == int(c):
print('YES')
else:
print('NO')
``` | vfc_134670 | {
"difficulty": "easy",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/75/A",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5\n4\n",
"output": "YES\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "341781108\n784147010\n",
"output": "NO\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "501871728\n725074574\n",
"output": "NO\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "868\n463\n",
"output": "YES\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "10502405\n40403403\n",
"output": "YES\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/maximum-weight-difference5036/1 | Solve the following coding problem using the programming language python:
Given an array arr of size n. The task is to choose k numbers from the array such that the absolute difference between the sum of chosen numbers and the sum of remaining numbers is maximum.
Example 1:
Input:
n = 5, k = 2
arr[] = {8, 4, 5, 2, 10}
Output: 17
Explanation: If we select 2 and 4,
then abs((2+4) - (8+5+10)) = 17.
Example 2:
Input:
n = 8, k = 3
arr[] = {1, 1, 1, 1, 1, 1, 1, 1}
Output: 2
Explanation:
We can select any 3 1's.
Your Task:
You don't need to read input or print anything. Your task is to complete the function solve() which takes the array of integers arr, n and k as parameters and returns an integer denoting the answer.
Expected Time Complexity: O(n*Logn)
Expected Auxiliary Space: O(1)
Constraints:
1 <= k <= n <=10^{5}
1 <= arr[i] <= 10^{6}
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def solve(self, arr, n, k):
# code here
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def solve(self, arr, n, k):
chosen_first = 0
chosen_last = 0
total = 0
arr.sort()
for num in arr:
total += num
for num in arr[0:k]:
chosen_first += num
for num in arr[n - k:]:
chosen_last += num
x = abs(total - chosen_first - chosen_first)
y = abs(total - chosen_last - chosen_last)
if x >= y:
return x
else:
return y
``` | vfc_134675 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/maximum-weight-difference5036/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "n = 5, k = 2\r\narr[] = {8, 4, 5, 2, 10}",
"output": "17",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "n = 8, k = 3\r\narr[] = {1, 1, 1, 1, 1, 1, 1, 1}",
"output": "2",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/CHEFPTNT | Solve the following coding problem using the programming language python:
Read problems statements in Mandarin chinese, Russian and Vietnamese as well.
Chef has decided to start a new firm called PatentChef. However, he's stuck with some big legal issues. Their firm has received offers from a lot of companies, so Chef told his friend Junior Chef to look over some patent cases and solve them as quickly as he can.
Junior Chef is very smart and has an eye for every little detail. He quickly found a case and went ahead to solve it. The patent case is as follows:
There are N patents to be filed for a company. Chef’s firm has the first M months of the year 2018 to finish this task. (The months in a year are numbered 1 through 12.) Chef's firm has K workers (including Junior Chef) available to work on this case. Each worker can prepare exactly one patent per month.
Junior Chef assigns work to workers starting from the first month of the year. He can have any workers work on this case on any month provided that they're chosen according to the following conditions:
Each worker can only work on this patent case for at most one month.
Each worker has to work either on an even month or an odd month of the year. You are given a string S with length K and the following meaning: for each valid i, if the i-th character of S is E, worker i has to work on an even month; if it's O, this worker has to work on an odd month.
At most X workers can work on this patent case each month.
Determine whether Chef’s firm and Junior Chef can finish this patent case in time.
------ Input ------
The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains four space-separated integers N, M, X and K.
The second line contains a single string S.
------ Output ------
For each test case, print a single line containing the string "yes" if it's possible to finish the patent case or "no" otherwise (without quotes).
------ Constraints ------
$1 ≤ T ≤ 1000$
$0 ≤ X ≤ 10^{6}$
$1 ≤ N, K ≤ 10^{6}$
$0 ≤ M ≤ 12$
$1 ≤ sum of K over all test cases ≤ 10^{7}$
$|S| = K$
$each character of S is either E or O$
------ Subtasks ------
Subtask #1 (20 points): 1 ≤ N ≤ M ≤ 12
Subtask #2 (80 points): original constraints
----- Sample Input 1 ------
2
4 4 2 4
EEOO
4 3 1 4
EEOO
----- Sample Output 1 ------
yes
no
----- explanation 1 ------
Example case 1: The firm can prepare 2 patents per month, so in the first month (odd), workers 3 and 4 can work on 2 patents, and in the second month (even), workers 1 and 2 can work on the remaining 2 patents.
Example case 2: There is no way for multiple workers to work on the same (even or odd) month. Hence, it's impossible to finish 4 patents in 3 months.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
for _ in range(int(input())):
(n, m, x, k) = map(int, input().split())
str1_list = list(input())
even = str1_list.count('E')
odd = k - even
if m % 2 == 0:
evenOutput = min(even, x * (m // 2))
oddOutput = min(odd, x * (m // 2))
else:
evenOutput = min(even, x * ((m - 1) // 2))
oddOutput = min(odd, x * ((m - 1) // 2) + x)
output = evenOutput + oddOutput
if output >= n:
print('yes')
else:
print('no')
``` | vfc_134676 | {
"difficulty": "medium",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/CHEFPTNT",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "2\n4 4 2 4\nEEOO\n4 3 1 4\nEEOO",
"output": "yes\nno",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Calvin is driving his favorite vehicle on the 101 freeway. He notices that the check engine light of his vehicle is on, and he wants to service it immediately to avoid any risks. Luckily, a service lane runs parallel to the highway. The length of the highway and the service lane is N units. The service lane consists of N segments of unit length, where each segment can have different widths.
Calvin can enter into and exit from any segment. Let's call the entry segment as index i and the exit segment as index j. Assume that the exit segment lies after the entry segment(j>i) and i ? 0. Calvin has to pass through all segments from index i to indexj (both inclusive).
Calvin has three types of vehicles - bike, car and truck, represented by 1, 2 and 3respectively. These numbers also denote the width of the vehicle. We are given an arraywidth[] of length N, where width[k] represents the width of kth segment of our service lane. It is guaranteed that while servicing he can pass through at most 1000 segments, including entry and exit segments.
If width[k] is 1, only the bike can pass through kth segment.
If width[k] is 2, the bike and car can pass through kth segment.
If width[k] is 3, any of the bike, car or truck can pass through kth segment.
Given the entry and exit point of Calvin's vehicle in the service lane, output the type of largest vehicle which can pass through the service lane (including the entry & exit segment)
Input Format
The first line of input contains two integers - N & T, where N is the length of the freeway, and T is the number of test cases. The next line has N space separated integers which represents the width array.
T test cases follow. Each test case contains two integers - i & j, where i is the index of segment through which Calvin enters the service lane and j is the index of the lane segment where he exits.
Output Format
For each test case, print the number that represents the largest vehicle type that can pass through the service lane.
Note: Calvin has to pass through all segments from index i to indexj (both inclusive).
Constraints
2 ≤ N ≤ 100000
1 ≤ T ≤ 1000
0 ≤ i < j < N
2 ≤ j-i+1 ≤ min(N,1000)
1 ≤ width[k] ≤ 3, where 0 ≤ k < N
SAMPLE INPUT
8 5
2 3 1 2 3 2 3 3
0 3
4 6
6 7
3 5
0 7
SAMPLE OUTPUT
1
2
3
2
1
Explanation
Below is the representation of lane.
|HIGHWAY|Lane| -> Width
0: | |--| 2
1: | |---| 3
2: | |-| 1
3: | |--| 2
4: | |---| 3
5: | |--| 2
6: | |---| 3
7: | |---| 3
(0, 3): Because width[2] = 1, only the bike represented as 1 can pass through it.
(4, 6): Here the largest allowed vehicle which can pass through the 5th segment is car and for the 4th and 6th segment it's the truck. Hence the largest vehicle allowed in these segments is a car.
(6, 7): In this example, the vehicle enters at the 6th segment and exits at the 7thsegment. Both segments allow even truck to pass through them. Hence truck is the answer.
(3, 5): width[3] = width[5] = 2. While 4th segment allow the truck, 3rd and 5th allow upto car. So 2 will be the answer here.
(0, 7): Bike is the only vehicle which can pass through the 2nd segment, which limits the strength of whole lane to 1.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n,q=input().split()
n=int(n)
q=int(q)
x=input()
a=list()
a=x.split()
b=list()
for i in range(len(a)):
b.append(int(a[i]))
for i in range(q):
x,y=input().split()
z=min(b[int(x):int(y)+1])
print(z)
``` | vfc_134681 | {
"difficulty": "unknown_difficulty",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "100000 1000\n2 3 3 2 2 3 3 2 2 3 3 2 3 2 3 2 2 3 3 2 2 2 2 2 2 3 2 3 3 2 2 3 3 3 3 3 3 2 2 3 3 3 3 2 3 3 3 2 2 2 2 2 2 3 2 2 2 2 3 3 2 3 2 2 3 3 3 2 2 3 3 3 2 3 2 2 2 3 2 2 3 2 2 3 3 2 3 3 3 2 2 3 3 2 3 2 2 3 2 2 2 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 3 3 2 2 3 3 2 3 3 3 3 3 2 2 3 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 2 2 2 3 3 3 2 3 2 2 2 3 2 3 3 3 2 2 3 3 2 2 3 3 2 2 2 2 3 3 2 3 3 2 2 2 3 2 3 3 3 2 2 3 3 3 3 3 3 2 2 3 3 2 3 3 2 3 3 3 2 3 2 3 3 2 3 3 3 3 2 2 3 3 3 2 2 2 2 2 3 2 3 2 2 2 3 3 3 3 2 3 2 3 2 2 3 2 3 2 3 3 2 2 2 2 2 3 2 2 3 3 3 2 3 3 3 3 2 2 2 3 2 2 2 2 2 3 2 3 2 3 3 2 3 3 2 2 2 3 2 3 2 3 2 2 3 3 3 3 3 3 2 3 3 2 2 2 2 2 3 2 2 2 2 3 2 3 3 2 2 2 2 2 3 2 2 2 3 3 2 2 2 2 2 2 3 2 2 3 2 3 3 2 2 3 2 2 2 3 2 2 3 2 3 3 2 3 3 3 3 3 2 3 2 2 3 3 2 3 2 2 3 3 3 3 2 3 3 3 2 3 3 2 2 2 3 2 2 2 2 3 2 2 2 2 2 2 3 2 3 3 2 2 2 3 3 2 2 2 3 3 2 3 3 2 3 2 2 3 2 2 2 2 2 3 2 2 3 3 2 2 2 3 3 2 2 2 3 3 3 2 2 3 3 3 3 3 3 3 2 3 2 3 2 2 2 2 3 3 2 3 3 2 2 2 3 3 3 2 2 2 2 2 3 2 3 2 3 2 3 3 3 3 2 3 2 2 2 3 3 2 2 3 2 3 3 3 2 2 3 2 2 2 2 3 2 3 3 3 3 3 2 2 2 3 2 2 3 2 3 3 2 2 2 2 3 3 2 3 2 3 3 2 3 3 2 3 2 3 2 3 2 3 3 3 2 3 3 3 3 3 2 3 3 2 2 2 2 2 3 2 3 2 2 3 3 2 2 3 2 3 2 2 2 3 3 2 3 3 3 2 2 2 2 3 2 2 3 2 2 2 2 3 2 3 2 3 3 3 2 3 2 2 2 2 3 3 2 2 2 3 3 2 3 3 3 2 3 2 2 3 2 3 2 2 2 3 3 3 2 3 3 2 3 3 2 3 2 2 3 3 3 2 3 3 3 3 3 2 3 3 3 2 2 2 2 2 3 2 2 3 3 3 3 3 2 3 2 2 3 3 3 2 2 3 3 3 2 2 2 3 2 3 3 2 3 2 3 2 2 3 3 3 2 2 2 2 3 2 2 2 2 2 3 2 3 2 3 3 3 3 2 3 3 2 3 2 2 2 3 2 3 2 3 3 3 2 3 2 2 2 3 2 2 2 2 3 2 2 2 3 3 2 2 2 2 2 3 2 2 2 2 2 2 2 3 3 2 3 2 2 3 3 3 3 3 3 2 3 3 2 3 3 2 3 3 3 3 2 3 2 2 2 2 3 2 3 2 2 2 2 2 3 3 3 2 2 3 2 2 2 2 3 3 3 2 3 2 2 3 3 2 2 3 2 3 3 3 3 3 2 2 2 3 3 3 2 2 2 2 2 3 3 3 2 2 3 3 2 3 3 3 3 3 3 3 2 2 3 3 2 3 3 2 2 3 3 2 3 2 3 3 3 2 2 3 2 3 3 2 3 2 3 2 3 2 2 3 3 3 2 3 2 2 3 2 3 2 3 2 2 2 3 3 2 3 3 2 2 2 2 3 2 3 2 3 2 2 2 3 3 2 2 3 2 3 3 3 3 2 3 2 2 2 3 2 3 2 2 2 2 2 3 2 2 3 3 2 3 3 3 2 2 3 3 2 2 3 2 3 3 3 3 2 2 3 2 3 3 3 3 2 3 3 2 3 2 2 3 2 3 2 2 3 3 2 2 3 3 2 3 2 3 2 2 3 3 3 3 3 2 2 3 3 3 3 3 2 3 2 2 3 3 2 2 2 2 2 2 3 2 3 2 2 3 2 3 3 3 2 2 3 3 3 3 2 2 2 2 2 2 2 3 3 3 3 2 3 2 2 3 2 3 3 2 2 3 2 3 3 2 3 2 3 2 3 2 3 3 2 3 2 3 2 3 2 3 3 3 3 3 2 2 2 3 2 3 3 2 2 2 3 2 2 2 2 2 2 3 3 3 2 3 2 2 3 2 2 2 3 3 2 2 3 2 3 2 3 2 2 2 2 3 2 3 2 2 3 2 2 2 3 2 2 3 3 3 3 3 3 3 2 3 3 2 2 2 2 3 3 2 3 3 2 3 2 2 2 3 2 2 2 2 2 2 3 3 3 3 2 2 2 3 2 3 3 2 3 3 3 2 3 3 2 3 2 2 3 2 2 2 2 2 2 3 2 3 2 3 2 3 3 2 2 3 3 3 3 3 2 3 3 3 2 3 3 2 2 2 3 2 2 3 2 2 2 2 2 3 3 2 2 2 3 2 2 2 3 3 3 3 2 3 2 2 2 3 3 2 2 2 2 2 3 2 3 2 2 3 3 3 3 3 2 2 3 2 3 2 3 2 3 2 3 3 2 2 3 3 2 3 3 3 3 3 3 2 3 2 3 2 3 3 3 3 3 2 3 2 2 3 3 3 3 2 2 3 2 3 3 3 2 2 2 2 3 3 2 2 3 2 2 3 3 3 2 2 3 2 3 3 3 2 2 2 2 3 3 3 2 2 2 3 3 2 3 2 3 3 3 3 3 3 2 2 3 2 3 2 2 2 2 3 2 2 3 2 3 3 3 2 3 3 3 2 3 2 3 3 3 2 2 3 3 2 3 2 2 2 3 3 2 3 2 2 3 2 3 3 3 2 3 2 2 2 3 3 2 2 2 3 2 2 2 3 2 2 2 3 2 3 2 3 2 2 3 3 2 2 2 3 3 3 2 3 3 3 2 3 3 3 3 3 3 3 2 2 3 2 3 2 3 3 3 3 3 2 3 2 3 3 3 2 3 3 3 2 2 3 2 3 2 3 2 2 2 3 2 2 3 3 2 3 2 3 2 3 3 3 3 2 3 3 2 2 2 3 2 3 3 2 2 3 3 3 3 2 2 3 2 3 2 2 2 2 3 3 2 2 2 3 3 3 2 3 3 3 3 2 2 2 2 2 3 2 3 3 2 3 2 2 3 3 2 3 3 3 2 3 3 3 3 2 2 3 2 2 2 3 2 2 3 2 3 2 2 2 3 2 2 2 2 3 3 2 2 2 3 3 2 3 2 3 3 2 2 3 2 3 2 2 3 3 3 2 2 3 3 3 2 3 3 2 2 2 3 2 3 2 3 3 3 3 2 3 2 2 2 2 3 3 3 3 2 2 3 2 3 2 2 3 3 3 2 3 2 3 2 3 3 3 2 3 3 2 2 3 2 2 3 2 3 2 3 2 2 2 2 2 3 2 3 2 2 3 2 2 2 2 3 2 3 3 3 2 3 3 3 3 3 3 3 3 3 2 3 2 3 3 2 2 2 3 2 2 3 2 2 3 2 3 3 2 2 2 2 3 3 2 2 3 3 2 2 2 2 3 2 3 2 2 3 2 2 2 2 3 2 2 2 3 3 2 3 3 2 3 2 2 3 3 3 2 3 3 3 3 2 3 3 2 2 2 3 2 2 3 3 3 2 3 2 3 3 3 3 2 2 3 2 2 2 3 2 3 2 3 3 2 3 2 2 3 2 3 3 3 3 2 2 3 3 2 2 3 3 3 3 3 3 3 3 3 2 2 3 2 3 2 2 2 2 2 3 2 2 2 3 3 2 3 2 2 3 2 3 2 2 2 2 3 3 3 2 3 3 3 3 3 3 3 3 3 3 3 2 3 3 3 2 2 3 2 2 2 3 3 3 3 3 3 2 2 2 2 3 2 2 2 3 3 3 2 3 2 3 3 2 3 2 2 3 3 3 3 2 2 2 3 3 3 2 3 3 2 3 2 2 3 2 3 3 3 2 2 3 3 3 3 2 3 2 3 3 3 2 3 3 2 2 2 3 2 3 2 2 2 2 3 2 2 2 3 3 2 3 3 2 2 2 2 3 2 2 2 3 2 3 2 3 3 2 2 3 3 3 2 2 3 3 2 2 3 3 3 2 2 2 2 2 3 2 3 3 2 2 3 3 3 3 2 3 2 2 2 3 3 2 3 3 3 3 3 3 2 2 3 2 3 3 2 2 3 2 3 2 2 2 3 3 2 3 2 2 3 3 3 3 3 3 2 3 2 3 2 3 3 3 3 2 2 2 2 3 2 2 3 2 2 2 3 2 3 2 2 3 3 2 2 2 3 2 3 3 3 3 2 2 2 2 3 2 2 3 2 2 3 3 2 2 2 3 2 3 3 3 2 2 3 2 3 2 2 2 3 3 2 2 2 2 2 3 3 2 2 3 2 2 2 2 2 2 3 2 2 3 3 2 3 2 3 2 2 3 3 2 3 3 2 3 3 2 2 2 2 2 3 2 2 2 2 2 2 3 3 2 2 2 3 2 3 2 2 3 3 3 3 2 2 3 3 2 3 3 2 3 2 2 3 2 2 3 3 2 3 2 3 3 2 2 3 3 2 2 3 3 3 2 2 2 2 3 2 3 3 2 3 3 2 2 3 2 2 2 3 3 2 2 2 3 2 2 2 2 2 3 3 3 2 3 3 2 3 3 3 2 2 2 3 2 3 2 3 3 3 2 2 3 2 2 2 2 2 3 2 2 2 3 2 2 3 3 2 2 3 2 2 3 2 3 3 3 3 2 2 2 2 2 2 2 3 2 2 3 3 2 2 2 2 2 2 3 3 3 3 2 3 3 3 3 2 3 3 3 3 3 2 2 2 2 2 3 2 3 2 2 3 2 2 3 2 2 2 2 3 3 2 2 2 2 2 2 3 3 2 2 2 2 2 2 2 3 3 2 2 2 2 3 2 2 3 3 3 3 3 2 3 3 3 3 3 3 2 2 2 2 3 2 2 3 3 2 2 2 2 2 2 3 2 3 3 3 2 2 2 3 3 3 2 2 3 2 3 3 2 3 3 3 3 3 3 2 3 3 3 3 2 3 2 2 2 2 3 2 2 3 3 3 3 2 3 2 2 2 3 2 3 2 2 3 3 3 3 2 2 2 3 2 2 2 2 2 2 3 3 2 3 2 2 2 2 3 2 2 2 3 3 3 3 3 2 2 2 2 2 2 2 3 3 2 3 3 3 3 3 2 2 2 2 2 2 3 3 2 3 3 3 2 3 2 3 3 2 3 3 2 2 2 3 3 2 3 2 3 2 3 3 2 3 2 2 3 3 2 3 2 3 2 3 2 2 2 2 2 2 3 2 2 3 2 3 2 3 3 3 3 3 3 2 2 3 2 2 2 2 3 2 2 2 3 2 2 2 2 3 2 2 3 2 3 3 3 3 2 2 3 2 3 2 2 2 3 2 2 3 3 3 3 3 3 3 3 2 3 2 3 3 2 2 3 3 2 2 3 2 2 2 2 2 2 2 2 3 3 2 2 2 3 3 3 3 2 2 3 3 2 2 2 2 2 3 2 2 3 3 3 2 3 3 2 3 2 2 2 3 2 2 3 3 3 2 2 2 2 3 3 3 3 2 3 2 3 3 2 3 2 3 3 3 3 3 2 3 3 2 2 3 2 3 2 2 3 3 2 3 2 3 2 2 3 2 2 3 3 2 2 2 2 3 3 3 2 2 2 3 2 2 2 3 3 2 3 2 3 3 3 2 2 2 2 2 2 2 3 3 2 3 3 2 2 3 3 2 3 3 3 3 3 3 2 2 3 3 2 3 2 2 3 3 2 3 3 2 3 2 3 3 3 3 2 3 2 3 3 3 3 2 2 2 3 3 3 2 2 3 3 3 3 2 2 3 3 3 3 2 3 3 2 2 2 2 3 2 3 2 3 2 2 3 3 3 2 2 3 3 3 3 2 3 3 2 2 3 3 2 3 2 3 3 2 3 3 3 3 3 3 2 3 3 3 2 2 2 2 3 3 2 2 3 3 2 3 3 3 2 3 2 2 2 2 2 3 3 3 2 2 2 2 2 3 2 2 3 2 3 2 3 3 3 2 2 3 3 3 2 3 3 2 3 3 2 3 3 2 2 3 2 2 2 2 3 2 3 2 2 2 2 3 3 3 3 3 2 2 2 2 3 3 3 2 3 3 3 2 3 3 3 2 3 3 2 2 3 3 2 3 3 2 2 2 2 3 3 2 3 2 3 2 3 2 2 2 3 3 2 3 2 2 3 3 3 3 3 3 3 3 2 2 2 3 3 2 2 2 2 2 2 3 2 2 3 3 2 3 2 3 2 3 3 3 2 2 2 2 3 3 3 2 2 3 3 3 3 3 3 2 3 2 3 2 2 3 3 2 2 3 3 2 2 2 3 3 3 3 3 2 3 2 2 3 2 3 2 3 3 3 3 2 3 3 2 3 2 3 2 2 3 3 2 3 2 3 2 3 2 3 3 3 2 2 2 2 3 2 3 2 2 3 3 3 2 3 3 2 3 3 2 2 2 2 3 2 3 2 3 3 3 3 3 3 3 3 3 2 2 3 3 2 2 2 3 2 3 2 2 2 3 2 2 2 2 2 2 3 2 2 2 3 3 3 3 2 3 2 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 2 2 3 2 3 2 2 3 3 3 2 2 2 2 3 3 2 2 2 2 3 2 3 2 3 2 3 3 2 3 2 2 2 2 2 3 2 2 2 3 3 2 2 2 2 3 3 2 2 3 2 3 3 2 3 3 2 2 2 2 2 2 2 2 2 2 3 3 2 3 2 2 3 2 2 3 2 3 3 2 2 2 3 2 2 3 3 2 3 3 3 3 3 3 3 2 2 2 3 2 3 3 2 2 2 2 3 2 3 3 2 2 3 3 2 3 2 3 3 2 2 2 3 3 2 3 3 2 3 2 2 3 2 3 3 2 3 3 2 3 2 2 3 3 3 3 2 2 2 3 2 2 2 3 3 2 2 3 2 2 3 2 3 3 3 2 3 3 3 3 2 3 3 3 2 3 2 2 3 2 2 3 2 2 2 3 2 3 2 2 3 2 2 2 3 2 2 3 3 2 2 3 3 2 2 2 3 2 2 2 2 2 3 2 2 3 3 2 2 2 2 3 2 3 3 3 3 2 2 2 2 3 3 3 3 3 3 2 3 2 2 3 2 3 3 3 2 2 3 3 2 2 2 2 3 2 2 2 2 2 2 2 3 3 3 2 2 3 2 3 3 2 2 3 3 3 2 2 3 2 3 2 2 3 2 3 3 2 3 3 3 3 3 2 2 3 3 2 2 3 3 3 2 3 2 3 2 3 3 3 3 2 3 3 2 2 2 3 2 3 3 3 2 2 2 2 3 3 2 3 2 3 2 2 2 3 2 2 2 3 3 3 2 3 2 2 3 2 3 3 3 2 3 3 3 3 3 2 2 3 2 2 2 2 3 2 3 3 2 3 2 3 2 2 2 2 2 3 2 2 3 3 2 2 2 3 3 3 2 3 2 2 3 2 2 2 2 2 3 2 3 2 2 2 2 2 2 3 2 3 3 3 2 3 3 3 3 2 2 3 3 3 3 2 3 3 3 2 3 2 2 3 2 2 3 3 3 3 2 3 2 3 2 3 2 3 2 3 3 2 2 2 3 3 2 3 3 3 3 2 2 3 3 2 2 2 3 3 2 3 2 2 2 2 3 3 3 3 2 2 2 3 2 3 2 2 2 3 2 3 2 2 3 3 2 3 2 2 2 2 3 2 2 2 2 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 3 2 3 2 2 2 3 2 2 3 2 3 3 2 3 3 2 2 2 2 2 3 3 2 2 2 3 3 2 2 3 3 2 2 3 2 2 2 2 2 3 2 2 2 3 3 3 3 2 3 3 2 2 3 2 2 3 2 3 3 2 2 2 2 3 3 3 3 3 3 2 2 2 2 2 3 3 3 2 3 2 2 3 2 3 2 2 2 2 2 2 2 2 2 2 3 2 3 3 3 3 3 2 3 3 2 2 2 2 2 2 2 2 3 3 3 3 3 2 3 3 2 3 2 2 2 3 2 3 2 2 2 3 2 3 2 2 3 3 2 2 3 3 2 2 2 2 2 3 2 3 3 2 3 3 2 3 2 3 2 3 3 3 2 3 2 3 3 2 2 2 2 3 3 2 3 3 2 3 2 2 3 3 2 2 2 3 3 3 2 3 2 3 2 2 2 3 3 3 3 3 3 3 2 2 3 2 3 2 3 2 2 2 3 3 2 2 2 3 3 2 3 3 3 3 3 3 2 3 2 3 2 2 2 3 2 2 3 2 2 2 2 2 3 3 3 3 3 3 3 2 3 2 3 2 3 3 3 2 2 2 3 2 2 2 3 2 2 2 2 2 3 2 2 2 2 2 3 3 3 2 2 3 2 3 3 2 2 3 2 2 3 3 3 3 3 2 3 3 3 2 3 2 2 2 2 2 2 3 2 3 2 2 2 2 3 2 2 2 3 2 2 2 2 3 3 3 2 2 3 3 2 2 3 3 2 3 3 2 2 3 2 2 3 2 3 3 2 3 3 3 2 3 3 2 3 2 3 3 3 2 2 3 3 3 2 3 2 2 2 2 3 2 3 3 2 2 2 3 3 3 2 3 2 2 3 3 2 3 2 3 3 2 3 2 3 3 2 3 3 2 2 3 2 3 2 2 3 2 3 2 3 2 2 3 2 3 3 2 2 3 2 2 2 3 2 3 2 2 3 2 2 3 3 2 2 3 3 3 3 2 3 2 2 3 3 2 3 2 3 3 2 3 3 2 2 2 3 2 2 2 2 3 3 3 3 3 2 2 2 3 3 2 3 3 3 3 2 2 3 3 2 3 2 3 3 2 3 3 2 2 3 3 3 3 2 2 2 3 3 3 2 2 3 2 3 2 3 3 3 2 2 3 2 2 2 3 2 2 2 3 2 2 2 3 3 2 3 3 3 2 2 2 2 3 2 2 2 3 3 3 3 2 2 3 2 2 3 3 2 3 2 2 3 2 3 2 2 3 2 2 3 2 2 2 2 2 2 2 3 3 3 2 3 3 2 2 3 3 3 2 2 3 2 2 3 2 2 3 3 2 3 2 3 3 2 3 3 2 3 2 2 2 3 3 3 3 3 2 2 2 2 2 3 3 2 2 3 3 2 2 3 3 2 2 3 3 3 2 3 2 3 3 2 2 3 3 3 2 2 3 3 2 3 2 3 2 3 2 3 3 2 2 2 3 3 3 2 2 2 3 3 3 3 3 3 2 3 2 2 3 3 3 3 2 2 3 2 3 3 3 2 2 2 2 3 3 2 3 3 2 2 2 3 3 2 2 3 3 2 2 2 3 3 3 3 3 2 2 2 2 3 2 2 3 3 3 2 3 2 2 3 2 2 2 2 2 2 3 3 2 3 3 3 3 3 2 2 3 2 3 3 2 3 3 3 2 2 2 3 2 2 2 3 2 2 3 3 2 2 2 2 2 2 3 3 3 2 3 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 3 2 3 2 3 3 3 2 2 3 2 3 2 3 3 3 3 2 2 3 2 2 3 2 2 3 2 2 3 2 3 3 3 2 3 2 3 2 2 3 3 2 3 3 3 2 3 3 2 3 2 3 3 3 3 3 2 2 2 2 2 3 3 3 3 2 2 2 3 2 2 2 3 3 2 2 3 3 3 2 2 2 3 3 3 2 2 2 2 2 2 3 3 3 2 2 3 2 3 2 3 3 3 2 2 3 2 3 2 2 3 2 2 2 3 3 3 3 3 3 2 3 2 3 2 3 2 2 3 3 2 2 2 3 2 2 2 3 3 2 3 3 2 3 3 3 2 2 3 2 2 3 3 2 2 2 3 2 2 3 3 2 3 3 2 2 3 2 3 3 3 2 2 3 3 3 3 3 2 2 3 2 3 3 2 3 3 2 2 3 3 3 3 2 3 3 2 2 2 3 3 3 3 3 2 2 3 3 2 3 3 3 3 2 2 3 2 3 3 2 2 2 3 2 3 2 3 3 3 3 3 2 2 2 2 3 3 3 2 3 2 2 2 3 2 3 2 2 2 2 2 3 2 2 3 3 2 2 3 3 2 2 2 2 2 2 3 3 3 2 2 3 2 3 2 2 2 2 3 2 2 3 3 3 3 2 2 2 3 3 3 3 3 3 3 2 3 3 3 2 3 2 3 3 3 3 3 3 2 2 3 2 2 3 3 3 3 2 3 2 3 3 3 3 2 3 3 2 2 2 2 3 2 2 2 3 3 3 2 3 2 2 2 2 3 3 3 2 3 3 3 3 2 2 2 2 3 3 2 3 3 3 2 2 3 2 3 2 2 2 2 2 2 2 2 3 3 3 3 3 2 2 2 2 3 2 3 2 3 3 2 2 2 2 2 3 3 2 2 3 2 2 3 2 2 3 3 3 2 2 2 3 3 2 3 2 2 2 2 3 2 2 2 2 3 2 2 2 2 2 3 2 2 2 2 2 3 3 3 3 2 2 2 3 2 2 3 3 2 3 2 2 2 2 3 3 3 3 3 3 3 2 2 3 2 2 3 3 2 2 2 2 2 3 3 3 3 2 2 3 3 2 2 3 3 3 2 2 2 3 3 3 3 3 2 3 2 3 2 2 3 3 2 2 2 3 3 3 3 3 2 2 3 2 2 2 3 2 2 3 2 2 2 3 3 2 3 3 3 3 3 3 2 3 3 2 2 2 3 3 3 2 2 2 2 2 3 2 2 3 3 2 3 2 2 3 2 3 2 2 2 2 3 3 3 2 3 2 2 3 3 3 3 3 3 3 3 2 3 2 2 3 2 3 3 2 2 3 3 3 3 2 3 2 3 2 2 2 2 2 3 2 3 2 3 3 2 3 3 3 3 3 2 3 3 3 2 3 3 3 2 2 3 3 3 2 2 3 3 2 2 2 2 3 3 2 2 3 3 2 2 2 3 3 3 2 2 3 2 3 3 2 2 2 2 3 3 2 2 2 2 2 2 3 2 3 3 2 2 2 2 3 2 2 2 3 2 2 3 2 2 2 3 2 2 3 3 3 3 3 3 3 2 2 2 2 3 3 2 2 3 3 3 3 3 3 3 3 3 2 2 3 2 3 3 2 2 2 2 3 2 3 2 2 3 3 2 3 2 2 3 2 3 2 3 2 3 2 2 2 2 2 2 2 3 3 3 3 2 3 2 2 2 2 2 2 3 2 3 2 2 2 2 2 2 3 2 3 2 2 3 2 2 3 3 3 3 2 2 3 3 2 3 3 3 3 3 2 3 2 2 3 2 2 3 2 2 2 3 2 2 3 2 3 2 3 2 3 3 3 2 2 3 3 2 2 2 3 3 3 2 3 3 2 2 2 3 2 2 2 2 3 3 2 2 2 2 2 3 3 3 2 2 3 3 2 3 2 3 2 3 3 2 2 2 2 3 3 2 3 3 2 2 3 2 2 3 2 3 2 2 2 2 2 3 2 2 3 2 3 3 3 3 3 2 3 3 3 2 3 2 3 3 3 2 2 3 3 2 2 2 2 3 2 2 2 2 2 3 2 2 3 2 3 2 2 2 2 3 2 3 3 3 3 2 2 3 2 3 3 2 3 2 3 2 2 2 2 3 3 3 3 2 3 2 3 3 2 3 2 2 2 3 3 3 2 3 2 2 3 2 2 2 2 2 2 2 2 3 3 3 2 2 2 3 2 3 2 2 2 2 2 2 3 2 2 3 3 2 3 2 2 2 3 2 2 3 3 2 2 2 3 2 3 3 3 3 2 3 2 2 3 2 3 3 2 3 2 2 3 2 2 2 2 3 2 2 3 3 2 3 2 3 2 3 3 3 2 3 3 2 2 2 3 3 3 3 2 2 3 3 2 2 3 2 3 2 2 2 3 2 2 3 3 2 2 2 3 3 2 2 3 2 3 2 3 2 2 3 2 3 2 2 3 2 2 3 2 2 3 3 2 3 3 2 3 3 2 3 2 2 3 2 2 2 2 3 3 2 2 3 2 3 2 3 3 2 2 3 3 2 2 3 3 3 3 3 3 2 2 3 2 3 3 3 2 2 2 3 2 3 2 2 2 2 2 3 3 2 2 2 2 2 3 2 2 3 3 3 3 3 2 3 2 2 2 2 2 3 3 2 2 2 3 2 2 3 3 3 3 3 3 2 3 3 2 3 2 3 2 3 2 3 2 2 3 3 3 3 2 2 3 2 2 2 2 3 3 3 2 3 2 2 3 3 3 3 3 3 2 3 2 2 2 2 2 3 3 3 2 3 2 2 3 2 2 3 3 2 2 2 3 2 2 2 2 3 3 3 2 3 2 2 3 3 2 3 2 2 3 3 3 3 3 3 3 3 2 3 3 3 3 2 3 3 2 3 2 3 2 2 2 3 2 3 2 2 3 2 2 2 3 2 3 2 3 2 2 3 3 3 2 2 2 2 3 2 3 3 2 2 3 2 3 3 2 3 2 3 3 2 3 3 2 2 3 3 2 3 3 2 3 3 2 3 3 2 3 3 3 3 3 3 2 2 2 2 3 2 3 2 3 2 3 3 2 3 3 2 2 2 2 3 3 3 2 3 3 2 2 2 3 3 3 3 3 2 3 2 2 2 2 3 2 2 3 2 3 2 3 3 2 3 3 3 2 3 2 3 3 2 2 3 3 3 2 2 3 2 2 2 2 3 3 3 3 2 3 2 2 2 3 2 2 2 2 2 2 2 2 3 3 2 2 2 3 3 3 3 3 3 3 3 2 2 2 3 3 2 3 3 2 3 2 2 3 2 3 3 2 3 3 3 3 3 2 2 2 3 3 3 2 2 3 3 3 3 2 2 3 2 3 2 3 3 2 2 3 3 2 2 2 3 3 3 2 3 2 3 2 3 2 3 2 3 2 3 3 2 3 2 2 2 2 3 2 3 2 3 2 2 3 3 3 3 2 3 2 2 2 3 2 3 2 2 2 2 3 3 2 2 2 3 2 2 2 2 3 2 2 2 2 3 3 3 2 3 3 3 2 3 2 2 2 2 2 3 2 3 2 2 3 2 3 3 3 2 2 2 2 2 2 3 3 3 2 2 3 3 3 3 3 3 3 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2 3 3 2 3 3 2 2 2 2 3 3 2 2 2 3 3 3 3 3 2 2 2 2 3 2 3 2 2 2 3 2 3 2 2 3 3 3 3 2 3 2 3 3 2 3 2 3 2 3 2 2 3 2 2 3 2 3 3 3 3 3 3 3 3 2 2 3 3 3 3 2 3 2 2 3 3 2 2 3 2 2 3 3 2 3 2 3 3 2 2 2 3 3 3 2 3 3 3 2 2 2 3 3 2 3 2 3 3 2 2 3 2 3 3 3 3 3 2 2 3 2 2 2 3 2 3 3 3 2 3 2 3 2 3 3 3 2 3 3 2 3 2 3 3 3 2 2 3 2 2 2 2 2 3 3 2 2 2 2 2 2 2 3 2 3 3 2 3 2 3 2 3 3 3 2 3 3 2 2 3 2 2 3 3 3 2 3 3 3 3 2 3 3 3 3 3 2 3 2 2 2 2 2 2 3 2 3 2 3 3 3 3 3 2 2 3 3 2 2 2 3 2 3 3 2 2 2 2 2 2 3 2 3 3 2 2 3 3 3 2 2 2 2 2 3 2 3 2 2 3 2 2 2 3 3 2 3 3 2 3 3 3 2 2 2 2 3 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 2 2 3 2 3 3 3 3 3 3 3 3 2 2 3 2 2 2 2 3 2 3 2 3 2 2 3 3 3 2 2 2 3 3 2 2 3 3 3 2 3 3 3 3 3 3 3 2 3 3 3 3 2 3 3 3 3 2 2 3 2 3 3 3 2 3 2 3 2 3 3 3 2 3 2 2 2 3 2 3 3 3 2 3 2 3 2 2 3 3 3 2 2 2 3 2 3 3 2 3 3 3 3 3 2 3 3 2 3 3 3 2 2 2 3 3 3 2 3 3 3 2 3 3 2 2 3 3 2 3 2 3 3 3 2 3 3 2 2 2 3 3 2 2 3 3 3 3 3 2 2 2 2 3 3 2 3 3 3 3 2 3 2 3 3 3 3 2 3 3 2 2 2 2 2 2 3 3 3 3 3 3 3 3 2 3 3 3 2 2 3 2 2 3 2 3 2 3 3 3 2 3 2 2 3 2 2 2 2 3 3 3 2 3 3 2 2 2 2 2 3 3 2 3 2 2 2 3 3 3 2 3 2 2 2 3 3 2 3 3 2 3 2 2 2 3 3 2 2 3 2 3 2 2 2 2 3 2 3 2 3 2 2 3 2 2 2 3 2 3 2 2 2 3 3 2 2 2 2 2 3 2 3 3 3 3 3 2 3 3 2 2 3 2 3 3 2 3 3 3 3 3 3 3 2 2 2 3 2 2 3 3 3 3 2 2 2 2 2 2 3 2 2 2 3 2 3 3 3 2 2 2 2 2 2 2 2 2 3 3 2 3 2 3 2 3 3 2 3 3 2 2 2 3 2 3 3 3 2 2 2 3 3 2 3 3 2 3 3 2 2 3 3 3 3 3 2 2 3 3 2 2 3 2 3 3 3 2 2 3 2 2 2 3 2 3 2 3 3 3 3 3 3 2 2 2 3 2 2 2 3 2 2 2 2 3 3 3 3 2 3 2 2 3 3 3 3 2 2 2 3 3 3 2 3 2 2 2 2 3 2 2 3 3 2 2 2 2 3 2 2 2 2 2 2 3 3 3 3 3 3 3 2 2 3 3 2 2 3 3 3 2 3 2 3 3 2 3 3 2 3 3 2 3 2 2 3 3 3 2 3 2 3 3 3 3 3 3 3 2 2 2 2 3 2 3 3 3 3 2 3 2 2 3 2 2 2 3 3 3 3 2 2 3 2 3 2 3 2 3 3 3 3 2 2 3 3 3 2 2 2 3 3 2 3 3 2 3 2 3 2 3 2 2 2 2 3 2 3 2 3 2 3 2 2 3 2 2 3 2 2 3 2 3 3 3 2 3 2 2 2 2 2 2 3 2 2 2 3 3 2 2 2 3 3 2 3 3 2 2 3 3 3 3 2 2 2 3 3 2 3 3 3 3 2 2 2 2 2 3 2 3 3 2 2 2 2 3 3 3 3 3 2 2 2 2 2 3 3 3 3 3 3 2 2 3 2 2 3 3 3 3 2 3 3 2 3 2 2 3 3 3 2 3 2 2 3 2 3 3 2 3 2 3 3 2 2 2 3 3 3 2 3 3 3 2 3 3 2 3 2 3 3 2 2 3 2 2 3 2 3 3 3 3 2 2 3 2 2 2 2 3 3 3 2 2 3 2 3 2 3 3 3 2 3 2 3 2 2 3 2 3 2 3 2 3 3 3 3 2 2 3 3 3 2 2 3 2 2 3 2 3 2 3 2 2 3 3 2 3 2 2 2 3 3 2 2 2 2 3 2 2 3 2 3 3 2 2 3 2 3 3 3 2 3 3 2 2 3 2 2 3 2 2 2 3 3 2 3 3 2 2 3 3 2 2 2 2 2 2 2 2 3 3 2 2 3 2 3 2 2 3 3 2 3 2 3 2 2 2 3 2 2 2 3 2 2 2 2 3 2 2 3 3 2 3 2 3 3 3 3 3 2 2 3 3 2 2 2 3 2 3 3 3 2 3 3 2 3 2 3 3 2 2 2 2 3 2 3 2 3 2 3 3 3 2 3 3 2 3 2 3 2 2 2 2 3 3 3 2 3 2 3 2 3 3 2 2 2 2 3 3 2 2 3 3 3 2 3 3 3 3 2 3 3 2 2 2 2 3 2 3 3 3 3 2 3 2 3 3 2 2 2 2 2 3 2 3 3 3 3 2 2 3 2 2 2 2 2 2 3 3 3 2 2 3 3 3 3 2 2 3 2 3 3 2 2 3 2 2 2 3 2 3 2 2 3 2 2 3 2 3 2 2 2 3 3 3 2 2 3 3 3 3 2 2 3 2 2 3 2 2 2 3 3 3 3 2 3 2 2 2 3 2 2 3 3 3 2 2 3 3 3 2 2 3 2 2 3 2 3 2 3 2 3 2 3 2 3 2 2 3 2 2 3 2 3 3 3 2 3 2 3 2 2 2 3 3 2 2 3 3 2 2 3 3 3 2 2 2 3 2 3 3 2 2 2 2 3 3 2 2 2 3 2 2 3 3 3 3 2 3 3 2 3 2 2 2 3 2 2 2 2 3 3 3 2 3 3 3 3 3 2 3 2 2 3 2 2 3 3 2 2 2 2 3 3 2 2 2 2 2 2 3 2 3 2 2 3 3 3 2 2 3 3 2 2 2 2 2 3 2 2 3 2 2 3 3 3 3 3 3 3 3 2 3 3 2 3 2 3 3 2 3 2 3 2 2 3 2 2 3 2 2 2 3 3 3 2 2 2 2 3 2 3 2 3 2 2 3 2 2 2 2 3 2 3 3 3 3 2 3 2 2 2 2 3 3 2 2 3 2 2 2 2 3 2 2 2 3 3 2 3 3 3 2 3 2 2 2 3 2 2 3 2 2 2 2 3 2 2 2 2 2 2 3 3 3 3 3 2 2 2 3 3 3 3 2 3 3 3 3 3 3 2 2 3 2 2 2 2 2 2 3 2 2 2 3 3 3 3 3 3 3 2 2 2 2 2 3 3 3 2 3 2 3 3 3 3 3 3 2 3 3 3 3 2 3 2 3 2 3 3 3 2 3 3 2 3 3 2 3 3 2 2 3 3 3 3 3 2 2 3 3 2 2 2 2 3 2 3 3 2 2 2 2 2 3 3 3 2 3 2 3 3 2 3 3 3 2 2 3 2 3 2 2 3 2 2 2 3 2 3 3 2 3 3 2 2 2 2 2 3 2 2 3 3 3 2 2 3 2 2 3 3 2 2 2 3 2 2 2 2 3 3 3 2 2 3 2 3 3 2 2 2 2 3 3 3 3 3 2 3 3 2 2 2 2 2 3 2 2 3 3 3 2 2 3 2 3 3 3 3 2 2 3 2 3 2 2 3 3 2 2 3 2 3 3 2 3 2 3 2 3 2 3 3 2 3 3 3 2 3 2 2 3 3 3 2 3 3 3 3 3 2 2 2 3 3 2 2 3 3 2 2 3 2 3 3 3 2 3 3 3 3 2 2 3 3 3 2 2 2 3 3 2 3 3 3 2 2 2 3 3 2 3 3 2 2 2 3 3 3 3 2 3 3 3 2 2 2 2 2 2 2 3 3 3 3 2 2 3 2 3 2 3 3 3 3 3 2 3 2 3 2 3 2 3 2 2 3 2 3 3 2 3 2 3 2 3 2 2 2 2 2 3 3 3 2 3 2 2 2 3 2 2 2 2 3 2 3 2 2 2 3 2 3 3 2 3 3 2 2 3 2 2 2 2 3 3 3 3 3 3 2 3 3 2 2 2 2 3 2 2 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 3 2 3 2 3 3 2 3 2 2 2 3 3 3 3 2 2 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 3 3 3 2 3 3 3 3 2 3 3 2 2 2 3 2 3 2 2 3 2 3 3 2 3 2 2 2 2 3 2 3 2 2 3 3 3 3 2 2 2 3 2 2 2 2 2 3 2 3 2 2 2 3 2 3 3 3 2 3 2 2 3 2 2 2 3 2 3 2 2 3 3 2 3 3 2 2 2 2 3 2 2 3 2 3 2 3 2 2 3 2 3 2 2 3 2 3 3 2 3 3 3 3 2 3 2 2 3 3 3 2 3 3 3 3 2 3 3 2 2 2 2 3 2 2 2 2 2 2 2 3 3 2 2 3 3 3 2 2 2 3 2 3 2 3 3 3 2 2 3 2 2 2 3 2 2 2 2 2 2 3 2 3 3 2 3 2 3 3 2 3 2 2 3 2 3 2 3 3 2 3 2 2 3 3 2 3 3 2 2 3 3 2 3 2 2 2 2 2 3 2 3 3 2 2 3 3 2 3 3 2 2 3 2 3 2 2 2 2 3 2 3 2 2 2 3 3 2 3 3 3 2 2 2 2 3 2 2 3 3 3 2 3 2 2 2 2 3 2 2 2 3 2 2 3 2 3 2 3 3 3 2 3 2 3 3 3 3 3 2 2 2 2 3 2 3 3 3 2 3 3 2 2 3 2 2 2 2 2 3 3 2 3 2 2 2 3 3 3 3 3 3 3 2 2 2 3 3 3 3 3 2 3 3 2 3 3 2 3 2 3 2 2 2 2 2 3 2 3 2 3 2 2 2 2 2 2 3 2 3 2 3 2 3 2 2 3 2 2 2 2 3 3 2 3 3 2 2 3 3 3 2 3 3 3 2 3 3 3 3 3 2 2 3 3 3 3 2 3 3 3 3 2 2 3 3 3 3 2 3 2 3 3 3 2 2 3 3 2 3 3 3 3 3 2 2 2 3 3 3 2 2 2 2 2 3 3 3 2 3 2 2 2 2 2 2 2 3 2 2 2 3 3 3 2 3 2 3 2 3 2 2 3 3 2 3 2 2 2 3 3 3 3 2 3 3 2 3 3 2 2 3 3 3 3 2 3 3 3 3 2 3 2 3 2 2 2 3 2 2 2 2 3 3 2 2 3 2 2 2 3 2 3 2 3 2 2 2 3 3 2 3 3 2 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"output": "3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3",
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12750\n11763 12762\n11775 12774\n11787 12786\n11799 12798\n11811 12810\n11823 12822\n11835 12834\n11847 12846\n11859 12858\n11871 12870\n11883 12882\n11895 12894\n11907 12906\n11919 12918\n11931 12930\n11943 12942\n11955 12954\n11967 12966\n11979 12978\n11991 12990\n12003 13002\n12015 13014\n12027 13026\n12039 13038\n12051 13050\n12063 13062\n12075 13074\n12087 13086\n12099 13098",
"output": "3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3\n3",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/sum-of-length3345/1 | Solve the following coding problem using the programming language python:
Given an array. Calculate the sum of lengths of contiguous subarrays having all distinct elements.
Example 1:
Input:
N=3
arr[] = { 1,2,3 }
Output: 10
Explanation:
{1, 2, 3} is a subarray of length 3 with
distinct elements. Total length of length
three = 3. {1, 2}, {2, 3} are 2 subarray
of length 2 with distinct elements. Total
length of lengths two = 2 + 2 = 4
{1}, {2}, {3} are 3 subarrays of length 1
with distinct element. Total lengths of
length one = 1 + 1 + 1 = 3
Sum of lengths = 3 + 4 + 3 = 10
Example 2:
Input:
N=1
arr[] = { 1 }
Output: 1
Explanation:
{1} is the only subarray with distinct
elements of length 1.
Your Task:
You don't need to read input or print anything. Your task is to complete the function sumoflength() that takes array arr and integer N as input parameters and returns the sum of lengths of contiguous subarrays having all distinct elements.
Expected Time Complexity: O(N).
Expected Auxiliary Space: O(N).
Constraints:
1 ≤ N ≤ 10^{6}
Write your solution by modifying this code:
```python
#User function Template for python3
class Solution:
def sumoflength(self,arr,n):
```
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
class Solution:
def sumoflength(self, arr, n):
store = set()
count = 0
i = 0
for item in arr:
while i < n and (not arr[i] in store):
store.add(arr[i])
i += 1
count += len(store) * (len(store) + 1) // 2
store.remove(item)
return count
``` | vfc_134687 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/sum-of-length3345/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "N=3\r\narr[] = { 1,2,3 }",
"output": "10",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "N=1\r\narr[] = { 1 }",
"output": "1",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/367/A | Solve the following coding problem using the programming language python:
Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q_1q_2... q_{k}. The algorithm consists of two steps:
Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2. Rearrange the letters of the found subsequence randomly and go to step 1.
Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string.
Sereja wants to test his algorithm. For that, he has string s = s_1s_2... s_{n}, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring s_{l}_{i}s_{l}_{i} + 1... s_{r}_{i} (1 ≤ l_{i} ≤ r_{i} ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (l_{i}, r_{i}) determine if the algorithm works correctly on this test or not.
-----Input-----
The first line contains non-empty string s, its length (n) doesn't exceed 10^5. It is guaranteed that string s only contains characters: 'x', 'y', 'z'.
The second line contains integer m (1 ≤ m ≤ 10^5) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers l_{i}, r_{i} (1 ≤ l_{i} ≤ r_{i} ≤ n).
-----Output-----
For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.
-----Examples-----
Input
zyxxxxxxyyz
5
5 5
1 3
1 11
1 4
3 6
Output
YES
YES
NO
YES
NO
-----Note-----
In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def main():
(l, xyz, res) = ([(0, 0, 0)], [0, 0, 0], [])
for c in input():
xyz[ord(c) - 120] += 1
l.append(tuple(xyz))
for _ in range(int(input())):
(a, b) = list(map(int, input().split()))
if b - a > 1:
xyz = [i - j for (i, j) in zip(l[b], l[a - 1])]
res.append(('NO', 'YES')[max(xyz) - min(xyz) < 2])
else:
res.append('YES')
print('\n'.join(res))
def __starting_point():
main()
__starting_point()
``` | vfc_134689 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/367/A",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "zyxxxxxxyyz\n5\n5 5\n1 3\n1 11\n1 4\n3 6\n",
"output": "YES\nYES\nNO\nYES\nNO\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "yxzyzxzzxyyzzxxxzyyzzyzxxzxyzyyzxyzxyxxyzxyxzyzxyzxyyxzzzyzxyyxyzxxy\n10\n17 67\n6 35\n12 45\n56 56\n14 30\n25 54\n1 1\n46 54\n3 33\n19 40\n",
"output": "NO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nYES\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "xxxxyyxyyzzyxyxzxyzyxzyyyzyzzxxxxzyyzzzzyxxxxzzyzzyzx\n5\n4 4\n3 3\n1 24\n3 28\n18 39\n",
"output": "YES\nYES\nNO\nNO\nNO\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "yzxyzxyzxzxzyzxyzyzzzyxzyz\n9\n4 6\n2 7\n3 5\n14 24\n3 13\n2 24\n2 5\n2 14\n3 15\n",
"output": "YES\nYES\nYES\nNO\nYES\nNO\nYES\nNO\nNO\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "zxyzxyzyyzxzzxyzxyzx\n15\n7 10\n17 17\n6 7\n8 14\n4 7\n11 18\n12 13\n1 1\n3 8\n1 1\n9 17\n4 4\n5 11\n3 15\n1 1\n",
"output": "NO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nYES\nNO\nYES\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://atcoder.jp/contests/abc167/tasks/abc167_c | Solve the following coding problem using the programming language python:
Takahashi, who is a novice in competitive programming, wants to learn M algorithms.
Initially, his understanding level of each of the M algorithms is 0.
Takahashi is visiting a bookstore, where he finds N books on algorithms.
The i-th book (1\leq i\leq N) is sold for C_i yen (the currency of Japan). If he buys and reads it, his understanding level of the j-th algorithm will increase by A_{i,j} for each j (1\leq j\leq M).
There is no other way to increase the understanding levels of the algorithms.
Takahashi's objective is to make his understanding levels of all the M algorithms X or higher. Determine whether this objective is achievable. If it is achievable, find the minimum amount of money needed to achieve it.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
f = lambda : map(int, input().split())
(n, m, x) = f()
(lc, ll) = ([], [])
for _ in range(n):
(c, *l) = f()
lc += [c]
ll += [l]
a = 10 ** 9
for i in range(2 ** n):
(lm, t) = ([0] * m, 0)
for j in range(n):
if i >> j & 1:
for k in range(m):
lm[k] += ll[j][k]
t += lc[j]
if all((s >= x for s in lm)):
a = min(a, t)
print([-1, a][a < 10 ** 9])
``` | vfc_134693 | {
"difficulty": "easy",
"memory_limit": "1024.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://atcoder.jp/contests/abc167/tasks/abc167_c",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3 3 10\n60 2 2 4\n70 8 7 9\n50 2 3 9\n",
"output": "120\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "3 3 10\n100 3 1 4\n100 1 5 9\n100 2 6 5\n",
"output": "-1\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "8 5 22\n100 3 7 5 3 1\n164 4 5 2 7 8\n334 7 2 7 2 9\n234 4 7 2 8 2\n541 5 4 3 3 6\n235 4 8 6 9 7\n394 3 6 1 6 2\n872 8 4 3 7 2\n",
"output": "1067\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "11 12 29799\n31069 1476 2003 8569 2842 5037 3403 5272 1385 3219 8518 470 1539\n54638 6079 2090 8362 4939 3471 108 4080 4363 402 2790 1367 2320\n13866 2404 2018 7220 3184 3519 9237 2095 1534 2621 3770 4420 7174\n42995 2907 8325 5257 6477 3107 4335 3284 1028 1849 5321 9771 1123\n94164 332 7324 1596 6623 3955 5021 5687 16 6636 199 4815 6685\n30398 5917 2635 8491 5553 9348 1500 4664 3342 886 9439 9629 4012\n48251 1822 3775 9454 5891 812 4057 2522 2628 5619 4083 1416 3474\n8783 2911 4440 7253 7960 9998 8546 2783 2293 1124 157 3096 2576\n1919 6022 628 1900 6300 8480 7942 3145 8984 8499 9372 5039 6809\n98988 6988 9030 850 248 8223 9867 7477 4125 2053 3073 4617 2814\n39829 3292 8341 5589 7882 7998 2722 8445 102 5404 4339 1512 7940\n",
"output": "464900\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "12 11 34203\n24515 7150 2848 7965 4930 3759 576 1065 8900 1787 8882 8420\n28650 8761 3488 5821 5136 5907 5258 844 824 3709 3128 7636\n75430 5768 2976 5964 2309 4630 4556 4918 5657 5807 4331 4438\n62434 9930 2739 9488 2700 9041 6546 4406 4080 7693 675 8638\n48521 4561 2887 8864 2969 1701 5140 993 4098 1973 2625 3299\n76651 1911 1692 3299 5963 494 9369 1070 4040 7164 4249 6716\n28617 3235 8958 8680 6648 5547 8276 7849 3057 2475 8333 2658\n25552 1126 191 2025 6940 4028 7016 4525 6588 5687 6358 3960\n85892 3791 5949 2539 5485 6344 9517 6150 1887 5862 2926 1491\n69328 7388 1097 7116 8694 8131 5328 114 7604 5646 4157 7122\n73040 1638 4241 4299 4866 3182 5610 1829 7007 5069 9515 109\n29237 6621 8454 129 2246 7505 8058 440 9678 461 5201 2301\n",
"output": "627867\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
View Russian Translation
One day Benny decides to miss lessons. But as she is a very good pig she will do her homework anyway. She doesn't know anything about the content of this lessons. Your task is to help her.
The problem is following: you are given a right triangle with coordinates of vertices (0, 0), (0, a), (b, 0) and integer p. You may choose any value val and draw a vertical line x = val. After that let area(val) be the area of the part of triangle at the right side of this line. Your task is to find such minimal val that area(val) is not more than p% of the area of the given triangle.
Input
The first and the only one line contains three integers a, b and p.
Output
Output in a single line answer to the problem with two signs after decimal point.
It is guaranteed that the answer+10^-6 and answer-10^-6 would produce the same rounded answer.
Constraints
1 ≤ a, b ≤ 10^7
1 ≤ p < 100
1 ≤ a, b ≤ 10^3 holds for test cases worth 20% of the problem's score.
SAMPLE INPUT
2 2 25
SAMPLE OUTPUT
1.00
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
# Hacker Earth
# Problem Statement
# Import libraries
import sys
import math
import pdb
def getOneLine(inFile):
if inFile != "":
return(inFile.readline().rstrip())
return(sys.stdin.readline().rstrip())
def main():
# Toggle mode of input
# inFile = open("input5.txt", 'r')
inFile = ""
# number of cases
input = getOneLine(inFile)
input = input.split(" ")
a = int(input[0])
b = int(input[1])
p = int(input[2]) / 100.0
requiredArea = 0.5 * a * b * p
result = b - math.sqrt(2*requiredArea * b / a)
result = round(result, 2)
print(("%.2f" % result))
main()
# Simple test case
# RRULDL
``` | vfc_134697 | {
"difficulty": "easy",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4742834 4535295 21",
"output": "258544.99",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "6165574 5038223 90",
"output": "0.28",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "279 81 31",
"output": "33985.19",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "4666515 6451293 1",
"output": "2032202.17",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5947083 343633 89",
"output": "11.27",
"type": "stdin_stdout"
}
]
} |
|
taco | verifiable_code | https://codeforces.com/problemset/problem/858/E | Solve the following coding problem using the programming language python:
The All-Berland National Olympiad in Informatics has just ended! Now Vladimir wants to upload the contest from the Olympiad as a gym to a popular Codehorses website.
Unfortunately, the archive with Olympiad's data is a mess. For example, the files with tests are named arbitrary without any logic.
Vladimir wants to rename the files with tests so that their names are distinct integers starting from 1 without any gaps, namely, "1", "2", ..., "n', where n is the total number of tests.
Some of the files contain tests from statements (examples), while others contain regular tests. It is possible that there are no examples, and it is possible that all tests are examples. Vladimir wants to rename the files so that the examples are the first several tests, all all the next files contain regular tests only.
The only operation Vladimir can perform is the "move" command. Vladimir wants to write a script file, each of the lines in which is "move file_1 file_2", that means that the file "file_1" is to be renamed to "file_2". If there is a file "file_2" at the moment of this line being run, then this file is to be rewritten. After the line "move file_1 file_2" the file "file_1" doesn't exist, but there is a file "file_2" with content equal to the content of "file_1" before the "move" command.
Help Vladimir to write the script file with the minimum possible number of lines so that after this script is run: all examples are the first several tests having filenames "1", "2", ..., "e", where e is the total number of examples; all other files contain regular tests with filenames "e + 1", "e + 2", ..., "n", where n is the total number of all tests.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5) — the number of files with tests.
n lines follow, each describing a file with test. Each line has a form of "name_i type_i", where "name_i" is the filename, and "type_i" equals "1", if the i-th file contains an example test, and "0" if it contains a regular test. Filenames of each file are strings of digits and small English letters with length from 1 to 6 characters. The filenames are guaranteed to be distinct.
-----Output-----
In the first line print the minimum number of lines in Vladimir's script file.
After that print the script file, each line should be "move file_1 file_2", where "file_1" is an existing at the moment of this line being run filename, and "file_2" — is a string of digits and small English letters with length from 1 to 6.
-----Examples-----
Input
5
01 0
2 1
2extra 0
3 1
99 0
Output
4
move 3 1
move 01 5
move 2extra 4
move 99 3
Input
2
1 0
2 1
Output
3
move 1 3
move 2 1
move 3 2
Input
5
1 0
11 1
111 0
1111 1
11111 0
Output
5
move 1 5
move 11 1
move 1111 2
move 111 4
move 11111 3
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
n = int(input())
t = [1] + [0] * n
(b, a) = d = ([], [])
(h, s) = ([], [])
for i in range(n):
(f, k) = input().split()
d[int(k)].append(f)
m = len(a)
for i in a:
if i.isdigit() and i[0] != '0':
j = int(i)
if 0 < j <= m:
t[j] = 1
elif m < j <= n:
t[j] = -1
else:
s.append(i)
else:
s.append(i)
for i in b:
if i.isdigit() and i[0] != '0':
j = int(i)
if m < j <= n:
t[j] = 1
elif 0 < j <= m:
t[j] = -1
else:
s.append(i)
else:
s.append(i)
x = [j for j in range(1, m + 1) if t[j] < 0]
y = [j for j in range(m + 1, n + 1) if t[j] < 0]
u = [j for j in range(1, m + 1) if not t[j]]
v = [j for j in range(m + 1, n + 1) if not t[j]]
if not s and (x or y):
s = ['0']
if y:
i = y.pop()
v.append(i)
else:
i = x.pop()
u.append(i)
h.append(str(i) + ' 0')
t[i] = 0
while x or y:
if v and x:
i = x.pop()
j = v.pop()
t[j] = 1
h.append(str(i) + ' ' + str(j))
u.append(i)
else:
(u, v, x, y) = (v, u, y, x)
k = 1
for j in s:
while t[k] == 1:
k += 1
h.append(j + ' ' + str(k))
k += 1
d = '\nmove '
print(str(len(h)) + d + d.join(h) if h else 0)
``` | vfc_134701 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/858/E",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "5\n01 0\n2 1\n2extra 0\n3 1\n99 0\n",
"output": "4\nmove 3 1\nmove 01 5\nmove 2extra 4\nmove 99 3\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "2\n1 0\n2 1\n",
"output": "3\nmove 1 07x45l\nmove 2 1\nmove 07x45l 2\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "5\n1 0\n11 1\n111 0\n1111 1\n11111 0\n",
"output": "5\nmove 1 5\nmove 11 1\nmove 1111 2\nmove 111 4\nmove 11111 3\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "4\nir7oz8 1\nvj4v5t 1\nkwkahb 1\nj5s8o1 0\n",
"output": "4\nmove ir7oz8 1\nmove vj4v5t 2\nmove kwkahb 3\nmove j5s8o1 4\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/SUMANDOR | Solve the following coding problem using the programming language python:
Chef has two integers X and S. He is interested in sequences of non-negative integers such that the sum of their elements is S and the [bitwise OR] of their elements is X. Chef would like to know the shortest possible length such a sequence can have.
Formally, find the smallest positive integer N such that there exists a sequence A = [A_{1}, A_{2}, \ldots, A_{N}] satisfying the following conditions:
Each A_{i} is a non-negative integer
A_{1} + A_{2} + \ldots + A_{N} = S
A_{1} \vee A_{2} \vee \ldots \vee A_{N} = X
where \vee denotes the bitwise OR operation.
If there does not exist any sequence which satisfies the above conditions, output -1. Otherwise, output the minimum possible length of such a sequence.
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- Each test case consists of a single line of input, containing two space-separated integers X and S — the required bitwise OR and sum, respectively.
------ Output Format ------
For each test case, output a single line containing one integer — the shortest possible length of such a sequence, or -1 if no such sequence exists.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ X ≤ S ≤ 10^{9}$
------ subtasks ------
Subtask #1 (100 points): Original constraints
----- Sample Input 1 ------
3
13 23
6 13
1 25
----- Sample Output 1 ------
3
-1
25
----- explanation 1 ------
Test Case $1$: One of the valid arrays of length $3$ is $[9,9,5]$. It can be shown that there is no valid array with length $< 3$.
Test Case $2$: There does not exist any sequence whose sum is $13$ and bitwise OR is $6$.
Test Case $3$: The valid array of minimum length contains $25$ ones.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
def binarize(num):
powers = []
power = 0
while num > 0:
bit = num % 2
num = num // 2
if bit:
powers.append(power)
power += 1
return list(reversed(powers))
def doit(x, s):
if s < x:
return -1
origs = s
repeats = 1
s = s - x
origs = s
binx = binarize(x)
mintimes = 0
for power in binx:
val = 2 ** power
times = s // val
s = s % val
mintimes = max(mintimes, times)
if s != 0:
return -1
return optimize(binx, origs, -1, mintimes) + repeats
def possible(binx, s, mpw):
for power in binx:
val = 2 ** power
times = min(s // val, mpw)
s = s - val * times
if s != 0:
return 0
return 1
def optimize(binx, s, mn, mx):
mid = (mn + mx) // 2
if mid == mn:
return mx
if possible(binx, s, mid):
return optimize(binx, s, mn, mid)
else:
return optimize(binx, s, mid, mx)
t = int(input())
for tt in range(t):
(x, s) = [int(xx) for xx in input().split(' ')]
print(str(doit(x, s)))
``` | vfc_134705 | {
"difficulty": "very_hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/SUMANDOR",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "3\n13 23\n6 13\n1 25\n",
"output": "3\n-1\n25\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://www.codechef.com/problems/UNICOLOR | Solve the following coding problem using the programming language python:
Read problem statements in [Russian], [Bengali], and [Mandarin Chinese] as well.
In a school, there are $C$ clubs (numbered $1$ through $C$) that students can participate in. Each student can be in zero, one or multiple clubs.
Tomorrow, there is a school assembly and $N$ students (numbered $1$ through $N$) will be standing in a line. Each student should wear one of $M$ single-colour shirts provided by the school. The only restriction is that for each club, students in this club must wear shirts with the same colour (possibly the same colour for multiple clubs).
Students in the same clubs try to stand near each other. Formally, for each club $i$, you are given $x_{i}$ segments — let's denote them by $[l_{1},r_{1}],[l_{2},r_{2}], \ldots, [l_{x_{i}},r_{x_{i}}]$, where $1 ≤ l_{1} ≤ r_{1} < l_{2} ≤ r_{2} < \ldots < l_{x_{i}} ≤ r_{x_{i}} ≤ N$ — such for each valid $j$, the students $l_{j}, l_{j}+1, \ldots, r_{j}$ belong to this club.
Determine the number of ways in which the students can choose the colours of their shirts. Two ways to choose colours are considered different if there is at least one student wearing a shirt with a different colour. Since this number can be very large, compute it modulo $998,244,353$.
------ Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first line of each test case contains three space-separated integers $C$, $N$ and $M$.
Then, $2C$ lines follow. For each valid $i$, the $(2i-1)$-th of these lines contains a single integer $x_{i}$ and the $2i$-th line contains $2 \cdot x_{i}$ space-separated integers $l_{1},r_{1},l_{2},r_{2}, \ldots, l_{x_{i}},r_{x_{i}}$.
------ Output ------
For each test case, print a single line containing one integer — the number of ways to choose colours modulo $998,244,353$.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N, M ≤ 10^{9}$
$1 ≤ C ≤ 1,000$
$1 ≤ x_{i} ≤ 100$ for each valid $i$
------ Subtasks ------
Subtask #1 (30 points): $N ≤ 1,000$
Subtask #2 (70 points): original constraints
----- Sample Input 1 ------
1
2 6 3
2
1 2 5 6
1
4 6
----- Sample Output 1 ------
9
----- explanation 1 ------
Example case 1: Students $5$ and $6$ are in both clubs, so these two clubs must wear shirts with the same colour. Student $3$ does not participate in any club and can wear a shirt with any colour out of the three available options.
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
from sys import stdin
input = stdin.readline
mod = 998244353
class dsu:
def __init__(self, n):
self.p = [0] * (n + 1)
self.rank = [0] * (n + 1)
for i in range(1, n + 1):
self.p[i] = i
def find(self, node):
if self.p[node] == node:
return node
self.p[node] = self.find(self.p[node])
return self.p[node]
def union(self, u, v):
(u, v) = (self.find(u), self.find(v))
if self.rank[u] == self.rank[v]:
self.p[v] = u
self.rank[u] += 1
elif self.rank[u] > self.rank[v]:
self.p[v] = u
else:
self.p[u] = v
def answer():
s = []
for i in range(1, c + 1):
x = int(input())
b = list(map(int, input().split()))
for j in range(0, 2 * x, 2):
(l, r) = (b[j], b[j + 1])
s.append([l, 1, i])
s.append([r + 1, -1, i])
s.sort()
(stack, extra, prev) = ([], 0, 1)
a = dsu(c)
for i in range(len(s)):
if s[i][1] == -1:
prev = s[i][0]
stack.pop()
continue
if len(stack):
a.union(stack[-1][2], s[i][2])
else:
extra += s[i][0] - prev
a.union(s[i][2], s[i][2])
stack.append(s[i])
extra += n - prev + 1
comp = set()
for i in range(1, c + 1):
comp.add(a.find(i))
total = len(comp)
ans = pow(m, total + extra, mod)
return ans
for T in range(int(input())):
(c, n, m) = map(int, input().split())
print(answer())
``` | vfc_134709 | {
"difficulty": "very_hard",
"memory_limit": "50000 bytes",
"memory_limit_bytes": null,
"problem_url": "https://www.codechef.com/problems/UNICOLOR",
"time_limit": "1 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "1\n2 6 3\n2\n1 2 5 6\n1\n4 6",
"output": "9",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://practice.geeksforgeeks.org/problems/delete-a-node-from-bst/1 | Solve the following coding problem using the programming language python:
Given a Binary Search Tree and a node value X. Delete the node with the given value X from the BST. If no node with value x exists, then do not make any change.
Example 1:
Input:
2
/ \
1 3
X = 12
Output: 1 2 3
Explanation: In the given input there
is no node with value 12 , so the tree
will remain same.
Example 2:
Input:
1
\
2
\
8
/ \
5 11
/ \ / \
4 7 9 12
X = 9
Output: 1 2 4 5 7 8 11 12
Explanation: In the given input tree after
deleting 9 will be
1
\
2
\
8
/ \
5 11
/ \ \
4 7 12
Your Task:
You don't need to read input or print anything. Your task is to complete the function deleteNode() which takes two arguments. The first being the root of the tree, and an integer 'X' denoting the node value to be deleted from the BST. Return the root of the BST after deleting the node with value X. Do not make any update if there's no node with value X present in the BST.
Note: The generated output will be the inorder traversal of the modified tree.
Expected Time Complexity: O(Height of the BST).
Expected Auxiliary Space: O(Height of the BST).
Constraints:
1 <= N <= 10^{5}
Write your solution by modifying this code:
```python
#User function Template for python3
#Function to delete a node from BST.
def deleteNode(root, X):
# code here.
```
Your solution should implemented in the function "deleteNode". The inputs will be passed to it and it should return the correct solution.
Now solve the problem and return the code. | ```python
def deleteNode(root, X):
if not root:
return None
if X < root.data:
root.left = deleteNode(root.left, X)
elif X > root.data:
root.right = deleteNode(root.right, X)
else:
if root.left is None and root.right is None:
return None
elif not root.left:
current = root.right
root = None
return current
elif not root.right:
current = root.left
root = None
return current
succParent = root
succ = root.right
while succ.left != None:
succParent = succ
succ = succ.left
if succParent != root:
succParent.left = succ.right
else:
succParent.right = succ.right
root.data = succ.data
return root
``` | vfc_134713 | {
"difficulty": "medium",
"memory_limit": null,
"memory_limit_bytes": null,
"problem_url": "https://practice.geeksforgeeks.org/problems/delete-a-node-from-bst/1",
"time_limit": "None"
} | {
"language": "python",
"test_cases": [
{
"fn_name": "deleteNode",
"input": "2\n / \\\n 1 3\nX = 12",
"output": "1 2 3",
"type": "function_call"
},
{
"fn_name": "deleteNode",
"input": "1\n \\\n 2\n \\\n 8 \n / \\\n 5 11\n / \\ / \\\n 4 7 9 12\nX = 9",
"output": "1 2 4 5 7 8 11 12",
"type": "function_call"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/199/B | Solve the following coding problem using the programming language python:
A renowned abstract artist Sasha, drawing inspiration from nowhere, decided to paint a picture entitled "Special Olympics". He justly thought that, if the regular Olympic games have five rings, then the Special ones will do with exactly two rings just fine.
Let us remind you that a ring is a region located between two concentric circles with radii r and R (r < R). These radii are called internal and external, respectively. Concentric circles are circles with centers located at the same point.
Soon a white canvas, which can be considered as an infinite Cartesian plane, had two perfect rings, painted with solid black paint. As Sasha is very impulsive, the rings could have different radii and sizes, they intersect and overlap with each other in any way. We know only one thing for sure: the centers of the pair of rings are not the same.
When Sasha got tired and fell into a deep sleep, a girl called Ilona came into the room and wanted to cut a circle for the sake of good memories. To make the circle beautiful, she decided to cut along the contour.
We'll consider a contour to be a continuous closed line through which there is transition from one color to another (see notes for clarification). If the contour takes the form of a circle, then the result will be cutting out a circle, which Iona wants.
But the girl's inquisitive mathematical mind does not rest: how many ways are there to cut a circle out of the canvas?
Input
The input contains two lines.
Each line has four space-separated integers xi, yi, ri, Ri, that describe the i-th ring; xi and yi are coordinates of the ring's center, ri and Ri are the internal and external radii of the ring correspondingly ( - 100 ≤ xi, yi ≤ 100; 1 ≤ ri < Ri ≤ 100).
It is guaranteed that the centers of the rings do not coinside.
Output
A single integer — the number of ways to cut out a circle from the canvas.
Examples
Input
60 60 45 55
80 80 8 32
Output
1
Input
60 60 45 55
80 60 15 25
Output
4
Input
50 50 35 45
90 50 35 45
Output
0
Note
Figures for test samples are given below. The possible cuts are marked with red dotted line.
<image> <image> <image>
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
(self.buffer.truncate(0), self.buffer.seek(0))
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda : self.buffer.read().decode('ascii')
self.readline = lambda : self.buffer.readline().decode('ascii')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
mod = int(1000000000.0) + 7
def lcm(a, b):
return a * b / gcd(a, b)
def power(k, n):
if n == 0:
return 1
if n % 2:
return power(k, n - 1) * k % mod
t = power(k, n // 2)
return t * t % mod
def intersect(a, b):
d2 = (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
r1r22 = (a[2] + b[2]) ** 2
dr1r22 = (a[2] - b[2]) ** 2
return dr1r22 < d2 < r1r22
def lies(a, b, c):
d2 = (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
return max(0, b[2] - a[2]) ** 2 < d2 <= max(0, c[2] - a[2]) ** 2
def main():
a = []
(x, y, r, R) = map(int, input().split())
a.append([x, y, r])
a.append([x, y, R])
(x, y, r, R) = map(int, input().split())
a.append([x, y, r])
a.append([x, y, R])
ans = 0
for i in range(4):
f = 0
for j in range(4):
if i != j and intersect(a[i], a[j]):
f = 1
if i < 2:
if lies(a[i], a[2], a[3]):
f = 1
elif lies(a[i], a[0], a[1]):
f = 1
if f == 0:
ans += 1
print(ans)
return
main()
``` | vfc_134714 | {
"difficulty": "hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/199/B",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "70 92 20 40\n99 92 9 49\n",
"output": "0",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "0 73 20 40\n59 73 39 100\n",
"output": "1",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | https://codeforces.com/problemset/problem/225/E | Solve the following coding problem using the programming language python:
Consider the following equation:
<image> where sign [a] represents the integer part of number a.
Let's find all integer z (z > 0), for which this equation is unsolvable in positive integers. The phrase "unsolvable in positive integers" means that there are no such positive integers x and y (x, y > 0), for which the given above equation holds.
Let's write out all such z in the increasing order: z1, z2, z3, and so on (zi < zi + 1). Your task is: given the number n, find the number zn.
Input
The first line contains a single integer n (1 ≤ n ≤ 40).
Output
Print a single integer — the number zn modulo 1000000007 (109 + 7). It is guaranteed that the answer exists.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
15
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
print([0, 1, 3, 15, 63, 4095, 65535, 262143, 73741816, 536396503, 140130950, 487761805, 319908070, 106681874, 373391776, 317758023, 191994803, 416292236, 110940209, 599412198, 383601260, 910358878, 532737550, 348927936, 923450985, 470083777, 642578561, 428308066, 485739298, 419990027, 287292016, 202484167, 389339971, 848994100, 273206869, 853092282, 411696552, 876153853, 90046024, 828945523, 697988359][int(input())])
``` | vfc_134722 | {
"difficulty": "hard",
"memory_limit": "256.0 megabytes",
"memory_limit_bytes": null,
"problem_url": "https://codeforces.com/problemset/problem/225/E",
"time_limit": "2.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "14\n",
"output": " 373391776\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "38\n",
"output": " 90046024\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "25\n",
"output": " 470083777\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "22\n",
"output": " 532737550\n",
"type": "stdin_stdout"
}
]
} |
taco | verifiable_code | Solve the following coding problem using the programming language python:
Problem statement
There is a rational number sequence $ X_0, X_1, X_2, ..., X_N $. Each term is defined as follows.
1. $ X_0 = 0 $
2. $ X_i = X_ {i-1} $ $ op_i $ $ Y_i $ ($ 1 \ leq i \ leq N $). However, $ op_i $ is $ + $, $ − $, $ × $, $ ÷ $ Either.
Find $ X_N $.
Constraint
* $ 1 \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq o_i \ leq 4 $
* $ -10 ^ 6 \ leq Y_i \ leq 10 ^ 6 $
* If $ o_i = 4 $, then $ Y_i \ neq 0 $
* $ X_N $ is an integer greater than or equal to $ -2 ^ {31} $ and less than $ 2 ^ {31} $.
input
Input follows the following format. All given numbers are integers.
$ N $
$ o_1 $ $ Y_1 $
$ o_2 $ $ Y_2 $
$ ... $
$ o_N $ $ Y_N $
When $ o_i = 1 $, $ op_i $ is +, when $ o_i = 2 $, $ op_i $ is −, when $ o_i = 3 $, $ op_i $ is ×, and when $ o_i = 4 $, $ op_i $ is ÷.
output
Print the value of $ X_N $ on one line.
Example
Input
4
1 1
4 2
2 4
3 4
Output
-14
The input will be stdin and you should print your solution to stdout
Now solve the problem and return the code. | ```python
m = 10000000019
x = 0
def mod(x):
return (x % m + m) % m
def pow(x, a):
ret = 1
while a:
if a & 1:
ret = mod(x * ret)
x = mod(x * x)
a >>= 1
return ret
for _ in range(int(input())):
(o, y) = map(int, input().split())
if o == 1:
x = (x + y) % m
elif o == 2:
x = (x - y) % m
elif o == 3:
x = mod(x * y)
else:
x = mod(x * pow(y, m - 2))
print(x if x < 1 << 31 else x - m)
``` | vfc_134726 | {
"difficulty": "unknown_difficulty",
"memory_limit": "134.217728 megabytes",
"memory_limit_bytes": null,
"problem_url": null,
"time_limit": "1.0 seconds"
} | {
"language": "python",
"test_cases": [
{
"fn_name": null,
"input": "4\n0 1\n4 2\n2 4\n3 4",
"output": "-16\n",
"type": "stdin_stdout"
},
{
"fn_name": null,
"input": "4\n0 1\n4 2\n2 1\n3 4",
"output": "-4\n",
"type": "stdin_stdout"
}
]
} |
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