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769_A. Year of University Entrance | There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university.
Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than x from the year of university entrance of this student, where x — some non-negative integer. A value x is not given, but it can be uniquely determined from the available data. Note that students don't join other groups.
You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.
Input
The first line contains the positive odd integer n (1 ≤ n ≤ 5) — the number of groups which Igor joined.
The next line contains n distinct integers a1, a2, ..., an (2010 ≤ ai ≤ 2100) — years of student's university entrance for each group in which Igor is the member.
It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.
Output
Print the year of Igor's university entrance.
Examples
Input
3
2014 2016 2015
Output
2015
Input
1
2050
Output
2050
Note
In the first test the value x = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016.
In the second test the value x = 0. Igor entered only the group which corresponds to the year of his university entrance. | ```python
n = int(input())
a = list(map(int, input().split()))
a.sort()
if n % 2:
print(a[n//2])
``` |
793_A. Oleg and shares | Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all n prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices.
Output
Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible.
Examples
Input
3 3
12 9 15
Output
3
Input
2 2
10 9
Output
-1
Input
4 1
1 1000000000 1000000000 1000000000
Output
2999999997
Note
Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3.
In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal.
In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time. | ```python
n,k=map(int,input().split())
l=list(map(int,input().split()))
K=min(l)
for i in range(n):
if (l[i]-K)%k!=0:
print(-1)
exit()
Sum=sum(l)
K=K*n
Sum=Sum-K
print(Sum//k)
``` |
813_D. Two Melodies | Alice is a beginner composer and now she is ready to create another masterpiece. And not even the single one but two at the same time!
Alice has a sheet with n notes written on it. She wants to take two such non-empty non-intersecting subsequences that both of them form a melody and sum of their lengths is maximal.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Subsequence forms a melody when each two adjacent notes either differs by 1 or are congruent modulo 7.
You should write a program which will calculate maximum sum of lengths of such two non-empty non-intersecting subsequences that both of them form a melody.
Input
The first line contains one integer number n (2 ≤ n ≤ 5000).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — notes written on a sheet.
Output
Print maximum sum of lengths of such two non-empty non-intersecting subsequences that both of them form a melody.
Examples
Input
4
1 2 4 5
Output
4
Input
6
62 22 60 61 48 49
Output
5
Note
In the first example subsequences [1, 2] and [4, 5] give length 4 in total.
In the second example subsequences [62, 48, 49] and [60, 61] give length 5 in total. If you choose subsequence [62, 61] in the first place then the second melody will have maximum length 2, that gives the result of 4, which is not maximal. | ```python
import sys
n = int(input()) + 1
a = [0] + list(map(int, input().split()))
mod7 = [x % 7 for x in a]
dp = [[0]*n for _ in range(n)]
maxnum = [0]*(10**5+10)
ans = 0
for i in range(n):
maxmod = [0]*7
for j in range(n):
maxnum[a[j]] = 0
for j in range(i):
maxnum[a[j]] = max(maxnum[a[j]], dp[j][i])
maxmod[mod7[j]] = max(maxmod[mod7[j]], dp[j][i])
for j in range(i+1, n):
dp[i][j] = max(
maxnum[a[j]-1],
maxnum[a[j]+1],
maxmod[mod7[j]],
dp[0][i]
) + 1
maxnum[a[j]] = max(maxnum[a[j]], dp[i][j])
maxmod[mod7[j]] = max(maxmod[mod7[j]], dp[i][j])
ans = max(ans, dp[i][j])
print(ans)
``` |
839_C. Journey | There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 ≤ n ≤ 100000) — number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number — the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | ```python
from queue import *
n = int(input())
g = [[] for x in range(n)]
for i in range(n-1):
a, b = map(int, input().split())
a -= 1
b -= 1
g[a].append(b)
g[b].append(a)
used = [0]*n
anw = 0
def solve(v, d, r):
global anw
q = Queue()
q.put((v, d, r))
while not q.empty():
v, d, r = q.get(v)
used[v] = True
for u in g[v]:
if not used[u]:
q.put((u, d+1, r*(len(g[v])-(v!=0))))
#print("put", u)
#print("so at", v, "len", len(g[v]))
if v != 0 and len(g[v]) == 1:
#print("At ", v, "is", d, r)
anw += d/r
solve(0, 0, 1)
print(anw)
``` |
908_G. New Year and Original Order | Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.
Given a number X, compute <image> modulo 109 + 7.
Input
The first line of input will contain the integer X (1 ≤ X ≤ 10700).
Output
Print a single integer, the answer to the question.
Examples
Input
21
Output
195
Input
345342
Output
390548434
Note
The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195. | ```python
# Problem G
num = input()
num_list = []
for i in range(len(num)):
num_list.append(int(num[i]))
myMod = (10 ** 9) + 7
length = len(num_list)
f = [0] * (length + 1)
t = [1] * (length + 1)
for i in range(length):
f[i+1] = (f[i] * 10 + 1) % myMod
t[i+1] = (t[i] * 10) % myMod
ans = 0
for i in range(1, 10):
dp = [0] * (length + 1)
for j in range(length):
dp[j+1] = (dp[j] * i + (10 - i) * (dp[j] * 10 + t[j])) % myMod
c = 0
ctr = 0
for k in num_list:
z = min(i, k)
o = k - z
ans += o * (dp[length-1-ctr] * t[c+1] + f[c+1] * t[length-1-ctr]) % myMod
ans += z * (dp[length-1-ctr] * t[c] + f[c] * t[length-1-ctr]) % myMod
ans %= myMod
c += k >= i
ctr += 1
ans += f[c]
if ans >= myMod:
ans -= myMod
print(ans)
``` |
930_A. Peculiar apple-tree | In Arcady's garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are n inflorescences, numbered from 1 to n. Inflorescence number 1 is situated near base of tree and any other inflorescence with number i (i > 1) is situated at the top of branch, which bottom is pi-th inflorescence and pi < i.
Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in a-th inflorescence gets to pa-th inflorescence. Apples that end up in first inflorescence are gathered by Arcady in exactly the same moment. Second peculiarity of this tree is that once two apples are in same inflorescence they annihilate. This happens with each pair of apples, e.g. if there are 5 apples in same inflorescence in same time, only one will not be annihilated and if there are 8 apples, all apples will be annihilated. Thus, there can be no more than one apple in each inflorescence in each moment of time.
Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest.
Input
First line of input contains single integer number n (2 ≤ n ≤ 100 000) — number of inflorescences.
Second line of input contains sequence of n - 1 integer numbers p2, p3, ..., pn (1 ≤ pi < i), where pi is number of inflorescence into which the apple from i-th inflorescence rolls down.
Output
Single line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest.
Examples
Input
3
1 1
Output
1
Input
5
1 2 2 2
Output
3
Input
18
1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4
Output
4
Note
In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won't be able to collect them.
In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it. | ```python
from collections import defaultdict,Counter,deque as dq
from sys import stdin
input=stdin.readline
n=int(input())
g=defaultdict(list)
w=list(map(int,input().strip().split()))
for i in range(len(w)):
g[w[i]-1].append(i+1)
g[i+1].append(w[i]-1)
# print(g)
q=dq([0])
d=[-1]*(n)
d[0]=1
cnt=defaultdict(int)
cnt[1]+=1
while q:
t=q.popleft()
for i in g[t]:
if d[i]==-1:
d[i]=d[t]+1
cnt[d[i]]+=1
q.append(i)
ans=0
for i in cnt:
if cnt[i]%2!=0:
ans+=1
print(ans)
``` |
984_A. Game | Two players play a game.
Initially there are n integers a_1, a_2, …, a_n written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. n - 1 turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after n - 1 turns if both players make optimal moves.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of numbers on the board.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print one number that will be left on the board.
Examples
Input
3
2 1 3
Output
2
Input
3
2 2 2
Output
2
Note
In the first sample, the first player erases 3 and the second erases 1. 2 is left on the board.
In the second sample, 2 is left on the board regardless of the actions of the players. | ```python
x = int(input())
l = [int(n) for n in input().split()]
l.sort()
i = 0
while len(l) > 1:
if i % 2 == 0:
l = l[0:len(l) - 1]
else:
l = l[1:len(l)]
i = i + 1
print(l[0])
``` |
p02569 AtCoder Library Practice Contest - Lazy Segment Tree | You are given a binary array A=(A_1,A_2,\cdots,A_N) of length N.
Process Q queries of the following types. The i-th query is represented by three integers T_i,L_i,R_i.
* T_i=1: Replace the value of A_j with 1-A_j for each L_i \leq j \leq R_i.
* T_i=2: Calculate the inversion(*) of the array A_{L_i},A_{L_i+1},\cdots,A_{R_i}．
Note：The inversion of the array x_1,x_2,\cdots,x_k is the number of the pair of integers i,j with 1 \leq i < j \leq k, x_i > x_j.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq A_i \leq 1
* 1 \leq Q \leq 2 \times 10^5
* 1 \leq T_i \leq 2
* 1 \leq L_i \leq R_i \leq N
* All values in Input are integer.
Input
Input is given from Standard Input in the following format:
N Q
A_1 A_2 \cdots A_N
T_1 L_1 R_1
T_2 L_2 R_2
\vdots
T_Q L_Q R_Q
Output
For each query with T_i=2, print the answer.
Example
Input
5 5
0 1 0 0 1
2 1 5
1 3 4
2 2 5
1 1 3
2 1 2
Output
2
0
1 | ```python
import sys
input = lambda: sys.stdin.readline().rstrip()
class LazySegmentTree():
def __init__(self, init, unitX, unitA, f, g, h):
self.f = f # (X, X) -> X
self.g = g # (X, A, size) -> X
self.h = h # (A, A) -> A
self.unitX = unitX
self.unitA = unitA
self.f = f
if type(init) == int:
self.n = init
# self.n = 1 << (self.n - 1).bit_length()
self.X = [unitX] * (self.n * 2)
self.size = [1] * (self.n * 2)
else:
self.n = len(init)
# self.n = 1 << (self.n - 1).bit_length()
self.X = [unitX] * self.n + init + [unitX] * (self.n - len(init))
self.size = [0] * self.n + [1] * len(init) + [0] * (self.n - len(init))
for i in range(self.n-1, 0, -1):
self.X[i] = self.f(self.X[i*2], self.X[i*2|1])
for i in range(self.n - 1, 0, -1):
self.size[i] = self.size[i*2] + self.size[i*2|1]
self.A = [unitA] * (self.n * 2)
def update(self, i, x):
i += self.n
self.X[i] = x
i >>= 1
while i:
self.X[i] = self.f(self.X[i*2], self.X[i*2|1])
i >>= 1
def calc(self, i):
return self.g(self.X[i], self.A[i], self.size[i])
def calc_above(self, i):
i >>= 1
while i:
self.X[i] = self.f(self.calc(i*2), self.calc(i*2|1))
i >>= 1
def propagate(self, i):
self.X[i] = self.g(self.X[i], self.A[i], self.size[i])
self.A[i*2] = self.h(self.A[i*2], self.A[i])
self.A[i*2|1] = self.h(self.A[i*2|1], self.A[i])
self.A[i] = self.unitA
def propagate_above(self, i):
H = i.bit_length()
for h in range(H, 0, -1):
self.propagate(i >> h)
def propagate_all(self):
for i in range(1, self.n):
self.propagate(i)
def getrange(self, l, r):
l += self.n
r += self.n
l0, r0 = l // (l & -l), r // (r & -r) - 1
self.propagate_above(l0)
self.propagate_above(r0)
al = self.unitX
ar = self.unitX
while l < r:
if l & 1:
al = self.f(al, self.calc(l))
l += 1
if r & 1:
r -= 1
ar = self.f(self.calc(r), ar)
l >>= 1
r >>= 1
return self.f(al, ar)
def getvalue(self, i):
i += self.n
self.propagate_above(i)
return self.calc(i)
def operate_range(self, l, r, a):
l += self.n
r += self.n
l0, r0 = l // (l & -l), r // (r & -r) - 1
self.propagate_above(l0)
self.propagate_above(r0)
while l < r:
if l & 1:
self.A[l] = self.h(self.A[l], a)
l += 1
if r & 1:
r -= 1
self.A[r] = self.h(self.A[r], a)
l >>= 1
r >>= 1
self.calc_above(l0)
self.calc_above(r0)
# Find r s.t. calc(l, ..., r-1) = True and calc(l, ..., r) = False
def max_right(self, l, z):
if l >= self.n: return self.n
l += self.n
s = self.unitX
while 1:
while l % 2 == 0:
l >>= 1
if not z(self.f(s, self.calc(l))):
while l < self.n:
l *= 2
if z(self.f(s, self.calc(l))):
s = self.f(s, self.calc(l))
l += 1
return l - self.n
s = self.f(s, self.calc(l))
l += 1
if l & -l == l: break
return self.n
# Find l s.t. calc(l, ..., r-1) = True and calc(l-1, ..., r-1) = False
def min_left(self, r, z):
if r <= 0: return 0
r += self.n
s = self.unitX
while 1:
r -= 1
while r > 1 and r % 2:
r >>= 1
if not z(self.f(self.calc(r), s)):
while r < self.n:
r = r * 2 + 1
if z(self.f(self.calc(r), s)):
s = self.f(self.calc(r), s)
r -= 1
return r + 1 - self.n
s = self.f(self.calc(r), s)
if r & -r == r: break
return 0
def debug(self):
X = self.X
print("X =", [self.calc(i) for i in range(self.n, self.n * 2)])
if False:
f = lambda x, y: (x[0] + y[0] + x[2] * y[1], x[1] + y[1], x[2] + y[2])
g = lambda x, a, s: (x[1] * x[2] - x[0], x[2], x[1]) if a else x
h = lambda a, b: a ^ b
unitX = (0, 0, 0)
unitA = 0
# (inversion, number of zeros, number of ones)
mm = (1 << 18) - 1
mmm = ((1 << 18) - 1) << 18
def f(x, y):
return x + y + ((x & mm) * (y & mmm) << 18)
def g(x, a, s):
x0, x1, x2 = x >> 36, (x >> 18) & mm, x & mm
return (x1 * x2 - x0 << 36) + (x2 << 18) + x1 if a else x
def h(a, b):
return a ^ b
unitX = 0
unitA = 0
N, Q = map(int, input().split())
A = [(1 << 18) if int(a) == 0 else 1 for a in input().split()]
st = LazySegmentTree(A, unitX, unitA, f, g, h)
for _ in range(Q):
t, l, r = map(int, input().split())
if t == 1:
st.operate_range(l - 1, r, 1)
else:
print(st.getrange(l - 1, r) >> 36)
``` |
p02700 AtCoder Beginner Contest 164 - Battle | Takahashi and Aoki will have a battle using their monsters.
The health and strength of Takahashi's monster are A and B, respectively, and those of Aoki's monster are C and D, respectively.
The two monsters will take turns attacking, in the order Takahashi's, Aoki's, Takahashi's, Aoki's, ... Here, an attack decreases the opponent's health by the value equal to the attacker's strength. The monsters keep attacking until the health of one monster becomes 0 or below. The person with the monster whose health becomes 0 or below loses, and the other person wins.
If Takahashi will win, print `Yes`; if he will lose, print `No`.
Constraints
* 1 \leq A,B,C,D \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C D
Output
If Takahashi will win, print `Yes`; if he will lose, print `No`.
Examples
Input
10 9 10 10
Output
No
Input
46 4 40 5
Output
Yes | ```python
a,b,c,d = map(int,input().split())
print('Yes' if (c+b-1)//b <= (a+d-1)//d else 'No')
``` |
p02829 AtCoder Beginner Contest 148 - Round One | Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3 | ```python
A = int(input())
B = int(input())
print(1+2+3 -A-B)
``` |
p02966 AtCoder Grand Contest 036 - Negative Cycle | We have a weighted directed graph with N vertices numbered 0 to N-1.
The graph initially has N-1 edges. The i-th edge (0 \leq i \leq N-2) is directed from Vertex i to Vertex i+1 and has a weight of 0.
Snuke will now add a new edge (i → j) for every pair i, j (0 \leq i,j \leq N-1,\ i \neq j). The weight of the edge will be -1 if i < j, and 1 otherwise.
Ringo is a boy. A negative cycle (a cycle whose total weight is negative) in a graph makes him sad. He will delete some of the edges added by Snuke so that the graph will no longer contain a negative cycle. The cost of deleting the edge (i → j) is A_{i,j}. He cannot delete edges that have been present from the beginning.
Find the minimum total cost required to achieve Ringo's objective.
Constraints
* 3 \leq N \leq 500
* 1 \leq A_{i,j} \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_{0,1} A_{0,2} A_{0,3} \cdots A_{0,N-1}
A_{1,0} A_{1,2} A_{1,3} \cdots A_{1,N-1}
A_{2,0} A_{2,1} A_{2,3} \cdots A_{2,N-1}
\vdots
A_{N-1,0} A_{N-1,1} A_{N-1,2} \cdots A_{N-1,N-2}
Output
Print the minimum total cost required to achieve Ringo's objective.
Examples
Input
3
2 1
1 4
3 3
Output
2
Input
4
1 1 1
1 1 1
1 1 1
1 1 1
Output
2
Input
10
190587 2038070 142162180 88207341 215145790 38 2 5 20
32047998 21426 4177178 52 734621629 2596 102224223 5 1864
41 481241221 1518272 51 772 146 8805349 3243297 449
918151 126080576 5186563 46354 6646 491776 5750138 2897 161
3656 7551068 2919714 43035419 495 3408 26 3317 2698
455357 3 12 1857 5459 7870 4123856 2402 258
3 25700 16191 102120 971821039 52375 40449 20548149 16186673
2 16 130300357 18 6574485 29175 179 1693 2681
99 833 131 2 414045824 57357 56 302669472 95
8408 7 1266941 60620177 129747 41382505 38966 187 5151064
Output
2280211 | ```python
N = int(input())
A = [[0] * N for _ in range(N)]
Al = [[0] * (N+1) for _ in range(N+1)]
Ar = [[0] * (N+1) for _ in range(N+1)]
for i in range(N) :
a = [int(i) for i in input().split()]
for j in range(N) :
if i < j :
A[i][j] = a[j-1]
elif i > j :
A[i][j] = a[j]
for i in range(N) :
for j in range(i+1, N) :
Al[j][i+1] = Al[j][i] + A[j][i]
Ar[i][j] = Ar[i-1][j] + A[i][j]
# for al in Al :
# print(al)
# print()
# for ar in Ar :
# print(ar)
dp = [[float('inf')] * (N+1) for _ in range(N+1)]
dp[0][0] = 0
for i in range(N+1) :
for j in range(i, N+1) :
l, r = 0, 0
for k in range(j+1, N+1) :
l += Al[k-1][i]
r += Ar[k-2][k-1] - Ar[j-1][k-1]
dp[j][k] = min(dp[j][k], dp[i][j] + l + r)
# print('i', i, 'j', j, 'k', k, l, r)
# print(dp)
print(min(dp[i][N] for i in range(N+1)))
``` |
p03101 AtCoder Beginner Contest 121 - White Cells | There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | ```python
#!/usr/bin/env python3
h, w, x, y = map(int, open(0).read().split())
print((h-x)*(w-y))
``` |
p03249 AtCoder Regular Contest 103 - Distance Sums | You are given a sequence D_1, D_2, ..., D_N of length N. The values of D_i are all distinct. Does a tree with N vertices that satisfies the following conditions exist?
* The vertices are numbered 1,2,..., N.
* The edges are numbered 1,2,..., N-1, and Edge i connects Vertex u_i and v_i.
* For each vertex i, the sum of the distances from i to the other vertices is D_i, assuming that the length of each edge is 1.
If such a tree exists, construct one such tree.
Constraints
* 2 \leq N \leq 100000
* 1 \leq D_i \leq 10^{12}
* D_i are all distinct.
Input
Input is given from Standard Input in the following format:
N
D_1
D_2
:
D_N
Output
If a tree with n vertices that satisfies the conditions does not exist, print `-1`.
If a tree with n vertices that satisfies the conditions exist, print n-1 lines. The i-th line should contain u_i and v_i with a space in between. If there are multiple trees that satisfy the conditions, any such tree will be accepted.
Examples
Input
7
10
15
13
18
11
14
19
Output
1 2
1 3
1 5
3 4
5 6
6 7
Input
2
1
2
Output
-1
Input
15
57
62
47
45
42
74
90
75
54
50
66
63
77
87
51
Output
1 10
1 11
2 8
2 15
3 5
3 9
4 5
4 10
5 15
6 12
6 14
7 13
9 12
11 13 | ```python
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(101010)
def dfs(v, adj_list, depth, visited):
visited[v] = True
x = depth
for w in adj_list[v]:
if not visited[w]:
x += dfs(w, adj_list, depth + 1, visited)
return x
def solve(n, d):
if n < 7:
print(-1)
return
d.sort()
w = [1] * n
edges = []
adj_list = [[] for _ in range(n)]
for j in range(n - 1, 0, -1):
di, i = d[j]
pdi = di - n + 2 * w[i]
p = None
lo, hi = 0, j
while lo < hi:
mid = (lo + hi) // 2
xdi, xi = d[mid]
if xdi == pdi:
p = xi
break
elif xdi < pdi:
lo = mid + 1
else:
hi = mid
if p is None:
print(-1)
return
u, v = i, p
if v < u:
u, v = v, u
edges.append((u + 1, v + 1))
adj_list[u].append(v)
adj_list[v].append(u)
w[p] += w[i]
d0, r = d[0]
visited = [False] * n
x = dfs(r, adj_list, 0, visited)
if x != d0:
print(-1)
return
edges.sort()
for uv in edges:
u, v = uv
print('{} {}'.format(u, v))
def main():
n = input()
n = int(n)
d = []
for i in range(n):
di = input()
di = int(di)
d.append((di, i))
solve(n, d)
if __name__ == '__main__':
main()
``` |
p03400 AtCoder Beginner Contest 092 - Chocolate | Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces.
Find the number of chocolate pieces prepared at the beginning of the camp.
Constraints
* 1 \leq N \leq 100
* 1 \leq D \leq 100
* 1 \leq X \leq 100
* 1 \leq A_i \leq 100 (1 \leq i \leq N)
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
D X
A_1
A_2
:
A_N
Output
Find the number of chocolate pieces prepared at the beginning of the camp.
Examples
Input
3
7 1
2
5
10
Output
8
Input
2
8 20
1
10
Output
29
Input
5
30 44
26
18
81
18
6
Output
56 | ```python
n=int(input())
d,x=map(int,input().split())
a=[int(input()) for _ in range(n)]
c=0
for ai in a:
c+=1+(d-1)//ai
print(c+x)
``` |
p03565 AtCoder Beginner Contest 076 - Dubious Document 2 | E869120 found a chest which is likely to contain treasure.
However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.
He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`.
One more thing he found is a sheet of paper with the following facts written on it:
* Condition 1: The string S contains a string T as a contiguous substring.
* Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.
Print the string S.
If such a string does not exist, print `UNRESTORABLE`.
Constraints
* 1 \leq |S'|, |T| \leq 50
* S' consists of lowercase English letters and `?`.
* T consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
T'
Output
Print the string S.
If such a string does not exist, print `UNRESTORABLE` instead.
Examples
Input
?tc????
coder
Output
atcoder
Input
??p??d??
abc
Output
UNRESTORABLE | ```python
S = input()
T = input()
ls = len(S)
lt = len(T)
for i in range(ls-lt,-1,-1):
for j in range(lt):
if S[i+j] != T[j] and S[i+j] != '?':
break
else:
print((S[:i] + T + S[i+lt:]).replace('?','a'))
break
else:
print('UNRESTORABLE')
``` |
p03720 AtCoder Beginner Contest 061 - Counting Roads | There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2≤N,M≤50
* 1≤a_i,b_i≤N
* a_i ≠ b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1≤i≤N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | ```python
n,m = map(int,input().split())
l = [0]*n
for i in range(0,m):
for j in list(map(int,input().split())):
l[j-1]+=1
for i in range(0,n):
print(l[i])
``` |
p03880 CODE FESTIVAL 2016 Grand Final(Parallel) - Cheating Nim | A cheetah and a cheater are going to play the game of Nim. In this game they use N piles of stones. Initially the i-th pile contains a_i stones. The players take turns alternately, and the cheetah plays first. In each turn, the player chooses one of the piles, and takes one or more stones from the pile. The player who can't make a move loses.
However, before the game starts, the cheater wants to cheat a bit to make sure that he can win regardless of the moves by the cheetah. From each pile, the cheater takes zero or one stone and eats it before the game. In case there are multiple ways to guarantee his winning, he wants to minimize the number of stones he eats.
Compute the number of stones the cheater will eat. In case there is no way for the cheater to win the game even with the cheating, print `-1` instead.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ a_i ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the answer.
Examples
Input
3
2
3
4
Output
3
Input
3
100
100
100
Output
-1 | ```python
from functools import reduce
from operator import xor
N = int(input())
As = [int(input()) for _ in range(N)]
dXORs = set([A ^ (A-1) for A in As])
XOR = reduce(xor, As)
ans = 0
for i in reversed(range(30)):
if XOR & (1<<i):
d = ((1<<(i+1)) - 1)
if d in dXORs:
XOR ^= d
ans += 1
print(ans if XOR == 0 else -1)
``` |
p04041 AtCoder Regular Contest 058 - Iroha and Haiku | Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
Iroha is looking for X,Y,Z-Haiku (defined below) in integer sequences.
Consider all integer sequences of length N whose elements are between 1 and 10, inclusive. Out of those 10^N sequences, how many contain an X,Y,Z-Haiku?
Here, an integer sequence a_0, a_1, ..., a_{N-1} is said to contain an X,Y,Z-Haiku if and only if there exist four indices x, y, z, w (0 ≦ x < y < z < w ≦ N) such that all of the following are satisfied:
* a_x + a_{x+1} + ... + a_{y-1} = X
* a_y + a_{y+1} + ... + a_{z-1} = Y
* a_z + a_{z+1} + ... + a_{w-1} = Z
Since the answer can be extremely large, print the number modulo 10^9+7.
Constraints
* 3 ≦ N ≦ 40
* 1 ≦ X ≦ 5
* 1 ≦ Y ≦ 7
* 1 ≦ Z ≦ 5
Input
The input is given from Standard Input in the following format:
N X Y Z
Output
Print the number of the sequences that contain an X,Y,Z-Haiku, modulo 10^9+7.
Examples
Input
3 5 7 5
Output
1
Input
4 5 7 5
Output
34
Input
37 4 2 3
Output
863912418
Input
40 5 7 5
Output
562805100 | ```python
from heapq import heappush, heappop
from collections import deque,defaultdict,Counter
import itertools
from itertools import permutations,combinations
import sys
import bisect
import string
#import math
#import time
#import random
def I():
return int(input())
def MI():
return map(int,input().split())
def LI():
return [int(i) for i in input().split()]
def LI_():
return [int(i)-1 for i in input().split()]
def StoI():
return [ord(i)-97 for i in input()]
def show(*inp,end='\n'):
if show_flg:
print(*inp,end=end)
YN=['Yes','No']
mo=10**9+7
#ts=time.time()
#sys.setrecursionlimit(10**6)
input=sys.stdin.readline
show_flg=False
#show_flg=True
n,X,Y,Z=MI()
t=1<<(X+Y+Z)
haiku=( (1<<(X+Y+Z)) + (1<<(Y+Z)) + (1<<Z) )>>1
dp=[[0]*t for _ in range(n+1)]
dp[0][0]=1
for i in range(n):
for k in range(t):
for d in range(10):
m=((k<<(d+1))+(1<<d))
if m&haiku == haiku:
continue
else:
dp[i+1][int(m)%t]+=dp[i][k]
dp[i+1][int(m)%t]%=mo
sub=0
for i in range(t):
sub+=dp[n][i]
ans=pow(10,n,mo)-sub
print(ans%mo)
``` |
p00122 Summer of Pyonkichi | The University of Aizu has a park covered with grass, and there are no trees or buildings that block the sunlight. On sunny summer days, sprinklers installed in the park operate to sprinkle water on the lawn. The frog Pyonkichi lives in this park. Pyonkichi is not good at hot weather, and on summer days when the sun is strong, he will dry out and die if he is not exposed to the water of the sprinkler. The sprinklers installed in the park are supposed to sprinkle only one sprinkler at a time to save water, so Pyonkichi must move according to the operation of the sprinklers.
<image>
---
(a) Map of the park
<image> | <image>
--- | ---
(b) Pyonkichi's jump range | (c) Sprinkler watering range
The park is as shown in (a) of the above figure, the location in the park is represented by the coordinates 0 to 9 in each of the vertical and horizontal directions, and the black ● is the sprinkler, which indicates the order in which the numbers operate. This is just an example, and the location and order of operation of each sprinkler changes daily.
Pyonkichi is clumsy, so he can only jump a certain distance to move. Pyonkichi's jumpable range is as shown in (b) above, and he cannot move to any other location. Moreover, one jump consumes a lot of physical strength, so you have to rest in the water for a while.
The range in which the sprinkler can sprinkle water is as shown in the above figure (c), including the coordinates of the sprinkler itself. Each sprinkler will stop after a period of watering, and the next sprinkler will start working immediately. Pyonkichi shall jump only once at this time, and shall not jump until the next watering stops. Also, since the park is surrounded on all sides by scorching asphalt, it is assumed that you will not jump in a direction that would cause you to go out of the park.
This summer was extremely hot. Was Pyonkichi able to survive? One day, read the initial position of Pyonkichi, the position of the sprinkler and the operation order, and if there is a movement path that Pyonkichi can survive, "OK", how? If you die even if you do your best, create a program that outputs "NA". However, the maximum number of sprinklers is 10, and Pyonkichi will jump from the initial position at the same time as the first sprinkler is activated.
Input
Given multiple datasets. Each dataset is given in the following format:
px py
n
x1 y1 x2 y2 ... xn yn
The first line gives the abscissa px and ordinate py of the initial position of Pyonkichi. The second line gives the number of sprinklers n. The third line gives the abscissa xi and the ordinate yi of the sprinkler that operates third.
The input ends with two zero lines. The number of datasets does not exceed 20.
Output
For each dataset, print OK if it is alive, or NA on one line otherwise.
Example
Input
6 1
10
6 4 3 3 1 2 0 5 4 6 1 8 5 9 7 7 8 6 8 3
6 1
10
6 4 3 3 1 2 0 5 4 6 1 8 5 9 7 7 8 6 9 0
0 0
Output
OK
NA | ```python
move = [[-1, 2], [0, 2], [1, 2],
[-1, -2], [0, -2], [1, -2],
[2, 1], [2, 0], [2, -1],
[-2, 1], [-2, 0], [-2, -1]]
spraing_range = [[-1, 1], [0, 1], [1, 1],
[-1, 0], [0, 0], [1, 0],
[-1, -1], [0, -1], [1, -1]]
def main():
while True:
px, py = map(int, input().split())
if px is 0 and py is 0:
break
n = int(input())
tmp = list(map(int, input().split()))
sprinklers_x = tmp[::2]
sprinklers_y = tmp[1::2]
fields = make_fields(sprinklers_x, sprinklers_y)
if bfs(fields, px, py, n):
print("OK")
else:
print("NA")
def make_fields(sprinklers_x, sprinklers_y):
fields = []
for x, y in zip(sprinklers_x, sprinklers_y):
field = [[0] * 10 for _ in range(10)]
for dx, dy in spraing_range:
nx = x + dx
ny = y + dy
if nx < 0 or ny < 0 or nx > 9 or ny > 9:
continue
field[ny][nx] = 1
fields.append(field)
return fields
def bfs(fields, init_px, init_py, n):
q = [[init_px, init_py, 0]] # [px, py, count]
while len(q) is not 0:
px, py, count = q.pop(0)
if count is n:
return True
for dx, dy in move:
nx = px + dx
ny = py + dy
if nx < 0 or ny < 0 or nx > 9 or ny > 9 or \
fields[count][ny][nx] is 0:
continue
q.append([nx, ny, count + 1])
return False
if __name__ == '__main__':
main()
``` |
p00255 Salary for a Plumber | I am a pipe tie craftsman. As long as you get the joints and pipes that connect the pipes, you can connect any pipe. Every day, my master gives me pipes and joints, which I connect and give to my master. But if you have too many pipes, you can't connect them all in one day. Even in such a case, the master smiles and gives me a salary.
By the way, I noticed something strange at one point. Often, the salary is higher when all the pipes are not connected than when they are all connected. It's so weird that one day when my boss came, I secretly saw a memo that describes how to calculate my salary. Then, how
"The salary is paid by" the number of pipes x the total length of pipes ". However, if they are connected by a joint and become one, it is regarded as one pipe. "
It was written. Now I understand why the salary may be cheaper if you connect them all. For example, as shown in the figure below, if you connect all three pipes of length 1 and two joints of length 2 to make one pipe of length 1 + 2 + 1 + 2 + 1 = 7, 1 × (7) ) = 7. However, if you use only one joint and make two pipes with a length of 1 + 2 + 1 = 4 and a pipe with a length of 1, 2 × (4 + 1) = 10, so you will get more salary than connecting all of them. ..
<image>
I don't know why my boss decides my salary this way, but I know how I can get more salary!
Now, create a program that calculates the maximum amount of salary you can get given the number of pipes.
input
The input consists of multiple datasets. The end of the input is indicated by a single zero line. Each dataset is given in the following format.
n
p1 ... pn
j1 ... jn-1
The number of pipes n (2 ≤ n ≤ 65000) is given in the first line. The second line consists of n integers separated by a single space. pi (1 ≤ pi ≤ 1000) indicates the length of the i-th pipe. The third line consists of n-1 integers separated by one space. ji (1 ≤ ji ≤ 1000) indicates the length of the i-th joint.
The i-th joint can connect only the i-th and i + 1-th pipes. The length of the connected pipe is pi + ji + pi + 1.
The number of datasets does not exceed 100.
output
For each dataset, print the maximum amount of salary you can get on one line. For datasets given as input, the output value must always fall within the range of 32-bit unsigned integers.
Example
Input
3
1 1 1
3 3
4
3 3 3 3
1 1 1
5
1 2 3 4 5
4 3 2 1
0
Output
12
48
76 | ```python
# -*- coding: utf-8 -*-
"""
Salary for a Plumber
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0260
"""
import sys
def solve(n):
pipes = [int(p) for p in input().split()]
joints = sorted([int(j) for j in input().split()])
total_length = sum(pipes)
record = total_length * n
while joints:
total_length += joints.pop()
n -= 1
if total_length * n > record:
record = total_length * n
else:
break
return record
def main(args):
while True:
n = int(input())
if n == 0:
break
ans = solve(n)
print(ans)
if __name__ == '__main__':
main(sys.argv[1:])
``` |
p00441 The Oldest Site | problem
Once upon a time there were settlements and many people lived there. People built buildings of various shapes and sizes. But those buildings have already been lost, only the literature and the pillars found in the ruins. Was a clue to the location of the building.
There is a description of the temple in the literature. The temple was exactly square when viewed from above, and there were pillars at its four corners. It is unknown which direction the temple was facing. Also, on the sides and inside. I don't know if there were any pillars. Archaeologists believed that among the pillars found in the ruins, the one with the largest square area must be the temple.
Since the coordinates of the position of the pillar are given, find the square with the largest area among the four pillars and write a program to output the area. Note that the sides of the square are not always parallel to the coordinate axes. Be careful.
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
On the first line, the number n of pillars found in the ruins is written.
In each of the n lines from the 2nd line to the n + 1st line, the x and y coordinates of the column are written separated by blanks.
A pillar never appears more than once.
n is an integer that satisfies 1 ≤ n ≤ 3000, and the x and y coordinates of the column are integers greater than or equal to 0 and less than or equal to 5000.
Of the scoring data, 30% of the points satisfy 1 ≤ n ≤ 100, and 60% of the points satisfy 1 ≤ n ≤ 500.
The number of datasets does not exceed 10.
output
Outputs one integer for each dataset. If there is a square with four columns, it outputs the area of the largest of those squares, and if no such square exists. Output 0.
Examples
Input
10
9 4
4 3
1 1
4 2
2 4
5 8
4 0
5 3
0 5
5 2
10
9 4
4 3
1 1
4 2
2 4
5 8
4 0
5 3
0 5
5 2
0
Output
10
10
Input
None
Output
None | ```python
def main():
while True:
n = int(input())
if not n:
break
ps = [list(map(int,input().split())) for i in range(n)]
dic = set()
for t in ps:
dic.add((t[0],t[1]))
ans = 0
for i in range(n):
for j in range(n):
p1 = ps[i]
p2 = ps[j]
p1x = p1[0]
p1y = p1[1]
p2x = p2[0]
p2y = p2[1]
vx = p2x - p1x
vy = p2y - p1y
if (p1x + vy, p1y - vx) in dic and (p2x + vy, p2y - vx) in dic:
ans = max(ans, vx ** 2 + vy ** 2)
print(ans)
main()
``` |
p00632 Ghost Buster! | The city is full of ghosts, something the average person doesn't know about. Most of them are harmless, but the trouble is that there are quite a few evil spirits that curse people.
There was a girl who fought against such evil spirits. She goes to high school with a faceless face during the day, but at night she walks around the city, finding a wandering soul and making her a Buddhahood.
In order for a ghost to become a Buddhahood, you must first approach the ghost, but that is not an easy task. Because ghosts can slip through things. Ghosts invade private houses through walls and exit from their own doors, or cross boulevards without traffic lights. It's very difficult to track down ghosts because she can't do that as a human being.
She knew from experience that ghosts behave regularly. So she thought of using this information to track down ghosts.
The city where she is located is represented as a grid of (H x W) squares. Both she and the ghost take the following five actions once a minute
* Go east one square
* Go west one square
* Go south one square
* Go north one square
* Stay on the spot
You can do any one of them. However, she and the ghosts can't get out of the city because she has set up a barrier around the city. Also, if you try to do that, you will stay there instead.
Each square is represented by the letters'#' or'.', The former represents a square that only ghosts can enter, and the latter represents a square that both can enter.
The ghost's behavior pattern is given as a sequence of L behaviors. The ghost executes these actions in order from the beginning, and when all the actions are completed, it returns to the first action and continues to execute in order.
If she and the ghost were in the same square N minutes later, it can be said that she and the ghost met at time N. She wants to catch the ghost as soon as possible because she doesn't know what kind of damage the ghost will do if left alone. So she wants you, a dependable senior, to write a program that asks for the earliest time and place where she and ghosts can meet.
Now, help her and lead her wandering soul to Sukhavati Jodo!
Input
Multiple datasets are given. The format of each dataset is shown below.
H W (number of cells in the north-south direction of the grid, number of cells in the east-west direction of the grid: two integers separated by blanks)
Letters H x W (mass type:'.','#','A', or'B')
pattern (ghost behavior pattern: character string)
'A' indicates the initial position of the girl and'B' indicates the initial position of the ghost. The initial positions of both are above the'.' Square.
pattern is a string consisting of '5', '2', '4', '6', and '8', and the meaning of each character is as follows:
'5' | Stay on the spot
--- | ---
'8' | Go north one square
'6' | Go east one square
'4' | Go west one square
'2' | Go south one square
For example, if the pattern is "8544", the ghost repeats the action patterns of "advancing 1 square to the north", "staying there", "advancing 1 square to the west", and "advancing 1 square to the west".
H and W are 1 or more and 20 or less, and the pattern length is 10 or less.
When both H and W are 0, the input ends.
Output
If the girl can encounter a ghost, output the earliest encounter time, north-south coordinates, and east-west coordinates of the encounter location on one line, separated by a blank. The northwesternmost part of the grid is the coordinate (0, 0), and the southeasternmost part is the coordinate (H-1, W-1).
If you can never encounter it no matter how you act, output "impossible".
Example
Input
4 7
A#...#B
.#.#.#.
.#.#.#.
...#...
5
4 7
A#...#B
.#.#.#.
.#.#.#.
...#...
2
4 7
A#...#B
.#.#.#.
.#.#.#.
...#...
4
4 7
A#...#B
.#.#.#.
.#.#.#.
...#...
442668
1 10
A#.......B
55555554
1 10
A#.......B
55555555
0 0
Output
18 0 6
15 3 6
6 0 0
14 0 4
72 0 0
impossible | ```python
from collections import deque
while True:
h, w = map(int, input().split())
if h == 0:
break
mp = [list("X" + input() + "X") for _ in range(h)]
mp.insert(0, ["X"] * (w + 2))
mp.append(["X"] * (w + 2))
for y in range(h + 2):
for x in range(w + 2):
if mp[y][x] == "A":
ax, ay = x, y
mp[y][x] = "."
if mp[y][x] == "B":
bx, by = x, y
mp[y][x] = "."
INF = 1000
dist_mp = [[INF] * (w + 2) for _ in range(h + 2)]
dist_mp[ay][ax] = 0
que = deque()
que.append((0, ax, ay))
vec = ((1, 0), (0, -1), (-1, 0), (0, 1))
while que:
d, x, y = que.popleft()
for dx, dy in vec:
nx, ny = x + dx, y + dy
if mp[ny][nx] == "." and dist_mp[ny][nx] == INF:
que.append((d + 1, nx, ny))
dist_mp[ny][nx] = d + 1
limit = d
pattern = input()
length = len(pattern)
t = 0
dic = {}
while True:
if t >= dist_mp[by][bx]:
print(t, by - 1, bx - 1)
break
if t > limit:
if (t % length, bx, by) in dic:
print("impossible")
break
else:
dic[(t % length, bx, by)] = True
c = pattern[t % length]
if c == "5": pass
elif c == "8" and mp[by - 1][bx] != "X": by -= 1
elif c == "6" and mp[by][bx + 1] != "X": bx += 1
elif c == "4" and mp[by][bx - 1] != "X": bx -= 1
elif c == "2" and mp[by + 1][bx] != "X": by += 1
t += 1
else:
print("impossible")
``` |
p00776 Encryption System | Encryption System
A programmer developed a new encryption system. However, his system has an issue that two or more distinct strings are `encrypted' to the same string.
We have a string encrypted by his system. To decode the original string, we want to enumerate all the candidates of the string before the encryption. Your mission is to write a program for this task.
The encryption is performed taking the following steps. Given a string that consists only of lowercase letters ('a' to 'z').
1. Change the first 'b' to 'a'. If there is no 'b', do nothing.
2. Change the first 'c' to 'b'. If there is no 'c', do nothing.
...
3. Change the first 'z' to 'y'. If there is no 'z', do nothing.
Input
The input consists of at most 100 datasets. Each dataset is a line containing an encrypted string. The encrypted string consists only of lowercase letters, and contains at least 1 and at most 20 characters.
The input ends with a line with a single '#' symbol.
Output
For each dataset, the number of candidates n of the string before encryption should be printed in a line first, followed by lines each containing a candidate of the string before encryption. If n does not exceed 10, print all candidates in dictionary order; otherwise, print the first five and the last five candidates in dictionary order.
Here, dictionary order is recursively defined as follows. The empty string comes the first in dictionary order. For two nonempty strings x = x1 ... xk and y = y1 ... yl, the string x precedes the string y in dictionary order if
* x1 precedes y1 in alphabetical order ('a' to 'z'), or
* x1 and y1 are the same character and x2 ... xk precedes y2 ... yl in dictionary order.
Sample Input
enw
abc
abcdefghijklmnopqrst
z
Output for the Sample Input
1
fox
5
acc
acd
bbd
bcc
bcd
17711
acceeggiikkmmooqqssu
acceeggiikkmmooqqstt
acceeggiikkmmooqqstu
acceeggiikkmmooqrrtt
acceeggiikkmmooqrrtu
bcdefghijklmnopqrrtt
bcdefghijklmnopqrrtu
bcdefghijklmnopqrssu
bcdefghijklmnopqrstt
bcdefghijklmnopqrstu
0
Example
Input
enw
abc
abcdefghijklmnopqrst
z
#
Output
1
fox
5
acc
acd
bbd
bcc
bcd
17711
acceeggiikkmmooqqssu
acceeggiikkmmooqqstt
acceeggiikkmmooqqstu
acceeggiikkmmooqrrtt
acceeggiikkmmooqrrtu
bcdefghijklmnopqrrtt
bcdefghijklmnopqrrtu
bcdefghijklmnopqrssu
bcdefghijklmnopqrstt
bcdefghijklmnopqrstu
0 | ```python
from itertools import chain
alph = "abcdefghijklmnopqrstuvwxyz"
# s = input()
def solve1(s):
cands = [s]
for c in reversed(alph[:-1]):
cands = chain.from_iterable(
[candidates(s, c) for s in cands]
)
cands = list(cands)
cands.sort()
print(len(cands))
if len(cands) > 10:
for s in cands[:5]:
print(s)
for s in cands[-5:]:
print(s)
else:
for s in cands:
print(s)
def next_char(c):
return chr(ord(c)+1)
def candidates(string, key):
nex = next_char(key)
flag = False
for i, c in enumerate(string):
if c == nex:
flag = True
break
if c != key:
continue
yield string[:i] + nex + string[i+1:]
if not flag:
yield string
while True:
s = input()
if s == "#":
break
solve1(s)
``` |
p00907 Find the Outlier | Professor Abacus has just built a new computing engine for making numerical tables. It was designed to calculate the values of a polynomial function in one variable at several points at a time. With the polynomial function f(x) = x2 + 2x + 1, for instance, a possible expected calculation result is 1 (= f(0)), 4 (= f(1)), 9 (= f(2)), 16 (= f(3)), and 25 (= f(4)).
It is a pity, however, the engine seemingly has faulty components and exactly one value among those calculated simultaneously is always wrong. With the same polynomial function as above, it can, for instance, output 1, 4, 12, 16, and 25 instead of 1, 4, 9, 16, and 25.
You are requested to help the professor identify the faulty components. As the first step, you should write a program that scans calculation results of the engine and finds the wrong values.
Input
The input is a sequence of datasets, each representing a calculation result in the following format.
d
v0
v1
...
vd+2
Here, d in the first line is a positive integer that represents the degree of the polynomial, namely, the highest exponent of the variable. For instance, the degree of 4x5 + 3x + 0.5 is five and that of 2.4x + 3.8 is one. d is at most five.
The following d + 3 lines contain the calculation result of f(0), f(1), ... , and f(d + 2) in this order, where f is the polynomial function. Each of the lines contains a decimal fraction between -100.0 and 100.0, exclusive.
You can assume that the wrong value, which is exactly one of f(0), f(1), ... , and f(d+2), has an error greater than 1.0. Since rounding errors are inevitable, the other values may also have errors but they are small and never exceed 10-6.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output i in a line when vi is wrong.
Example
Input
2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0
Output
2
1
1
6 | ```python
from itertools import combinations
def build(X, Y):
A = []
for x in X:
res = 1
for xi in X:
if x == xi:
continue
res *= x - xi
A.append(Y[x] / res)
return A
def calc(X, A, x):
base = 1
for xi in X:
base *= i - xi
return sum(base / (i - x) * a for x, a in zip(X, A))
while 1:
d = int(input())
if d == 0:
break
N = d+3
Y = [float(input()) for i in range(N)]
cnts = [0]*N
for X in combinations(range(N), d+1):
U = [0]*N
for x in X:
U[x] = 1
A = build(X, Y)
for i in range(N):
if U[i]:
continue
res = calc(X, A, i)
if abs(Y[i] - res) > 0.5:
cnts[i] += 1
print(cnts.index(max(cnts)))
``` |
p01479 Chicken or the Egg | Time Limit: 8 sec / Memory Limit: 64 MB
Example
Input
eggchickenegg
Output
egg | ```python
s=input()
i=c=p=0;b=''
while i<len(s):
if s[i]==b:
if c > p:r=b;p=c
c=0
b=s[i]
i+=[3,7][s[i]=='c']
c+=1
if c>p:r=b
print(['chicken','egg'][r=='e'])
``` |
p02206 Prize | Prize
Segtree entered a programming contest with a team of $ N $ and won a $ K $ yen prize! I'm trying to distribute this prize now.
Each $ N $ team member, including Segtree, is numbered from $ 1 $ to $ N $ in order of ability. Segtree is $ 1 $.
If the prize amount of $ i $'s teammate $ (i \ geq 2) $ is less than "$ i-half of the prize amount of $ i's teammate rounded down to an integer", that person Get angry.
When distributing the $ K $ Yen prize so that no one gets angry, find the maximum prize that Segtree can get.
input
Input is given from standard input in the following format.
N K
output
Please output the maximum prize money that Segtree can receive.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N, K \ leq 10 ^ {18} $
* All inputs are integers.
Input example 1
1 1
Output example 1
1
Input example 2
819875141880895728 349993004923078537
Output example 2
174996502461539284
Example
Input
1 1
Output
1 | ```python
n,k = map(int,input().split())
ok,ng = 0,k+1
while ng-ok>1:
mid = (ng+ok)//2
s = 0
m = mid
for i in range(n):
s += m
m //= 2
if m == 0:
break
if s <= k:
ok = mid
else:
ng = mid
print(ok)
``` |
p02360 The Maximum Number of Overlaps | Given a set of $N$ axis-aligned rectangular seals, find the number of overlapped seals on the region which has the maximum number of overlapped seals.
Constraints
* $ 1 \leq N \leq 100000 $
* $ 0 \leq x1_i < x2_i \leq 1000 $
* $ 0 \leq y1_i < y2_i \leq 1000 $
* $ x1_i, y1_i, x2_i, y2_i$ are given in integers
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left and the bottom-right corner of the $i$-th seal respectively.
Output
Print the maximum number of overlapped seals in a line.
Examples
Input
2
0 0 3 2
2 1 4 3
Output
2
Input
2
0 0 2 2
2 0 4 2
Output
1
Input
3
0 0 2 2
0 0 2 2
0 0 2 2
Output
3 | ```python
#!/usr/bin/env python3
# DSL_5_B: The Maximum Number of Overlaps
from itertools import accumulate
import sys
if __name__ == '__main__':
n = int(input())
ys = [0] * 1001
rects = [None] * 2 * n
i = 0
for line in sys.stdin:
x1, y1, x2, y2 = [int(j) for j in line.split()]
rects[i] = (x2, -1, y1, y2)
rects[i+n] = (x1, 1, y1, y2)
i += 1
rects.sort(key=lambda x: x[0])
max_overlap = 0
for x, t, y1, y2 in rects:
if t > 0:
ys[y1] += 1
ys[y2] -= 1
else:
overlap = max(accumulate(ys))
if overlap > max_overlap:
max_overlap = overlap
ys[y1] -= 1
ys[y2] += 1
print(max_overlap)
``` |
1045_B. Space Isaac | Everybody seems to think that the Martians are green, but it turns out they are metallic pink and fat. Ajs has two bags of distinct nonnegative integers. The bags are disjoint, and the union of the sets of numbers in the bags is \{0,1,…,M-1\}, for some positive integer M. Ajs draws a number from the first bag and a number from the second bag, and then sums them modulo M.
What are the residues modulo M that Ajs cannot obtain with this action?
Input
The first line contains two positive integer N (1 ≤ N ≤ 200 000) and M (N+1 ≤ M ≤ 10^{9}), denoting the number of the elements in the first bag and the modulus, respectively.
The second line contains N nonnegative integers a_1,a_2,…,a_N (0 ≤ a_1<a_2< …< a_N<M), the contents of the first bag.
Output
In the first line, output the cardinality K of the set of residues modulo M which Ajs cannot obtain.
In the second line of the output, print K space-separated integers greater or equal than zero and less than M, which represent the residues Ajs cannot obtain. The outputs should be sorted in increasing order of magnitude. If K=0, do not output the second line.
Examples
Input
2 5
3 4
Output
1
2
Input
4 1000000000
5 25 125 625
Output
0
Input
2 4
1 3
Output
2
0 2
Note
In the first sample, the first bag and the second bag contain \{3,4\} and \{0,1,2\}, respectively. Ajs can obtain every residue modulo 5 except the residue 2: 4+1 ≡ 0, 4+2 ≡ 1, 3+0 ≡ 3, 3+1 ≡ 4 modulo 5. One can check that there is no choice of elements from the first and the second bag which sum to 2 modulo 5.
In the second sample, the contents of the first bag are \{5,25,125,625\}, while the second bag contains all other nonnegative integers with at most 9 decimal digits. Every residue modulo 1 000 000 000 can be obtained as a sum of an element in the first bag and an element in the second bag. | ```python
import sys
input = sys.stdin.readline
def main():
n, m = map(int, input().split())
a = list(map(int, input().split())) + [0]*500000
ans_S = 0
a[n] = a[0] + m
s = [0]*600600
for i in range(n):
s[i] = a[i + 1] - a[i]
s[n] = -1
for i in range(n):
s[2*n - i] = s[i]
for i in range(2*n + 1, 3*n + 1):
s[i] = s[i - n]
l, r = 0, 0
z = [0]*600600
for i in range(1, 3*n + 1):
if i < r:
z[i] = z[i - l]
while i + z[i] <= 3*n and (s[i + z[i]] == s[z[i]]):
z[i] += 1
if i + z[i] > r:
l = i
r = i + z[i]
ans = []
for i in range(n + 1, 2*n + 1):
if z[i] < n:
continue
ans_S += 1
ans.append((a[0] + a[2*n - i + 1]) % m)
ans.sort()
print(ans_S)
print(*ans)
return
if __name__=="__main__":
main()
``` |
1068_B. LCM | Ivan has number b. He is sorting through the numbers a from 1 to 10^{18}, and for every a writes ([a, b])/(a) on blackboard. Here [a, b] stands for least common multiple of a and b. Ivan is very lazy, that's why this task bored him soon. But he is interested in how many different numbers he would write on the board if he would finish the task. Help him to find the quantity of different numbers he would write on the board.
Input
The only line contains one integer — b (1 ≤ b ≤ 10^{10}).
Output
Print one number — answer for the problem.
Examples
Input
1
Output
1
Input
2
Output
2
Note
In the first example [a, 1] = a, therefore ([a, b])/(a) is always equal to 1.
In the second example [a, 2] can be equal to a or 2 ⋅ a depending on parity of a. ([a, b])/(a) can be equal to 1 and 2. | ```python
b=int(input())
out=[1]
n=b
i=0
while n%2==0:
i=i+1
out.append(2**i)
n=int(n/2)
out1=[]
for i in range (1,int(n**0.5)+1,2):
if n%i==0:
out1.append(i)
out1.append(int(n/i))
out2=set()
for i in out:
for j in out1:
out2.add(i*j)
#print (out2)
print (len(out2))
``` |
1139_A. Even Substrings | You are given a string s=s_1s_2... s_n of length n, which only contains digits 1, 2, ..., 9.
A substring s[l ... r] of s is a string s_l s_{l + 1} s_{l + 2} … s_r. A substring s[l ... r] of s is called even if the number represented by it is even.
Find the number of even substrings of s. Note, that even if some substrings are equal as strings, but have different l and r, they are counted as different substrings.
Input
The first line contains an integer n (1 ≤ n ≤ 65000) — the length of the string s.
The second line contains a string s of length n. The string s consists only of digits 1, 2, ..., 9.
Output
Print the number of even substrings of s.
Examples
Input
4
1234
Output
6
Input
4
2244
Output
10
Note
In the first example, the [l, r] pairs corresponding to even substrings are:
* s[1 ... 2]
* s[2 ... 2]
* s[1 ... 4]
* s[2 ... 4]
* s[3 ... 4]
* s[4 ... 4]
In the second example, all 10 substrings of s are even substrings. Note, that while substrings s[1 ... 1] and s[2 ... 2] both define the substring "2", they are still counted as different substrings. | ```python
n = int(input())
s = input()
k = 0
for i in range(1, n + 1):
d = int(s[i - 1])
if d % 2 == 0:
k += i
print(k)
``` |
1157_B. Long Number | You are given a long decimal number a consisting of n digits from 1 to 9. You also have a function f that maps every digit from 1 to 9 to some (possibly the same) digit from 1 to 9.
You can perform the following operation no more than once: choose a non-empty contiguous subsegment of digits in a, and replace each digit x from this segment with f(x). For example, if a = 1337, f(1) = 1, f(3) = 5, f(7) = 3, and you choose the segment consisting of three rightmost digits, you get 1553 as the result.
What is the maximum possible number you can obtain applying this operation no more than once?
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of digits in a.
The second line contains a string of n characters, denoting the number a. Each character is a decimal digit from 1 to 9.
The third line contains exactly 9 integers f(1), f(2), ..., f(9) (1 ≤ f(i) ≤ 9).
Output
Print the maximum number you can get after applying the operation described in the statement no more than once.
Examples
Input
4
1337
1 2 5 4 6 6 3 1 9
Output
1557
Input
5
11111
9 8 7 6 5 4 3 2 1
Output
99999
Input
2
33
1 1 1 1 1 1 1 1 1
Output
33 | ```python
n = int(input())
a = list(input())
a = [int(x) for x in a]
d = {}
replacements = list(map(int, input().split()))
for i in range(9):
d[i+1] = replacements[i]
b = [d[x] for x in a]
ans = ""
flag = False
i = 0
j = 0
k = 0
for i in range(len(a)):
if(a[i] >= b[i]):
ans += str(a[i])
else:
flag = True
break
if(flag):
flag = False
for j in range(i, len(a)):
if(b[j] >= a[j]):
ans += str(b[j])
else:
flag = True
break
if(flag):
for k in range(j, len(a)):
ans += str(a[k])
print(ans)
``` |
1197_E. Culture Code | There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2. | ```python
from sys import stdin, stdout
mod = 10**9+7
n = int(input())
dolls = []
for i in range(n):
o, i = map(int, stdin.readline().split())
dolls.append((o, i))
dolls.sort()
dolls = [(i, o) for (o, i) in dolls]
#print(dolls)
def bin_search(i):
lo = -1
hi = n-1
while lo+1 < hi:
mid = (lo+hi-1)//2 + 1
top = dolls[mid][1]
if top <= i:
lo = mid
else:
hi = mid
return hi
counts = [1]*(n+1)
dp = [0]*(n+1)
for k in range(n):
i, o = dolls[k]
m_prev = dp[k]+o-(dolls[k-1][1] if k > 0 else 0)
kk = bin_search(i)
m_with = dp[kk] + i-(dolls[kk-1][1] if kk > 0 else 0)
if m_prev < m_with:
dp[k+1] = m_prev
counts[k+1] = counts[k]
elif m_prev > m_with:
dp[k+1] = m_with
counts[k+1] = counts[kk]
else:
dp[k+1] = m_prev
counts[k+1] = (counts[k]+counts[kk]) % mod
#print(dp)
#print(counts)
best = 10**9+10
best_count = 0
maximal = max([i for i, o in dolls])
for i in range(bin_search(maximal), n):
ii = bin_search(dolls[i][0])
cur = dp[ii] + dolls[i][0] - (dolls[ii-1][1] if ii > 0 else 0)
#print(cur, "via", ii)
if cur < best:
best = cur
best_count = counts[ii]
elif cur == best:
best_count += counts[ii]
best_count %= mod
print(best_count)
``` |
1237_H. Balanced Reversals | You have two strings a and b of equal even length n consisting of characters 0 and 1.
We're in the endgame now. To finally make the universe perfectly balanced, you need to make strings a and b equal.
In one step, you can choose any prefix of a of even length and reverse it. Formally, if a = a_1 a_2 … a_n, you can choose a positive even integer p ≤ n and set a to a_p a_{p-1} … a_1 a_{p+1} a_{p+2} … a_n.
Find a way to make a equal to b using at most n + 1 reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized.
Input
The first line contains a single integer t (1 ≤ t ≤ 2000), denoting the number of test cases.
Each test case consists of two lines. The first line contains a string a of length n, and the second line contains a string b of the same length (2 ≤ n ≤ 4000; n mod 2 = 0). Both strings consist of characters 0 and 1.
The sum of n over all t test cases doesn't exceed 4000.
Output
For each test case, if it's impossible to make a equal to b in at most n + 1 reversals, output a single integer -1.
Otherwise, output an integer k (0 ≤ k ≤ n + 1), denoting the number of reversals in your sequence of steps, followed by k even integers p_1, p_2, …, p_k (2 ≤ p_i ≤ n; p_i mod 2 = 0), denoting the lengths of prefixes of a to be reversed, in chronological order.
Note that k doesn't have to be minimized. If there are many solutions, output any of them.
Example
Input
4
0100011011
1101011000
10101010
10101010
0011
1001
100011
110010
Output
3
6 4 10
0
-1
7
2 6 2 6 2 2 6
Note
In the first test case, string a changes as follows:
* after the first reversal: 1000101011;
* after the second reversal: 0001101011;
* after the third reversal: 1101011000. | ```python
import os
import sys
from io import BytesIO, IOBase
# region fastio
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")
def solve_case():
a = [int(x) for x in input()]
b = [int(x) for x in input()]
n = len(a)
a_rev = []
b_rev = []
def rev_a(x):
if not x:
return
nonlocal a
a = a[:x][::-1] + a[x:]
a_rev.append(x)
def rev_b(x):
if not x:
return
nonlocal b
b = b[:x][::-1] + b[x:]
b_rev.append(x)
def answer():
#cleaning up
nonlocal a_rev
nonlocal b_rev
b_rev.reverse()
a_rev += b_rev
while True:
final = []
i = 0
repl = False
while i < len(a_rev):
if i < len(a_rev) - 1 and a_rev[i] == a_rev[i + 1]:
repl = True
i += 2
else:
final.append(a_rev[i])
i += 1
a_rev = final
if not repl:
break
print(len(a_rev))
print(*a_rev)
a_occ = [[0, 0], [0, 0]]
b_occ = [[0, 0], [0, 0]]
for i in range(0, n, 2):
a_occ[a[i]][a[i + 1]] += 1
for i in range(0, n, 2):
b_occ[b[i]][b[i + 1]] += 1
if a_occ[0][0] != b_occ[0][0] or a_occ[1][1] != b_occ[1][1]:
print(-1)
return
balanced = a_occ[0][1] == b_occ[1][0]
if not balanced:
zero, one = 0, 0
for i in range(0, n, 2):
if a[i] + a[i + 1] == 1:
zero, one = zero + a[i], one + a[i + 1]
if zero + (a_occ[0][1] - one) == b_occ[1][0]:
balanced = True
rev_a(i + 2)
break
if not balanced:
zero, one = 0, 0
for i in range(0, n, 2):
if b[i] + b[i + 1] == 1:
zero, one = zero + b[i], one + b[i + 1]
if zero + (b_occ[0][1] - one) == a_occ[1][0]:
balanced = True
rev_b(i + 2)
break
for i in range(0, n, 2):
for j in range(i, n, 2):
if a[j] == b[n - i - 1] and a[j + 1] == b[n - i - 2]:
rev_a(j)
rev_a(j + 2)
break
answer()
def main():
for _ in range(int(input())):
solve_case()
main()
``` |
1256_A. Payment Without Change | You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | ```python
from fractions import gcd
from datetime import date, timedelta
from heapq import*
import math
from collections import defaultdict, Counter, deque
import sys
from bisect import *
import itertools
import copy
sys.setrecursionlimit(10 ** 7)
MOD = 10 ** 9 + 7
def main():
q = int(input())
for i in range(q):
a, b, n, s = map(int, input().split())
if s - n * a >= 0:
s -= n * a
else:
s = s - (s//n)*n
if s <= b:
print("YES")
else:
print("NO")
if __name__ == '__main__':
main()
``` |
127_A. Wasted Time | Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000 | ```python
n,k = list(map(int,input().split()))
ans = 0
for i in range(n):
if i==0:
x,y= list(map(int,input().split()))
else:
X,Y = list(map(int,input().split()))
d = ((X-x)**2+(Y-y)**2)**(1/2)
x,y=X,Y
ans += d
ans=((ans*k)/50)
print('%.9f'%ans)
``` |
1323_B. Count Subrectangles | You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image> | ```python
n, m, k = map(int, input().split())
def intervals(arr):
res = []
curr = 0
for e in arr:
if e == 1:
curr += 1
if e == 0:
if curr != 0:
res.append(curr)
curr = 0
if curr != 0:
res.append(curr)
return res
a = intervals(list(map(int, input().split())))
b = intervals(list(map(int, input().split())))
factors = []
for x in range(1, int(k**(1/2))+1):
if k % x == 0:
factors.append(x)
def count(ivs, x):
total = 0
for iv in ivs:
total += max(iv-x+1, 0)
return total
total = 0
for f in factors:
one = count(a, f) * count(b, k//f)
if f == k//f:
total += one
continue
two = count(a, k//f) * count(b, f)
total += one
total += two
print(total)
``` |
1342_A. Road To Zero | You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money. | ```python
def to_list(s):
return list(map(lambda x: int(x), s.split(' ')))
def solve(x,y,a,b):
cost = 0
if b <= 2*a:
min_val = min(x,y)
cost += b*min_val
x -= min_val
y -= min_val
max_val = max(x,y)
cost += max_val*a
else:
cost = (x+y)*a
print(cost)
t = int(input())
for i in range(t):
x, y = to_list(input())
a, b = to_list(input())
solve(x,y,a,b)
``` |
1364_A. XXXXX | Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1. | ```python
t = int(input())
for tc in range(t):
n = int(input())
a = list(map(int, input().split()))
s = set(a)
print(len(s))
``` |
1384_B2. Koa and the Beach (Hard Version) | The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | ```python
import sys
import os,io
input = sys.stdin.readline
# input_all = sys.stdin.read
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# input_all = io.BytesIO(os.read(0,os.fstat(0).st_size)).read
def read_int():
return map(int, input().split())
def read_list():
return list(map(int, input().split()))
def print_list(l):
print(' '.join(map(str,l)))
# import heapq as hq
# import bisect as bs
# from collections import deque as dq
# from collections import defaultdict as dc
# import math
# f = open('test.py')
# input = f.readline
# input_all = f.read
N = int(input())
for _ in range(N):
n, k, l = read_list()
d = read_list()
if max(d)>l:
print('No')
else:
# print(dic)
flag = True
tt = 2*k
now = tt
for i in range(n):
if d[i]+k<=l:
tmp = tt
else:
tm = d[i]+k-l
tmp = [tm,2*k-tm]
# print(tmp)
# tmp = dic[d[i]]
if tmp==tt:
now = tt
elif now==tt:
now = tmp
else:
now = [max(now[0]+1,tmp[0]),min(now[1]+1,tmp[1])]
if now[0]>now[1]:
flag = False
break
if flag:
print('Yes')
else:
print('No')
``` |
1405_D. Tree Tag | Alice and Bob are playing a fun game of tree tag.
The game is played on a tree of n vertices numbered from 1 to n. Recall that a tree on n vertices is an undirected, connected graph with n-1 edges.
Initially, Alice is located at vertex a, and Bob at vertex b. They take turns alternately, and Alice makes the first move. In a move, Alice can jump to a vertex with distance at most da from the current vertex. And in a move, Bob can jump to a vertex with distance at most db from the current vertex. The distance between two vertices is defined as the number of edges on the unique simple path between them. In particular, either player is allowed to stay at the same vertex in a move. Note that when performing a move, a player only occupies the starting and ending vertices of their move, not the vertices between them.
If after at most 10^{100} moves, Alice and Bob occupy the same vertex, then Alice is declared the winner. Otherwise, Bob wins.
Determine the winner if both players play optimally.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains five integers n,a,b,da,db (2≤ n≤ 10^5, 1≤ a,b≤ n, a≠ b, 1≤ da,db≤ n-1) — the number of vertices, Alice's vertex, Bob's vertex, Alice's maximum jumping distance, and Bob's maximum jumping distance, respectively.
The following n-1 lines describe the edges of the tree. The i-th of these lines contains two integers u, v (1≤ u, v≤ n, u≠ v), denoting an edge between vertices u and v. It is guaranteed that these edges form a tree structure.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output a single line containing the winner of the game: "Alice" or "Bob".
Example
Input
4
4 3 2 1 2
1 2
1 3
1 4
6 6 1 2 5
1 2
6 5
2 3
3 4
4 5
9 3 9 2 5
1 2
1 6
1 9
1 3
9 5
7 9
4 8
4 3
11 8 11 3 3
1 2
11 9
4 9
6 5
2 10
3 2
5 9
8 3
7 4
7 10
Output
Alice
Bob
Alice
Alice
Note
In the first test case, Alice can win by moving to vertex 1. Then wherever Bob moves next, Alice will be able to move to the same vertex on the next move.
<image>
In the second test case, Bob has the following strategy to win. Wherever Alice moves, Bob will always move to whichever of the two vertices 1 or 6 is farthest from Alice.
<image> | ```python
for t in range(int(input())):
n, a, b, da, db = map(int, input().split())
graph = [set() for i in range(n + 1)]
for i in range(n - 1):
u, v = map(int, input().split())
graph[u].add(v)
graph[v].add(u)
visited = {1}
last_visits = set()
visit_now = {1}
while len(visited) < n:
while visit_now:
i = visit_now.pop()
last_visits |= graph[i] - visited
visited |= graph[i]
visit_now = last_visits
last_visits = set()
last = visit_now.pop()
visited = {last}
last_visits = set()
visit_now = {last}
N = 1
while len(visited) < n:
N += 1
while visit_now:
i = visit_now.pop()
last_visits |= graph[i] - visited
visited |= graph[i]
visit_now = last_visits
last_visits = set()
visited = {a}
last_visits = set()
visit_now = {a}
dist = 0
while not b in visit_now:
dist += 1
while visit_now:
i = visit_now.pop()
last_visits |= graph[i] - visited
visited |= graph[i]
visit_now = last_visits
last_visits = set()
# print(f'N: {N}, Last: {last}, Dist AB: {dist}')
if db < (2*da + 1):
print('Alice')
elif (2*da + 1) >= N:
print('Alice')
elif dist <= da:
print('Alice')
else:
print('Bob')
``` |
1426_F. Number of Subsequences | You are given a string s consisting of lowercase Latin letters "a", "b" and "c" and question marks "?".
Let the number of question marks in the string s be k. Let's replace each question mark with one of the letters "a", "b" and "c". Here we can obtain all 3^{k} possible strings consisting only of letters "a", "b" and "c". For example, if s = "ac?b?c" then we can obtain the following strings: ["acabac", "acabbc", "acabcc", "acbbac", "acbbbc", "acbbcc", "accbac", "accbbc", "accbcc"].
Your task is to count the total number of subsequences "abc" in all resulting strings. Since the answer can be very large, print it modulo 10^{9} + 7.
A subsequence of the string t is such a sequence that can be derived from the string t after removing some (possibly, zero) number of letters without changing the order of remaining letters. For example, the string "baacbc" contains two subsequences "abc" — a subsequence consisting of letters at positions (2, 5, 6) and a subsequence consisting of letters at positions (3, 5, 6).
Input
The first line of the input contains one integer n (3 ≤ n ≤ 200 000) — the length of s.
The second line of the input contains the string s of length n consisting of lowercase Latin letters "a", "b" and "c" and question marks"?".
Output
Print the total number of subsequences "abc" in all strings you can obtain if you replace all question marks with letters "a", "b" and "c", modulo 10^{9} + 7.
Examples
Input
6
ac?b?c
Output
24
Input
7
???????
Output
2835
Input
9
cccbbbaaa
Output
0
Input
5
a???c
Output
46
Note
In the first example, we can obtain 9 strings:
* "acabac" — there are 2 subsequences "abc",
* "acabbc" — there are 4 subsequences "abc",
* "acabcc" — there are 4 subsequences "abc",
* "acbbac" — there are 2 subsequences "abc",
* "acbbbc" — there are 3 subsequences "abc",
* "acbbcc" — there are 4 subsequences "abc",
* "accbac" — there is 1 subsequence "abc",
* "accbbc" — there are 2 subsequences "abc",
* "accbcc" — there are 2 subsequences "abc".
So, there are 2 + 4 + 4 + 2 + 3 + 4 + 1 + 2 + 2 = 24 subsequences "abc" in total. | ```python
n=int(input())
s=input()
a=[[0,0,0] for i in range(4)]
k=s.count('?')
b=[1]
for i in range(k):
b.append(b[i]*3%1000000007)
for i in range(n):
if s[i]=='a':
a[0][0]+=1
elif s[i]=='b':
a[0][1]+=a[0][0]
a[1][1]+=a[1][0]
elif s[i]=='c':
a[0][2]+=a[0][1]
a[1][2]+=a[1][1]
a[2][2]+=a[2][1]
else:
a[3][2]+=a[2][1]
a[2][2]+=a[1][1]
a[2][1]+=a[1][0]
a[1][2]+=a[0][1]
a[1][1]+=a[0][0]
a[1][0]+=1
z=0
for i in range(min(k+1,4)):
z+=a[i][2]*b[k-i]
print(z%1000000007)
``` |
1473_E. Minimum Path | You are given a weighted undirected connected graph consisting of n vertices and m edges. It is guaranteed that there are no self-loops or multiple edges in the given graph.
Let's define the weight of the path consisting of k edges with indices e_1, e_2, ..., e_k as ∑_{i=1}^{k}{w_{e_i}} - max_{i=1}^{k}{w_{e_i}} + min_{i=1}^{k}{w_{e_i}}, where w_i — weight of the i-th edge in the graph.
Your task is to find the minimum weight of the path from the 1-st vertex to the i-th vertex for each i (2 ≤ i ≤ n).
Input
The first line contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ m ≤ 2 ⋅ 10^5) — the number of vertices and the number of edges in the graph.
Following m lines contains three integers v_i, u_i, w_i (1 ≤ v_i, u_i ≤ n; 1 ≤ w_i ≤ 10^9; v_i ≠ u_i) — endpoints of the i-th edge and its weight respectively.
Output
Print n-1 integers — the minimum weight of the path from 1-st vertex to the i-th vertex for each i (2 ≤ i ≤ n).
Examples
Input
5 4
5 3 4
2 1 1
3 2 2
2 4 2
Output
1 2 2 4
Input
6 8
3 1 1
3 6 2
5 4 2
4 2 2
6 1 1
5 2 1
3 2 3
1 5 4
Output
2 1 4 3 1
Input
7 10
7 5 5
2 3 3
4 7 1
5 3 6
2 7 6
6 2 6
3 7 6
4 2 1
3 1 4
1 7 4
Output
3 4 2 7 7 3 | ```python
import io
import os
# import __pypy__
def dijkstra(*args):
# return dijkstraHeap(*args) # 2979 ms
return dijkstraHeapComparatorWrong(*args) # 2823 ms
# return dijkstraHeapComparator(*args) # 2370 ms
# return dijkstraSegTree(*args) # 2417 ms with inf=float('inf), 2995 ms with inf=-1
# return dijkstraSortedList(*args) # 2995 ms
def dijkstraHeap(source, N, getAdj):
# Heap of (dist, node)
# Use float for dist because max dist for this problem doesn't fit in 32-bit
# Then node has to be a float too, because `(float, int)` will use `W_SpecialisedTupleObject_oo` but we want `W_SpecialisedTupleObject_ff`
from heapq import heappop, heappush
inf = float("inf")
dist = [inf] * N
dist[source] = 0.0
queue = [(0.0, float(source))]
# print(__pypy__.internal_repr(queue[0])) # W_SpecialisedTupleObject_ff
# print(__pypy__.strategy(dist)) # FloatListStrategy
while queue:
d, u = heappop(queue)
u = int(u)
if dist[u] == d:
for v, w in getAdj(u):
cost = d + w
if cost < dist[v]:
dist[v] = cost
heappush(queue, (cost, float(v)))
return dist
def dijkstraHeapComparatorWrong(source, N, getAdj):
# Heap of nodes, sorted with a comparator
# This implementation is actually incorrect but kept for reference since it performs well when using a SPFA-like heuristic
# Note: normal SPFA will TLE since there's a uphack for it in testcase #62
inf = float("inf")
dist = [inf] * N
dist[source] = 0.0
inQueue = [0] * N
inQueue[source] = 1
queue = [source]
# print(__pypy__.strategy(queue)) # IntegerListStrategy
def cmp_lt(u, v):
return dist[u] < dist[v]
heappush, heappop, _ = import_heapq(cmp_lt)
while queue:
u = heappop(queue)
d = dist[u]
inQueue[u] = 0
for v, w in getAdj(u):
cost = d + w
if cost < dist[v]:
dist[v] = cost
if not inQueue[v]:
heappush(queue, v)
inQueue[v] = 1
# If v is already in the queue, we were suppose to bubble it to fix heap invariant
return dist
def dijkstraHeapComparator(source, N, getAdj):
# Same above, except correctly re-bubbling the key after updates
inf = float("inf")
dist = [inf] * N
dist[source] = 0.0
def cmp_lt(u, v):
return dist[u] < dist[v]
heappush, heappop, _siftdown = import_heapq(cmp_lt)
class ListWrapper:
# Exactly like a regular list except with fast .index(x) meant to be used with heapq
# Not general purpose and relies on the exact heapq implementation for correctness (swaps only, added via append, deleted via pop)
def __init__(self, maxN):
self.arr = []
self.loc = [-1] * maxN
def append(self, x):
arr = self.arr
arr.append(x)
self.loc[x] = len(arr) - 1
def pop(self):
ret = self.arr.pop()
self.loc[ret] = -1
return ret
def index(self, x):
return self.loc[x]
def __setitem__(self, i, x):
self.arr[i] = x
self.loc[x] = i
def __getitem__(self, i):
return self.arr[i]
def __len__(self):
return len(self.arr)
queue = ListWrapper(N)
queue.append(source)
# print(__pypy__.strategy(queue.arr)) # IntegerListStrategy
while queue:
u = heappop(queue)
d = dist[u]
for v, w in getAdj(u):
cost = d + w
if cost < dist[v]:
dist[v] = cost
heapIndex = queue.index(v)
if heapIndex == -1:
heappush(queue, v)
else:
_siftdown(queue, 0, heapIndex)
return dist
def dijkstraSegTree(start, n, getAdj):
# From pajenegod: https://github.com/cheran-senthil/PyRival/pull/55
# Modifications:
# Use floats instead of ints for inf/_min
# Fix typo: m -> self.m
# Fix python 3 compatibility: __getitem__
# Cache self.data
# Remove parent pointers
if False:
inf = -1
def _min(a, b):
return a if b == inf or inf != a < b else b
else:
inf = float("inf")
_min = min
class DistanceKeeper:
def __init__(self, n):
m = 1
while m < n:
m *= 2
self.m = m
self.data = 2 * m * [inf]
self.dist = n * [inf]
def __getitem__(self, x):
return self.dist[x]
def __setitem__(self, ind, x):
data = self.data
self.dist[ind] = x
ind += self.m
data[ind] = x
ind >>= 1
while ind:
data[ind] = _min(data[2 * ind], data[2 * ind + 1])
ind >>= 1
def trav(self):
m = self.m
data = self.data
dist = self.dist
while data[1] != inf:
x = data[1]
ind = 1
while ind < m:
ind = 2 * ind + (data[2 * ind] != x)
ind -= m
self[ind] = inf
dist[ind] = x
yield ind
# P = [-1] * n
D = DistanceKeeper(n)
D[start] = 0.0
for node in D.trav():
for nei, weight in getAdj(node):
new_dist = D[node] + weight
if D[nei] == inf or new_dist < D[nei]:
D[nei] = new_dist
# P[nei] = node
# print(__pypy__.strategy(D.dist))
# print(__pypy__.strategy(D.data))
return D.dist
def dijkstraSortedList(source, N, getAdj):
# Just for completeness
# COPY AND PASTE from https://github.com/cheran-senthil/PyRival/blob/master/pyrival/data_structures/SortedList.py
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [
values[i : i + _load] for i in range(0, _len, _load)
]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError("{0!r} not in list".format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (
value for _list in reversed(self._lists) for value in reversed(_list)
)
def __repr__(self):
"""Return string representation of sorted list."""
return "SortedList({0})".format(list(self))
# END COPY AND PASTE #####################################
inf = float("inf")
dist = [inf] * N
dist[source] = 0.0
queue = SortedList([(0.0, float(source))])
while queue:
negD, u = queue.pop(-1)
d = -negD
u = int(u)
for v, w in getAdj(u):
prevCost = dist[v]
cost = d + w
if cost < prevCost:
if prevCost != inf:
queue.discard((-prevCost, float(v)))
dist[v] = cost
queue.add((-cost, float(v)))
return dist
def import_heapq(cmp_lt):
# Python 2 has a heapq.cmp_lt but python 3 removed it
# Add it back for pypy3 submissions
import sys
if sys.version_info < (3,):
# Python 2
import heapq
from heapq import heappush, heappop, _siftdown
heapq.cmp_lt = cmp_lt
else:
# Python 3
# COPY AND PASTE python 2.7 heapq from https://github.com/python/cpython/blob/2.7/Lib/heapq.py
def heappush(heap, item):
"""Push item onto heap, maintaining the heap invariant."""
heap.append(item)
_siftdown(heap, 0, len(heap) - 1)
def heappop(heap):
"""Pop the smallest item off the heap, maintaining the heap invariant."""
lastelt = heap.pop() # raises appropriate IndexError if heap is empty
if heap:
returnitem = heap[0]
heap[0] = lastelt
_siftup(heap, 0)
return returnitem
return lastelt
def _siftdown(heap, startpos, pos):
newitem = heap[pos]
# Follow the path to the root, moving parents down until finding a place
# newitem fits.
while pos > startpos:
parentpos = (pos - 1) >> 1
parent = heap[parentpos]
if cmp_lt(newitem, parent):
heap[pos] = parent
pos = parentpos
continue
break
heap[pos] = newitem
def _siftup(heap, pos):
endpos = len(heap)
startpos = pos
newitem = heap[pos]
# Bubble up the smaller child until hitting a leaf.
childpos = 2 * pos + 1 # leftmost child position
while childpos < endpos:
# Set childpos to index of smaller child.
rightpos = childpos + 1
if rightpos < endpos and not cmp_lt(heap[childpos], heap[rightpos]):
childpos = rightpos
# Move the smaller child up.
heap[pos] = heap[childpos]
pos = childpos
childpos = 2 * pos + 1
# The leaf at pos is empty now. Put newitem there, and bubble it up
# to its final resting place (by sifting its parents down).
heap[pos] = newitem
_siftdown(heap, startpos, pos)
# END COPY AND PASTE ###############################
return heappush, heappop, _siftdown
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
N, M = [int(x) for x in input().split()]
graph = [[] for i in range(N)]
for i in range(M):
u, v, w = [int(x) for x in input().split()]
u -= 1
v -= 1
graph[u].append((v, w))
graph[v].append((u, w))
# Want shortest path except one edge is worth 0 and one edge is worth 2x
# Track this with 2 bits of extra state
def getAdj(node):
u = node >> 2
state = node & 3
for v, w in graph[u]:
vBase = v << 2
# Regular edge
yield vBase | state, w
if not state & 1:
# Take max edge, worth 0
yield vBase | state | 1, 0
if not state & 2:
# Take min edge, worth double
yield vBase | state | 2, 2 * w
if not state & 3:
# Take both min and max edge, worth normal
yield vBase | state | 3, w
dist = dijkstra(0, 4 * N, getAdj)
print(" ".join(str(int(dist[(u << 2) | 3])) for u in range(1, N)))
``` |
1521_E. Nastia and a Beautiful Matrix | You like numbers, don't you? Nastia has a lot of numbers and she wants to share them with you! Isn't it amazing?
Let a_i be how many numbers i (1 ≤ i ≤ k) you have.
An n × n matrix is called beautiful if it contains all the numbers you have, and for each 2 × 2 submatrix of the original matrix is satisfied:
1. The number of occupied cells doesn't exceed 3;
2. The numbers on each diagonal are distinct.
Make a beautiful matrix of minimum size.
Input
The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.
The first line of each test case contains 2 integers m and k (1 ≤ m, k ≤ 10^5) — how many numbers Nastia gave you and the length of the array a, respectively.
The second line of each test case contains k integers a_1, a_2, …, a_{k} (0 ≤ a_i ≤ m, a_1 + a_2 + … + a_{k} = m), where a_i is how many numbers i you have.
It's guaranteed that the sum of m and k in one test doesn't exceed 2 ⋅ 10^5.
Output
For each t test case print a single integer n — the size of the beautiful matrix.
In the next n lines print n integers b_{i, j} (0 ≤ b_{i, j} ≤ k; if position is empty, print b_{i, j} = 0) — the beautiful matrix b you made up.
Example
Input
2
3 4
2 0 0 1
15 4
2 4 8 1
Output
2
4 1
0 1
5
3 0 0 2 2
3 2 3 3 0
0 1 0 4 0
3 0 0 0 0
2 1 3 3 3
Note
Note that 0 in this problem represents a blank, not a number.
Examples of possible answers for the first test case:
\begin{array}{cc} 1 & 1 \\\ 4 & 0 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 1 & 4 \\\ 1 & 0 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 4 & 0 \\\ 1 & 1 \\\ \end{array}
Examples of not beautiful matrices for the first test case:
\begin{array}{cc} 1 & 0 \\\ 4 & 1 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 4 & 1 \\\ 7 & 1 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 1 & 0 \\\ 4 & 0 \\\ \end{array}
The example of the not beautiful matrix for the second test case:
\begin{array}{cc} 3 & 4 & 0 & 2 & 2 \\\ 3 & 2 & 3 & 3 & 0 \\\ 0 & 1 & 0 & 0 & 0 \\\ 3 & 0 & 0 & 0 & 0 \\\ 2 & 1 & 3 & 3 & 3 \\\ \end{array}
Everything is okay, except the left-top submatrix contains 4 numbers. | ```python
def cheak(x):
return x**2-(x//2)**2>=m and x*(x//2+(1 if x%2!=0 else 0))>=mx
for test in range(int(input())):
m,k=(int(i) for i in input().split())
a=[int(i) for i in input().split()]
mx=max(a)
z=0;y=m*4
while z!=y:
x=(z+y)//2
if cheak(x):
y=x
else:
z=x+1
else:
x=z
a=sorted(list(map(list,zip(a,range(1,len(a)+1)))))
def get():
i=len(a)
while i!=0:
i-=1
while a[i][0]>0:
a[i][0]-=1
yield a[i][1]
yield 0
mt=[[0 for i in range(x)] for j in range(x)]
t=1
it=get()
for i in range(0,x,2):
if t==0:break
for j in range(1,x,2):
t=next(it)
if t:mt[i][j]=t
else:break
for i in range(0,x,2):
if t==0:break
for j in range(0,x,2):
t=next(it)
if t:mt[i][j]=t
else:break
for i in range(1,x,2):
if t==0:break
for j in range(0,x,2):
t=next(it)
if t:mt[i][j]=t
else:break
print(len(mt))
for i in mt:
print(*i)
``` |
1550_E. Stringforces | You are given a string s of length n. Each character is either one of the first k lowercase Latin letters or a question mark.
You are asked to replace every question mark with one of the first k lowercase Latin letters in such a way that the following value is maximized.
Let f_i be the maximum length substring of string s, which consists entirely of the i-th Latin letter. A substring of a string is a contiguous subsequence of that string. If the i-th letter doesn't appear in a string, then f_i is equal to 0.
The value of a string s is the minimum value among f_i for all i from 1 to k.
What is the maximum value the string can have?
Input
The first line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 17) — the length of the string and the number of first Latin letters used.
The second line contains a string s, consisting of n characters. Each character is either one of the first k lowercase Latin letters or a question mark.
Output
Print a single integer — the maximum value of the string after every question mark is replaced with one of the first k lowercase Latin letters.
Examples
Input
10 2
a??ab????b
Output
4
Input
9 4
?????????
Output
2
Input
2 3
??
Output
0
Input
15 3
??b?babbc??b?aa
Output
3
Input
4 4
cabd
Output
1
Note
In the first example the question marks can be replaced in the following way: "aaaababbbb". f_1 = 4, f_2 = 4, thus the answer is 4. Replacing it like this is also possible: "aaaabbbbbb". That way f_1 = 4, f_2 = 6, however, the minimum of them is still 4.
In the second example one of the possible strings is "aabbccdda".
In the third example at least one letter won't appear in the string, thus, the minimum of values f_i is always 0. | ```python
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
s = input()[:-1]
left, right = 0, n
while left < right:
mid = right - (right - left) // 2
A = [[0] * (n + 2) for _ in range(k)]
for c in range(k):
A[c][n] = A[c][n + 1] = n + 1
L = 0
for i in range(n - 1, -1, -1):
if s[i] == '?' or ord(s[i]) - ord('a') == c: L += 1
else: L = 0
A[c][i] = i + mid if L >= mid else A[c][i + 1]
dp = [n + 1] * (1 << k)
dp[0] = 0
for mask in range(1 << k):
for i in range(k):
if mask >> k & 1: continue
t = mask | 1 << i
dp[t] = min(dp[t], A[i][dp[mask]])
if dp[-1] <= n: left = mid
else: right = mid - 1
print(left)
``` |
177_B1. Rectangular Game | The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has n pebbles. He arranges them in a equal rows, each row has b pebbles (a > 1). Note that the Beaver must use all the pebbles he has, i. e. n = a·b.
<image> 10 pebbles are arranged in two rows, each row has 5 pebbles
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, b pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of a and b) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers c1, ..., ck, where:
* c1 = n
* ci + 1 is the number of pebbles that the Beaver ends up with after the i-th move, that is, the number of pebbles in a row after some arrangement of ci pebbles (1 ≤ i < k). Note that ci > ci + 1.
* ck = 1
The result of the game is the sum of numbers ci. You are given n. Find the maximum possible result of the game.
Input
The single line of the input contains a single integer n — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
* 2 ≤ n ≤ 50
The input limitations for getting 100 points are:
* 2 ≤ n ≤ 109
Output
Print a single number — the maximum possible result of the game.
Examples
Input
10
Output
16
Input
8
Output
15
Note
Consider the first example (c1 = 10). The possible options for the game development are:
* Arrange the pebbles in 10 rows, one pebble per row. Then c2 = 1, and the game ends after the first move with the result of 11.
* Arrange the pebbles in 5 rows, two pebbles per row. Then c2 = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. c3 = 1, and the game ends with the result of 13.
* Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to c2 = 5, c3 = 1, and the game ends with the result of 16 — the maximum possible result. | ```python
n=int(input())
r=n+1
i=2
s=n**0.5
while i<=s:
if n%i==0:
r+=n//i
n//=i
s=n**0.5
i=1
i+=1
print(r)
``` |
221_A. Little Elephant and Function | The Little Elephant enjoys recursive functions.
This time he enjoys the sorting function. Let a is a permutation of an integers from 1 to n, inclusive, and ai denotes the i-th element of the permutation. The Little Elephant's recursive function f(x), that sorts the first x permutation's elements, works as follows:
* If x = 1, exit the function.
* Otherwise, call f(x - 1), and then make swap(ax - 1, ax) (swap the x-th and (x - 1)-th elements of a).
The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to n, such that after performing the Little Elephant's function (that is call f(n)), the permutation will be sorted in ascending order.
Input
A single line contains integer n (1 ≤ n ≤ 1000) — the size of permutation.
Output
In a single line print n distinct integers from 1 to n — the required permutation. Numbers in a line should be separated by spaces.
It is guaranteed that the answer exists.
Examples
Input
1
Output
1
Input
2
Output
2 1 | ```python
n=int(input())
l=list(range(1,n+1))
l=sorted(l)
l.insert(0,l[-1])
l.pop()
print(*l)
``` |
245_B. Internet Address | Vasya is an active Internet user. One day he came across an Internet resource he liked, so he wrote its address in the notebook. We know that the address of the written resource has format:
<protocol>://<domain>.ru[/<context>]
where:
* <protocol> can equal either "http" (without the quotes) or "ftp" (without the quotes),
* <domain> is a non-empty string, consisting of lowercase English letters,
* the /<context> part may not be present. If it is present, then <context> is a non-empty string, consisting of lowercase English letters.
If string <context> isn't present in the address, then the additional character "/" isn't written. Thus, the address has either two characters "/" (the ones that go before the domain), or three (an extra one in front of the context).
When the boy came home, he found out that the address he wrote in his notebook had no punctuation marks. Vasya must have been in a lot of hurry and didn't write characters ":", "/", ".".
Help Vasya to restore the possible address of the recorded Internet resource.
Input
The first line contains a non-empty string that Vasya wrote out in his notebook. This line consists of lowercase English letters only.
It is guaranteed that the given string contains at most 50 letters. It is guaranteed that the given string can be obtained from some correct Internet resource address, described above.
Output
Print a single line — the address of the Internet resource that Vasya liked. If there are several addresses that meet the problem limitations, you are allowed to print any of them.
Examples
Input
httpsunrux
Output
http://sun.ru/x
Input
ftphttprururu
Output
ftp://http.ru/ruru
Note
In the second sample there are two more possible answers: "ftp://httpruru.ru" and "ftp://httpru.ru/ru". | ```python
s = input()
if s[:3] == "ftp":
pre = "ftp://"
s = s[3:]
elif s[:4] == "http":
pre = "http://"
s = s[4:]
index = s[1:].index("ru")+1
if index == len(s)-2:
s = pre+s[:index]+"."+s[index:]
else:
s = pre+s[:index]+"."+s[index:index+2]+"/"+s[index+2:]
print(s)
``` |
26_E. Multithreading | 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. | ```python
import sys
from array import array # noqa: F401
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()
``` |
340_E. Iahub and Permutations | Iahub is so happy about inventing bubble sort graphs that he's staying all day long at the office and writing permutations. Iahubina is angry that she is no more important for Iahub. When Iahub goes away, Iahubina comes to his office and sabotage his research work.
The girl finds an important permutation for the research. The permutation contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n). She replaces some of permutation elements with -1 value as a revenge.
When Iahub finds out his important permutation is broken, he tries to recover it. The only thing he remembers about the permutation is it didn't have any fixed point. A fixed point for a permutation is an element ak which has value equal to k (ak = k). Your job is to proof to Iahub that trying to recover it is not a good idea. Output the number of permutations which could be originally Iahub's important permutation, modulo 1000000007 (109 + 7).
Input
The first line contains integer n (2 ≤ n ≤ 2000). On the second line, there are n integers, representing Iahub's important permutation after Iahubina replaces some values with -1.
It's guaranteed that there are no fixed points in the given permutation. Also, the given sequence contains at least two numbers -1 and each positive number occurs in the sequence at most once. It's guaranteed that there is at least one suitable permutation.
Output
Output a single integer, the number of ways Iahub could recover his permutation, modulo 1000000007 (109 + 7).
Examples
Input
5
-1 -1 4 3 -1
Output
2
Note
For the first test example there are two permutations with no fixed points are [2, 5, 4, 3, 1] and [5, 1, 4, 3, 2]. Any other permutation would have at least one fixed point. | ```python
#lahub and Permutations
import sys
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0
mod = 10**9+7
def pow(n,p,mod=mod): #繰り返し二乗法(nのp乗)
res = 1
while p > 0:
if p % 2 == 0:
n = n ** 2 % mod
p //= 2
else:
res = res * n % mod
p -= 1
return res % mod
def factrial_memo(n=10**5,mod=mod):
fact = [1, 1]
for i in range(2, n + 1):
fact.append((fact[-1] * i) % mod)
return fact
fact = factrial_memo()
def permutation(n,r): #nPr
return fact[n]*pow(fact[n-r],mod-2)%mod
def combination(n,r): #nCr
return permutation(n,r)*pow(fact[r],mod-2)%mod
#return fact[n]*pow(fact[n-r],mod-2)*pow(fact[r],mod-2)
def homogeneous(n,r): #nHr
return combination(n+r-1,r)%mod
#return fact[n+m-1]*pow(fact[n-1],mod-2)*pow(fact[r],mod-2)
n = int(readline())
lst1 = list(map(int,readline().split()))
ct = 0
lst2 = [0]*2_001
lst3 = [0]*2_001
for i in range(n):
if i+1 == lst1[i]:
print(0)
exit()
if lst1[i] == -1:
ct += 1
lst3[i+1] = 1
else:
lst2[lst1[i]] = 1
ct2 = 0
for i in range(1,n+1):
if lst3[i] == 1 and lst2[i] == 0:
ct2 += 1
#lst2:その場所が埋まってないindex
#lst3:その数字が使われてるindex
#何個入れちゃいけない位置に入れるかで数え上げる
#入れちゃいけないものはct2個って、
#ct-ct2個のものはどこに入れても要件を満たす
ans = 0
for i in range(ct2+1):
ans += pow(-1,i)*combination(ct2,i)*fact[ct-i]
ans %= mod
print(ans)
``` |
388_A. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | ```python
I=lambda:list(map(int,input().split()))
n,=I()
l=I()
l.sort()
ans=0
i=0
k=1
while i<n:
if l[i]<i//k:
k+=1
i+=1
print(k)
``` |
408_D. Long Path | One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | ```python
n = int(input())
a = list(map(int,input().split()))
values = [0 for k in range(n)]
for k in range(n) :
back = a[k]
if a[k] == k+1 :
values[k] = 2
else :
values[k] += 2
for t in range(a[k]-1,k) :
values [k] += values[t]
values[k] %= int(1e9 +7)
print(int(sum(values)%(1e9 + 7)))
``` |
459_A. Pashmak and Garden | Pashmak has fallen in love with an attractive girl called Parmida since one year ago...
Today, Pashmak set up a meeting with his partner in a romantic garden. Unfortunately, Pashmak has forgotten where the garden is. But he remembers that the garden looks like a square with sides parallel to the coordinate axes. He also remembers that there is exactly one tree on each vertex of the square. Now, Pashmak knows the position of only two of the trees. Help him to find the position of two remaining ones.
Input
The first line contains four space-separated x1, y1, x2, y2 ( - 100 ≤ x1, y1, x2, y2 ≤ 100) integers, where x1 and y1 are coordinates of the first tree and x2 and y2 are coordinates of the second tree. It's guaranteed that the given points are distinct.
Output
If there is no solution to the problem, print -1. Otherwise print four space-separated integers x3, y3, x4, y4 that correspond to the coordinates of the two other trees. If there are several solutions you can output any of them.
Note that x3, y3, x4, y4 must be in the range ( - 1000 ≤ x3, y3, x4, y4 ≤ 1000).
Examples
Input
0 0 0 1
Output
1 0 1 1
Input
0 0 1 1
Output
0 1 1 0
Input
0 0 1 2
Output
-1 | ```python
#!/usr/bin/env python3
import atexit
import io
import sys
import bisect
_I_B = sys.stdin.read().splitlines()
input = iter(_I_B).__next__
_O_B = io.StringIO()
sys.stdout = _O_B
@atexit.register
def write():
sys.__stdout__.write(_O_B.getvalue())
def main():
x1,y1,x2,y2=map(int,input().split())
if x1==x2:
d=abs(y1-y2)
print(x1+d,y1,x2+d,y2)
elif y1==y2:
d=abs(x1-x2)
print(x1,y1+d,x2,y2+d)
elif abs(x1-x2)==abs(y1-y2):
d=abs(x1-x2)
if x2<x1:
x1,x2=x2,x1
y1,y2=y2,y1
if y1>y2:
print(x1,y1-d,x2,y2+d)
else:
print(x1,y1+d,x2,y2-d)
else:
print(-1)
if __name__=='__main__':
main()
``` |
47_E. Cannon | Bertown is under siege! The attackers have blocked all the ways out and their cannon is bombarding the city. Fortunately, Berland intelligence managed to intercept the enemies' shooting plan. Let's introduce the Cartesian system of coordinates, the origin of which coincides with the cannon's position, the Ox axis is directed rightwards in the city's direction, the Oy axis is directed upwards (to the sky). The cannon will make n more shots. The cannon balls' initial speeds are the same in all the shots and are equal to V, so that every shot is characterized by only one number alphai which represents the angle at which the cannon fires. Due to the cannon's technical peculiarities this angle does not exceed 45 angles (π / 4). We disregard the cannon sizes and consider the firing made from the point (0, 0).
The balls fly according to the known physical laws of a body thrown towards the horizon at an angle:
vx(t) = V·cos(alpha) vy(t) = V·sin(alpha) – g·t x(t) = V·cos(alpha)·t y(t) = V·sin(alpha)·t – g·t2 / 2
Think of the acceleration of gravity g as equal to 9.8.
Bertown defends m walls. The i-th wall is represented as a vertical segment (xi, 0) - (xi, yi). When a ball hits a wall, it gets stuck in it and doesn't fly on. If a ball doesn't hit any wall it falls on the ground (y = 0) and stops. If the ball exactly hits the point (xi, yi), it is considered stuck.
Your task is to find for each ball the coordinates of the point where it will be located in the end.
Input
The first line contains integers n and V (1 ≤ n ≤ 104, 1 ≤ V ≤ 1000) which represent the number of shots and the initial speed of every ball. The second line contains n space-separated real numbers alphai (0 < alphai < π / 4) which represent the angles in radians at which the cannon will fire. The third line contains integer m (1 ≤ m ≤ 105) which represents the number of walls. Then follow m lines, each containing two real numbers xi and yi (1 ≤ xi ≤ 1000, 0 ≤ yi ≤ 1000) which represent the wall’s coordinates. All the real numbers have no more than 4 decimal digits. The walls may partially overlap or even coincide.
Output
Print n lines containing two real numbers each — calculate for every ball the coordinates of its landing point. Your answer should have the relative or absolute error less than 10 - 4.
Examples
Input
2 10
0.7853
0.3
3
5.0 5.0
4.0 2.4
6.0 1.9
Output
5.000000000 2.549499369
4.000000000 0.378324889
Input
2 10
0.7853
0.3
2
4.0 2.4
6.0 1.9
Output
10.204081436 0.000000000
4.000000000 0.378324889 | ```python
import sys
from array import array # noqa: F401
from math import pi, sin, cos
from bisect import bisect_left
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, v = map(int, input().split())
v = float(v)
alpha = [float(input()) for _ in range(n)]
m = int(input())
wall = sorted(tuple(map(float, input().split())) for _ in range(m)) + [(1e9, 1e9)]
max_angle = pi / 4
eps = 1e-9
a = [0.0] * m + [max_angle + eps]
for i in range(m):
ng_angle, ok_angle = 0.0, max_angle + eps
for _ in range(50):
mid_angle = (ok_angle + ng_angle) / 2
t = wall[i][0] / (v * cos(mid_angle))
if (v * sin(mid_angle) * t - 9.8 * t**2 / 2) - eps <= wall[i][1]:
ng_angle = mid_angle
else:
ok_angle = mid_angle
a[i] = max(a[i], ng_angle)
a[i + 1] = max(a[i], a[i + 1])
ans = [''] * n
for i in range(n):
wi = bisect_left(a, alpha[i])
ok, ng = 0.0, 1e7
sin_a = sin(alpha[i])
for _ in range(50):
t = (ok + ng) / 2
if v * sin_a * t - 9.8 * t**2 / 2 >= 0.0:
ok = t
else:
ng = t
x = v * cos(alpha[i]) * ok
if x < wall[wi][0]:
ans[i] = f'{x:.8f} {0:.8f}'
else:
ok, ng = 0.0, 1e7
cos_a = cos(alpha[i])
for _ in range(50):
t = (ok + ng) / 2
if v * cos_a * t <= wall[wi][0]:
ok = t
else:
ng = t
y = v * sin_a * ok - 9.8 * ok**2 / 2
ans[i] = f'{wall[wi][0]:.8f} {y:.8f}'
print('\n'.join(ans))
``` |
554_A. Kyoya and Photobooks | Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input
The first line of input will be a single string s (1 ≤ |s| ≤ 20). String s consists only of lowercase English letters.
Output
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Examples
Input
a
Output
51
Input
hi
Output
76
Note
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | ```python
s = input()
l = list(s)
a = 'abcdefghijklmnopqrstuvwxyz'
for i in range(len(s)+1):
for j in a:
l.insert(i, j)
print(len(l)-2*len(s))
``` |
624_B. Making a String | You are given an alphabet consisting of n letters, your task is to make a string of the maximum possible length so that the following conditions are satisfied:
* the i-th letter occurs in the string no more than ai times;
* the number of occurrences of each letter in the string must be distinct for all the letters that occurred in the string at least once.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 26) — the number of letters in the alphabet.
The next line contains n integers ai (1 ≤ ai ≤ 109) — i-th of these integers gives the limitation on the number of occurrences of the i-th character in the string.
Output
Print a single integer — the maximum length of the string that meets all the requirements.
Examples
Input
3
2 5 5
Output
11
Input
3
1 1 2
Output
3
Note
For convenience let's consider an alphabet consisting of three letters: "a", "b", "c". In the first sample, some of the optimal strings are: "cccaabbccbb", "aabcbcbcbcb". In the second sample some of the optimal strings are: "acc", "cbc". | ```python
n = int(input())
m = list(map(int, input().split()))
m.sort(reverse = True)
ans = m[0]
last = m[0]
for i in range(1, len(m)):
last = max(min(last - 1, m[i]), 0)
ans += last
print(ans)
``` |
672_D. Robin Hood | We all know the impressive story of Robin Hood. Robin Hood uses his archery skills and his wits to steal the money from rich, and return it to the poor.
There are n citizens in Kekoland, each person has ci coins. Each day, Robin Hood will take exactly 1 coin from the richest person in the city and he will give it to the poorest person (poorest person right after taking richest's 1 coin). In case the choice is not unique, he will select one among them at random. Sadly, Robin Hood is old and want to retire in k days. He decided to spend these last days with helping poor people.
After taking his money are taken by Robin Hood richest person may become poorest person as well, and it might even happen that Robin Hood will give his money back. For example if all people have same number of coins, then next day they will have same number of coins too.
Your task is to find the difference between richest and poorest persons wealth after k days. Note that the choosing at random among richest and poorest doesn't affect the answer.
Input
The first line of the input contains two integers n and k (1 ≤ n ≤ 500 000, 0 ≤ k ≤ 109) — the number of citizens in Kekoland and the number of days left till Robin Hood's retirement.
The second line contains n integers, the i-th of them is ci (1 ≤ ci ≤ 109) — initial wealth of the i-th person.
Output
Print a single line containing the difference between richest and poorest peoples wealth.
Examples
Input
4 1
1 1 4 2
Output
2
Input
3 1
2 2 2
Output
0
Note
Lets look at how wealth changes through day in the first sample.
1. [1, 1, 4, 2]
2. [2, 1, 3, 2] or [1, 2, 3, 2]
So the answer is 3 - 1 = 2
In second sample wealth will remain the same for each person. | ```python
import sys
sys.stderr = sys.stdout
def hood(n, k, C):
C.sort()
m, r = divmod(sum(C), n)
m1 = (m + 1) if r else m
c_lo = C[0]
k_lo = k
for i, c in enumerate(C):
if c_lo == m:
break
c_m = min(c, m)
dc = c_m - c_lo
dk = i * dc
if k_lo >= dk:
k_lo -= dk
c_lo = c_m
else:
dc = k_lo // i
c_lo += dc
break
c_hi = C[-1]
k_hi = k
for i, c in enumerate(reversed(C)):
if c_hi == m1:
break
c_m1 = max(c, m1)
dc = c_hi - c_m1
dk = i * dc
if k_hi >= dk:
k_hi -= dk
c_hi = c_m1
else:
dc = k_hi // i
c_hi -= dc
break
return c_hi - c_lo
def main():
n, k = readinti()
C = readintl()
print(hood(n, k, C))
##########
def readint():
return int(input())
def readinti():
return map(int, input().split())
def readintt():
return tuple(readinti())
def readintl():
return list(readinti())
def readinttl(k):
return [readintt() for _ in range(k)]
def readintll(k):
return [readintl() for _ in range(k)]
def log(*args, **kwargs):
print(*args, **kwargs, file=sys.__stderr__)
if __name__ == '__main__':
main()
``` |
763_B. Timofey and rectangles | One of Timofey's birthday presents is a colourbook in a shape of an infinite plane. On the plane n rectangles with sides parallel to coordinate axes are situated. All sides of the rectangles have odd length. Rectangles cannot intersect, but they can touch each other.
Help Timofey to color his rectangles in 4 different colors in such a way that every two rectangles touching each other by side would have different color, or determine that it is impossible.
Two rectangles intersect if their intersection has positive area. Two rectangles touch by sides if there is a pair of sides such that their intersection has non-zero length
<image> The picture corresponds to the first example
Input
The first line contains single integer n (1 ≤ n ≤ 5·105) — the number of rectangles.
n lines follow. The i-th of these lines contains four integers x1, y1, x2 and y2 ( - 109 ≤ x1 < x2 ≤ 109, - 109 ≤ y1 < y2 ≤ 109), that means that points (x1, y1) and (x2, y2) are the coordinates of two opposite corners of the i-th rectangle.
It is guaranteed, that all sides of the rectangles have odd lengths and rectangles don't intersect each other.
Output
Print "NO" in the only line if it is impossible to color the rectangles in 4 different colors in such a way that every two rectangles touching each other by side would have different color.
Otherwise, print "YES" in the first line. Then print n lines, in the i-th of them print single integer ci (1 ≤ ci ≤ 4) — the color of i-th rectangle.
Example
Input
8
0 0 5 3
2 -1 5 0
-3 -4 2 -1
-1 -1 2 0
-3 0 0 5
5 2 10 3
7 -3 10 2
4 -2 7 -1
Output
YES
1
2
2
3
2
2
4
1 | ```python
n = int(input())
coordinates = []
for i in range(n):
coordinates.append([int(x) % 2 for x in input().split()])
print('YES')
for coordinate in coordinates:
x1, y1, x2, y2 = coordinate
print(2*x2 + y2 + 1)
``` |
808_C. Tea Party | Polycarp invited all his friends to the tea party to celebrate the holiday. He has n cups, one for each of his n friends, with volumes a1, a2, ..., an. His teapot stores w milliliters of tea (w ≤ a1 + a2 + ... + an). Polycarp wants to pour tea in cups in such a way that:
* Every cup will contain tea for at least half of its volume
* Every cup will contain integer number of milliliters of tea
* All the tea from the teapot will be poured into cups
* All friends will be satisfied.
Friend with cup i won't be satisfied, if there exists such cup j that cup i contains less tea than cup j but ai > aj.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input
The first line contains two integer numbers n and w (1 ≤ n ≤ 100, <image>).
The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 100).
Output
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Examples
Input
2 10
8 7
Output
6 4
Input
4 4
1 1 1 1
Output
1 1 1 1
Input
3 10
9 8 10
Output
-1
Note
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available. | ```python
import math
import re
n, w = map(int, input().split())
a = list(map(int, input().split()))
a1 = [math.ceil(el/2) for el in a]
s = w - sum(a1)
if s < 0:
print(-1)
elif s == 0:
print(' '.join(map(str, a1)))
else:
for i in range(n):
if a[i] == 1:
a1[i] = 0
a[i] = 0
a2 = sorted(a, reverse=True)
#print(a)
while s > 0:
i = a.index(a2[0])
#print(a1)
#print(i)
if a[i] - a1[i] < s:
s -= (a[i] - a1[i])
a1[i] = a[i]
else:
a1[i] += s
s = 0
del (a2[0])
a[i] = -1
for i in range(n):
if a1[i] == 0:
a1[i] = 1
print(' '.join(map(str, a1)))
# n, m = map(int, input().split())
# w = []
# c = []
# for i in range(n):
# x, y = map(int, input().split())
# w.append(x)
# c.append(y)
#
# A = [[0] * (m + 1) for i in range(n)]
#
#
# for k in range(n):
# for s in range(1, m + 1):
# if s >= w[k]:
# A[k][s] = max(A[k - 1][s], A[k - 1][s - w[k]] + c[k])
# else:
# A[k][s] = A[k-1][s]
#
# print(A[n - 1][m])
# arr = list(map(int, input().split()))
# res = 0
# a = {math.pow(2, i) for i in range(35)}
# for i in range(n-1):
# for j in range(i+1,n):
# if arr[i] + arr[j] % 2 % 2 % 2 % 2 % 2 in a:
# res += 1
#
# print(res)
# arr = list(map(int, input().split()))
# m = int(input())
# spis = list(map(int, input().split()))
#
# arr1 = sorted(arr, reverse=True)
# a = [n - arr1.index(arr[el - 1]) for el in spis]
# print(' '.join(map(str, a)))
``` |
832_D. Misha, Grisha and Underground | Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other.
The boys decided to have fun and came up with a plan. Namely, in some day in the morning Misha will ride the underground from station s to station f by the shortest path, and will draw with aerosol an ugly text "Misha was here" on every station he will pass through (including s and f). After that on the same day at evening Grisha will ride from station t to station f by the shortest path and will count stations with Misha's text. After that at night the underground workers will wash the texts out, because the underground should be clean.
The boys have already chosen three stations a, b and c for each of several following days, one of them should be station s on that day, another should be station f, and the remaining should be station t. They became interested how they should choose these stations s, f, t so that the number Grisha will count is as large as possible. They asked you for help.
Input
The first line contains two integers n and q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of stations and the number of days.
The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi ≤ n). The integer pi means that there is a route between stations pi and i. It is guaranteed that it's possible to reach every station from any other.
The next q lines contains three integers a, b and c each (1 ≤ a, b, c ≤ n) — the ids of stations chosen by boys for some day. Note that some of these ids could be same.
Output
Print q lines. In the i-th of these lines print the maximum possible number Grisha can get counting when the stations s, t and f are chosen optimally from the three stations on the i-th day.
Examples
Input
3 2
1 1
1 2 3
2 3 3
Output
2
3
Input
4 1
1 2 3
1 2 3
Output
2
Note
In the first example on the first day if s = 1, f = 2, t = 3, Misha would go on the route 1 <image> 2, and Grisha would go on the route 3 <image> 1 <image> 2. He would see the text at the stations 1 and 2. On the second day, if s = 3, f = 2, t = 3, both boys would go on the route 3 <image> 1 <image> 2. Grisha would see the text at 3 stations.
In the second examle if s = 1, f = 3, t = 2, Misha would go on the route 1 <image> 2 <image> 3, and Grisha would go on the route 2 <image> 3 and would see the text at both stations. | ```python
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
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")
sys.setrecursionlimit(3*10**5)
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
MA=10**5+1
level=20
tree=[[] for i in range(MA)]
depth=[0 for i in range(MA)]
parent=[[0 for i in range(level)] for j in range(MA)]
@bootstrap
def dfs(cur,prev):
depth[cur] = depth[prev] + 1
parent[cur][0] = prev
for i in range(len(tree[cur])):
if (tree[cur][i] != prev):
yield dfs(tree[cur][i], cur)
yield
def precomputeSparseMatrix(n):
for i in range(1,level):
for node in range(1,n+1):
if (parent[node][i-1] != -1):
parent[node][i] =parent[parent[node][i-1]][i-1]
def lca(u,v):
if (depth[v] < depth[u]):
u,v=v,u
diff = depth[v] - depth[u]
for i in range(level):
if ((diff >> i) & 1):
v = parent[v][i]
if (u == v):
return u
i=level-1
while(i>=0):
if (parent[u][i] != parent[v][i]):
u = parent[u][i]
v = parent[v][i]
i+=-1
return parent[u][0]
def add(a,b):
tree[a].append(b)
tree[b].append(a)
def res(s,t,f):
p=lca(t,f)
q=lca(s,f)
if p==q:
r=depth[lca(s,t)]-depth[p]
return min(depth[p] + depth[f] - 2 * depth[p] + 1+r, depth[q] + depth[f] - 2 * depth[q] + 1+r)
return min(depth[p]+depth[f]-2*depth[p]+1,depth[q]+depth[f]-2*depth[q]+1)
n,q=map(int,input().split())
p=list(map(int,input().split()))
for j in range(n-1):
tree[p[j]].append(j+2)
tree[j+2].append(p[j])
dfs(1,0)
precomputeSparseMatrix(n)
for j in range(q):
a,b,c=map(int,input().split())
print(max(res(a,b,c),res(a,c,b),res(b,c,a)))
``` |
853_B. Jury Meeting | Country of Metropolia is holding Olympiad of Metrpolises soon. It mean that all jury members of the olympiad should meet together in Metropolis (the capital of the country) for the problem preparation process.
There are n + 1 cities consecutively numbered from 0 to n. City 0 is Metropolis that is the meeting point for all jury members. For each city from 1 to n there is exactly one jury member living there. Olympiad preparation is a long and demanding process that requires k days of work. For all of these k days each of the n jury members should be present in Metropolis to be able to work on problems.
You know the flight schedule in the country (jury members consider themselves important enough to only use flights for transportation). All flights in Metropolia are either going to Metropolis or out of Metropolis. There are no night flights in Metropolia, or in the other words, plane always takes off at the same day it arrives. On his arrival day and departure day jury member is not able to discuss the olympiad. All flights in Megapolia depart and arrive at the same day.
Gather everybody for k days in the capital is a hard objective, doing that while spending the minimum possible money is even harder. Nevertheless, your task is to arrange the cheapest way to bring all of the jury members to Metrpolis, so that they can work together for k days and then send them back to their home cities. Cost of the arrangement is defined as a total cost of tickets for all used flights. It is allowed for jury member to stay in Metropolis for more than k days.
Input
The first line of input contains three integers n, m and k (1 ≤ n ≤ 105, 0 ≤ m ≤ 105, 1 ≤ k ≤ 106).
The i-th of the following m lines contains the description of the i-th flight defined by four integers di, fi, ti and ci (1 ≤ di ≤ 106, 0 ≤ fi ≤ n, 0 ≤ ti ≤ n, 1 ≤ ci ≤ 106, exactly one of fi and ti equals zero), the day of departure (and arrival), the departure city, the arrival city and the ticket cost.
Output
Output the only integer that is the minimum cost of gathering all jury members in city 0 for k days and then sending them back to their home cities.
If it is impossible to gather everybody in Metropolis for k days and then send them back to their home cities, output "-1" (without the quotes).
Examples
Input
2 6 5
1 1 0 5000
3 2 0 5500
2 2 0 6000
15 0 2 9000
9 0 1 7000
8 0 2 6500
Output
24500
Input
2 4 5
1 2 0 5000
2 1 0 4500
2 1 0 3000
8 0 1 6000
Output
-1
Note
The optimal way to gather everybody in Metropolis in the first sample test is to use flights that take place on days 1, 2, 8 and 9. The only alternative option is to send jury member from second city back home on day 15, that would cost 2500 more.
In the second sample it is impossible to send jury member from city 2 back home from Metropolis. | ```python
from bisect import *
from sys import *
n,m,k=[int(i) for i in input().split()]
pln=[]
if m==0:
print(-1)
exit(0)
for i in range(m):
pln.append([int(i) for i in input().split()])
pln.sort()
grp=[[pln[0]]];gt=0;
for i in range(1,m):
if pln[i][0]!=pln[i-1][0]:
gt=gt+1
grp.append([])
grp[gt].append(pln[i])
xx=[]
for i in range(len(grp)):
xx.append(grp[i][0][0])
#print('grp',grp)
#print('xx',xx)
from math import inf
pre=[0]*len(xx)
ct=0
mincost=[inf]*(n+1);sumcost=inf
for i,x in enumerate(grp):
for di,fi,ti,ci in x:
if ti==0:
if mincost[fi]==inf:
ct+=1
if sumcost==inf:
mincost[fi]=min(mincost[fi],ci)
else:
sumcost=sumcost-mincost[fi]
mincost[fi]=min(mincost[fi],ci)
sumcost=sumcost+mincost[fi]
if ct==n and sumcost==inf:
sumcost=sum(mincost[1:])
pre[i]=sumcost
#print(pre)
sa=[0]*len(xx)
ct=0
mincost=[inf]*(n+1);sumcost=inf
grp.reverse()
for i,x in enumerate(grp):
for di,fi,ti,ci in x:
if fi==0:
if mincost[ti]==inf:
ct+=1
if sumcost==inf:
mincost[ti]=min(mincost[ti],ci)
else:
sumcost=sumcost-mincost[ti]
mincost[ti]=min(mincost[ti],ci)
sumcost=sumcost+mincost[ti]
if ct==n and sumcost==inf:
sumcost=sum(mincost[1:])
sa[i]=sumcost
sa.reverse()
#print(sa)
ans=inf
for l,xxi in enumerate(xx):
r=bisect_right(xx,xxi+k)
ansl=pre[l]
ansr= inf if r==len(xx) else sa[r]
ans=min(ans,ansl+ansr)
print(ans) if ans!=inf else print(-1)
``` |
877_F. Ann and Books | In Ann's favorite book shop are as many as n books on math and economics. Books are numbered from 1 to n. Each of them contains non-negative number of problems.
Today there is a sale: any subsegment of a segment from l to r can be bought at a fixed price.
Ann decided that she wants to buy such non-empty subsegment that the sale operates on it and the number of math problems is greater than the number of economics problems exactly by k. Note that k may be positive, negative or zero.
Unfortunately, Ann is not sure on which segment the sale operates, but she has q assumptions. For each of them she wants to know the number of options to buy a subsegment satisfying the condition (because the time she spends on choosing depends on that).
Currently Ann is too busy solving other problems, she asks you for help. For each her assumption determine the number of subsegments of the given segment such that the number of math problems is greaten than the number of economics problems on that subsegment exactly by k.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, - 109 ≤ k ≤ 109) — the number of books and the needed difference between the number of math problems and the number of economics problems.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 2), where ti is 1 if the i-th book is on math or 2 if the i-th is on economics.
The third line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the number of problems in the i-th book.
The fourth line contains a single integer q (1 ≤ q ≤ 100 000) — the number of assumptions.
Each of the next q lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) describing the i-th Ann's assumption.
Output
Print q lines, in the i-th of them print the number of subsegments for the i-th Ann's assumption.
Examples
Input
4 1
1 1 1 2
1 1 1 1
4
1 2
1 3
1 4
3 4
Output
2
3
4
1
Input
4 0
1 2 1 2
0 0 0 0
1
1 4
Output
10
Note
In the first sample Ann can buy subsegments [1;1], [2;2], [3;3], [2;4] if they fall into the sales segment, because the number of math problems is greater by 1 on them that the number of economics problems. So we should count for each assumption the number of these subsegments that are subsegments of the given segment.
Segments [1;1] and [2;2] are subsegments of [1;2].
Segments [1;1], [2;2] and [3;3] are subsegments of [1;3].
Segments [1;1], [2;2], [3;3], [2;4] are subsegments of [1;4].
Segment [3;3] is subsegment of [3;4]. | ```python
# ------------------- fast io --------------------
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")
# ------------------- fast io --------------------
from math import ceil
def prod(a, mod=10 ** 9 + 7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def gcd(x, y):
while y:
x, y = y, x % y
return x
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
#n = int(input())
n, k = map(int, input().split())
# a, b = map(int, input().split())
# c, d = map(int, input().split())
t = list(map(int, input().split()))
a = list(map(int, input().split()))
# s = input()
for i in range(n):
# math = eco + k
# math - eco = k
if t[i] == 2:
a[i] *= -1
count = [0] * (n + 1)
pre = [0]
for i in a:
pre += [pre[-1] + i]
index = {}
cc = list(set(pre))
for i in range(len(cc)):
index[cc[i]] = i
minusK = [-1]*(n + 1)
plusK = [-1] * (n + 1)
zero = [-1] * (n + 1)
for i in range(n + 1):
if pre[i] - k in index:
minusK[i] = index[pre[i] - k]
if pre[i] + k in index:
plusK[i] = index[pre[i] + k]
zero[i] = index[pre[i]]
BLOCK_SIZE = 320
blocks = [[] for i in range(BLOCK_SIZE)]
q = int(input())
ans = [0]*q
for i in range(q):
l, r = map(int, input().split())
blocks[l // BLOCK_SIZE] += [[l-1, r, i]]
for i in range(len(blocks)):
if not blocks[i]: continue
blocks[i] = sorted(blocks[i], key=lambda x: x[1])
left = right = BLOCK_SIZE * i
res = 0
count[zero[left]] += 1
for l, r, ind in blocks[i]:
while right < r:
right += 1
if minusK[right] != -1:
res += count[minusK[right]]
count[zero[right]] += 1
while left < l:
count[zero[left]] -= 1
if plusK[left] != -1:
res -= count[plusK[left]]
left += 1
while left > l:
left -= 1
if plusK[left] != -1:
res += count[plusK[left]]
count[zero[left]] += 1
ans[ind] = res
while left <= right:
count[zero[left]] -= 1
if plusK[left] != -1:
res -= count[plusK[left]]
left += 1
assert res == 0
for i in ans:
print(i)
``` |
901_B. GCD of Polynomials | Suppose you have two polynomials <image> and <image>. Then polynomial <image> can be uniquely represented in the following way:
<image>
This can be done using [long division](https://en.wikipedia.org/wiki/Polynomial_long_division). Here, <image> denotes the degree of polynomial P(x). <image> is called the remainder of division of polynomial <image> by polynomial <image>, it is also denoted as <image>.
Since there is a way to divide polynomials with remainder, we can define Euclid's algorithm of finding the greatest common divisor of two polynomials. The algorithm takes two polynomials <image>. If the polynomial <image> is zero, the result is <image>, otherwise the result is the value the algorithm returns for pair <image>. On each step the degree of the second argument decreases, so the algorithm works in finite number of steps. But how large that number could be? You are to answer this question.
You are given an integer n. You have to build two polynomials with degrees not greater than n, such that their coefficients are integers not exceeding 1 by their absolute value, the leading coefficients (ones with the greatest power of x) are equal to one, and the described Euclid's algorithm performs exactly n steps finding their greatest common divisor. Moreover, the degree of the first polynomial should be greater than the degree of the second. By a step of the algorithm we mean the transition from pair <image> to pair <image>.
Input
You are given a single integer n (1 ≤ n ≤ 150) — the number of steps of the algorithm you need to reach.
Output
Print two polynomials in the following format.
In the first line print a single integer m (0 ≤ m ≤ n) — the degree of the polynomial.
In the second line print m + 1 integers between - 1 and 1 — the coefficients of the polynomial, from constant to leading.
The degree of the first polynomial should be greater than the degree of the second polynomial, the leading coefficients should be equal to 1. Euclid's algorithm should perform exactly n steps when called using these polynomials.
If there is no answer for the given n, print -1.
If there are multiple answer, print any of them.
Examples
Input
1
Output
1
0 1
0
1
Input
2
Output
2
-1 0 1
1
0 1
Note
In the second example you can print polynomials x2 - 1 and x. The sequence of transitions is
(x2 - 1, x) → (x, - 1) → ( - 1, 0).
There are two steps in it. | ```python
"""
NTC here
"""
import sys
inp= sys.stdin.readline
input = lambda : inp().strip()
flush= sys.stdout.flush
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(2**25)
def iin(): return int(input())
def lin(): return list(map(int, input().split()))
# range = xrange
# input = raw_input
def main():
n = iin()
ans = [0]*(n+1)
ans[0]=1
ans1 = [0]*(n+1)
sol = 0
# print(ans, ans1)
for i in range(n):
su = ans1[:]
for j in range(n):
su[j+1]+=ans[j]
if su[j+1]>1:su[j+1]=0
# print(1, su, ans)
ans, ans1= su, ans
# print(ans, ans1, mx)
if sol:
print(-1)
else:
m1, m2=0, 0
for i in range(n+1):
if abs(ans[i]):
m1 = i+1
if abs(ans1[i]):
m2 = i+1
print(m1-1)
print( *ans[:m1])
print( m2-1)
print( *ans1[:m2])
main()
# threading.Thread(target=main).start()
``` |
952_A. Quirky Quantifiers |
Input
The input contains a single integer a (10 ≤ a ≤ 999).
Output
Output 0 or 1.
Examples
Input
13
Output
1
Input
927
Output
1
Input
48
Output
0 | ```python
b=int(input())
if b%2==0:
print(0)
else:
print(1)
``` |
979_C. Kuro and Walking Route | Kuro is living in a country called Uberland, consisting of n towns, numbered from 1 to n, and n - 1 bidirectional roads connecting these towns. It is possible to reach each town from any other. Each road connects two towns a and b. Kuro loves walking and he is planning to take a walking marathon, in which he will choose a pair of towns (u, v) (u ≠ v) and walk from u using the shortest path to v (note that (u, v) is considered to be different from (v, u)).
Oddly, there are 2 special towns in Uberland named Flowrisa (denoted with the index x) and Beetopia (denoted with the index y). Flowrisa is a town where there are many strong-scent flowers, and Beetopia is another town where many bees live. In particular, Kuro will avoid any pair of towns (u, v) if on the path from u to v, he reaches Beetopia after he reached Flowrisa, since the bees will be attracted with the flower smell on Kuro’s body and sting him.
Kuro wants to know how many pair of city (u, v) he can take as his route. Since he’s not really bright, he asked you to help him with this problem.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ x, y ≤ n, x ≠ y) - the number of towns, index of the town Flowrisa and index of the town Beetopia, respectively.
n - 1 lines follow, each line contains two integers a and b (1 ≤ a, b ≤ n, a ≠ b), describes a road connecting two towns a and b.
It is guaranteed that from each town, we can reach every other town in the city using the given roads. That is, the given map of towns and roads is a tree.
Output
A single integer resembles the number of pair of towns (u, v) that Kuro can use as his walking route.
Examples
Input
3 1 3
1 2
2 3
Output
5
Input
3 1 3
1 2
1 3
Output
4
Note
On the first example, Kuro can choose these pairs:
* (1, 2): his route would be 1 → 2,
* (2, 3): his route would be 2 → 3,
* (3, 2): his route would be 3 → 2,
* (2, 1): his route would be 2 → 1,
* (3, 1): his route would be 3 → 2 → 1.
Kuro can't choose pair (1, 3) since his walking route would be 1 → 2 → 3, in which Kuro visits town 1 (Flowrisa) and then visits town 3 (Beetopia), which is not allowed (note that pair (3, 1) is still allowed because although Kuro visited Flowrisa and Beetopia, he did not visit them in that order).
On the second example, Kuro can choose the following pairs:
* (1, 2): his route would be 1 → 2,
* (2, 1): his route would be 2 → 1,
* (3, 2): his route would be 3 → 1 → 2,
* (3, 1): his route would be 3 → 1. | ```python
def calculate_max_paths(edges, node, dest, par, cnt):
ans = 1
for child in edges.get(node, []):
if child != par:
ans += calculate_max_paths(edges, child, dest, node, cnt)
if dest == node:
cnt[0] = ans
return ans
def main():
from collections import defaultdict
n, flower, bee = list(map(int, input().split()))
edges = defaultdict(list)
for _ in range(n-1):
x, y = list(map(int, input().split()))
edges[x].append(y)
edges[y].append(x)
bees, flowers = [0], [0]
x = calculate_max_paths(edges, flower, bee, par=None, cnt=bees)
y = calculate_max_paths(edges, bee, flower, par=None, cnt=flowers)
print(x*(x-1) - bees[0]*flowers[0])
if __name__=="__main__":
import sys
import threading
sys.setrecursionlimit(10**6)
threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
``` |
999_D. Equalize the Remainders | You are given an array consisting of n integers a_1, a_2, ..., a_n, and a positive integer m. It is guaranteed that m is a divisor of n.
In a single move, you can choose any position i between 1 and n and increase a_i by 1.
Let's calculate c_r (0 ≤ r ≤ m-1) — the number of elements having remainder r when divided by m. In other words, for each remainder, let's find the number of corresponding elements in a with that remainder.
Your task is to change the array in such a way that c_0 = c_1 = ... = c_{m-1} = n/m.
Find the minimum number of moves to satisfy the above requirement.
Input
The first line of input contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ n). It is guaranteed that m is a divisor of n.
The second line of input contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), the elements of the array.
Output
In the first line, print a single integer — the minimum number of moves required to satisfy the following condition: for each remainder from 0 to m - 1, the number of elements of the array having this remainder equals n/m.
In the second line, print any array satisfying the condition and can be obtained from the given array with the minimum number of moves. The values of the elements of the resulting array must not exceed 10^{18}.
Examples
Input
6 3
3 2 0 6 10 12
Output
3
3 2 0 7 10 14
Input
4 2
0 1 2 3
Output
0
0 1 2 3 | ```python
from collections import deque
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = list(map(int, input().split()))
cnt = [0] * m
x = [[] for _ in range(m)]
for i in range(n):
ai = a[i] % m
cnt[ai] += 1
x[ai].append(i)
l = n // m
q = deque()
for i in range(m):
if cnt[i] < l:
q.append(i)
ans = 0
for i in range(m):
for j in x[i]:
if cnt[i] <= l:
break
while q[0] < i:
q.append(q.popleft() + m)
k = q[0] % m
cnt[k] += 1
cnt[i] -= 1
ans += (k - i) % m
a[j] += (k - i) % m
if cnt[k] == l:
q.popleft()
print(ans)
print(*a)
``` |
p02534 ACL Beginner Contest - Repeat ACL | You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them.
For example, if K = 3, print `ACLACLACL`.
Constraints
* 1 \leq K \leq 5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
K
Output
Print the string obtained by repeating the string `ACL` K times and concatenating them.
Example
Input
3
Output
ACLACLACL | ```python
K=int(input())
print("ACL"*K)
``` |
p02666 NOMURA Programming Competition 2020 - Urban Planning | There are N towns numbered 1, 2, \cdots, N.
Some roads are planned to be built so that each of them connects two distinct towns bidirectionally. Currently, there are no roads connecting towns.
In the planning of construction, each town chooses one town different from itself and requests the following: roads are built so that the chosen town is reachable from itself using one or more roads.
These requests from the towns are represented by an array P_1, P_2, \cdots, P_N. If P_i = -1, it means that Town i has not chosen the request; if 1 \leq P_i \leq N, it means that Town i has chosen Town P_i.
Let K be the number of towns i such that P_i = -1. There are (N-1)^K ways in which the towns can make the requests. For each way to make requests, find the minimum number of roads needed to meet all the requests, and print the sum of those (N-1)^K numbers, modulo (10^9+7).
Constraints
* 2 \leq N \leq 5000
* P_i = -1 or 1 \leq P_i \leq N.
* P_i \neq i
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
P_1 P_2 \cdots P_N
Output
For each way to make requests, find the minimum number of roads needed to meet all the requests, and print the sum of those (N-1)^K numbers, modulo (10^9+7).
Examples
Input
4
2 1 -1 3
Output
8
Input
2
2 1
Output
1
Input
10
2 6 9 -1 6 9 -1 -1 -1 -1
Output
527841 | ```python
def modfac(n, MOD):
f = 1
factorials = [1]
for m in range(1, n + 1):
f *= m
f %= MOD
factorials.append(f)
inv = pow(f, MOD - 2, MOD)
invs = [1] * (n + 1)
invs[n] = inv
for m in range(n, 1, -1):
inv *= m
inv %= MOD
invs[m - 1] = inv
return factorials, invs
def modnCr(n,r,mod,fac,inv): #上で求めたfacとinvsを引数に入れるべし(上の関数で与えたnが計算できる最大のnになる)
return fac[n] * inv[n-r] * inv[r] % mod
def uf_find(n,p):
ufl = []
while p[n] != n:
ufl.append(n)
n = p[n]
for i in ufl:
p[i] = n
return n
def uf_union(a,b,p,rank,tsize):
ap = uf_find(a,p)
bp = uf_find(b,p)
if ap == bp:
return True
else:
if rank[ap] > rank[bp]:
p[bp] = ap
tsize[ap] += tsize[bp]
elif rank[ap] < rank[bp]:
p[ap] = bp
tsize[bp] += tsize[ap]
else:
p[bp] = ap
rank[ap] += 1
tsize[ap] += tsize[bp]
return False
mod = 10**9+7
N = int(input())
P = list(map(int,input().split()))
fac,inv = modfac(N+10,mod)
ans = 0
p = [i for i in range(N)]
rank = [1] * N
tsize = [1] * N
for i in range(N):
if P[i] == -1:
continue
nex = P[i] - 1
uf_union(i,nex,p,rank,tsize)
able = [True] * N
c = []
d = []
for i in range(N):
if P[i] == -1:
nowp = uf_find(i,p)
c.append(tsize[nowp])
able[nowp] = False
for i in range(N):
if uf_find(i,p) == i and able[i]:
d.append(tsize[i])
ans = 0
for i in c:
ans += i
for i in d:
ans += i-1
ans *= pow(N-1,len(c),mod)
ans %= mod
#print (ans)
#自分を選んで減る場合
for i in c:
ans -= (i-1) * pow(N-1,len(c)-1,mod)
ans %= mod
#互いに選んで減る場合
dp = [0] * (len(c)+1)
dp[0] = 1
for i in c:
for j in range(len(c)-1,-1,-1):
dp[j+1] += dp[j] * i
#dp[j+1] %= mod
#print (c,dp)
for i in range(2,len(c)+1):
ans -= dp[i] * fac[i-1] * pow(N-1,len(c)-i,mod)
ans %= mod
print (ans % mod)
"""
連結成分で考える
全ての街が -1 以外を選んだ時、
最小数 + -1の数
-1同士が選ぶことを考える
-1同士が選んだ時のみ答えが1減る
そうでなければ変化なし？
連結成分に-1は1つ以下しかない？
→正しい
-1を含まない連結成分は分けておく
-1の連結成分同士が求め合えば、答えは-1される
a個含む & bこ含む
場合
a*b * (N-1)^(K-2)
引けばいい
自分の連結成分を求めるときも1減る
ループを作ると1減るんだ！
"""
``` |
p02795 Keyence Programming Contest 2020 - Painting | We have a grid with H rows and W columns, where all the squares are initially white.
You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions:
* Choose one row, then paint all the squares in that row black.
* Choose one column, then paint all the squares in that column black.
At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.
Constraints
* 1 \leq H \leq 100
* 1 \leq W \leq 100
* 1 \leq N \leq H \times W
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
H
W
N
Output
Print the minimum number of operations needed.
Examples
Input
3
7
10
Output
2
Input
14
12
112
Output
8
Input
2
100
200
Output
2 | ```python
h = int(input())
w = int(input())
n = int(input())
print(min([(n+h-1)//h, (n+w-1)//w]))
``` |
p02931 Japanese Student Championship 2019 Qualification - Card Collector | There are N cards placed on a grid with H rows and W columns of squares.
The i-th card has an integer A_i written on it, and it is placed on the square at the R_i-th row from the top and the C_i-th column from the left.
Multiple cards may be placed on the same square.
You will first pick up at most one card from each row.
Then, you will pick up at most one card from each column.
Find the maximum possible sum of the integers written on the picked cards.
Constraints
* All values are integers.
* 1 \leq N \leq 10^5
* 1 \leq H, W \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq R_i \leq H
* 1 \leq C_i \leq W
Input
Input is given from Standard Input in the following format:
N H W
R_1 C_1 A_1
R_2 C_2 A_2
\vdots
R_N C_N A_N
Output
Print the maximum possible sum of the integers written on the picked cards.
Examples
Input
6 2 2
2 2 2
1 1 8
1 1 5
1 2 9
1 2 7
2 1 4
Output
28
Input
13 5 6
1 3 35902
4 6 19698
4 6 73389
3 6 3031
3 1 4771
1 4 4784
2 1 36357
2 1 24830
5 6 50219
4 6 22645
1 2 30739
1 4 68417
1 5 78537
Output
430590
Input
1 100000 100000
1 1 1
Output
1 | ```python
import sys
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
N,H,W = map(int,readline().split())
RCA = [tuple(int(x) for x in line.split()) for line in readlines()]
RCA.sort(key = lambda x: -x[2])
RCA
root = list(range(H+W))
size = [0] * (H+W)
no_cycle = [True] * (H+W)
def find_root(x):
y = root[x]
if x == y:
return y
z = find_root(y)
root[x] = z
return z
def merge(x,y):
x,y = find_root(x),find_root(y)
sx,sy = size[x],size[y]
if sx < sy:
sx,sy = sy,sx
x,y = y,x
root[y] = x
no_cycle[x] = no_cycle[x] and no_cycle[y]
answer = 0
for R,C,A in RCA:
x,y = R-1,H+C-1
rx,ry = find_root(x),find_root(y)
if rx == ry:
if not no_cycle[rx]:
continue
no_cycle[rx] = False
else:
if (not no_cycle[rx]) and (not no_cycle[ry]):
continue
merge(rx,ry)
answer += A
print(answer)
``` |
p03067 Tenka1 Programmer Beginner Contest 2019 - On the Way | There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
Constraints
* 0\leq A,B,C\leq 100
* A, B and C are distinct integers.
Input
Input is given from Standard Input in the following format:
A B C
Output
Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
Examples
Input
3 8 5
Output
Yes
Input
7 3 1
Output
No
Input
10 2 4
Output
Yes
Input
31 41 59
Output
No | ```python
a, b, c = map(int, input().split())
if (c-a)*(c-b) < 0:
print("Yes")
else:
print("No")
``` |
p03210 AtCoder Beginner Contest 114 - 753 | Shichi-Go-San (literally "Seven-Five-Three") is a traditional event in a certain country to celebrate the growth of seven-, five- and three-year-old children.
Takahashi is now X years old. Will his growth be celebrated in Shichi-Go-San this time?
Constraints
* 1 ≤ X ≤ 9
* X is an integer.
Input
Input is given from Standard Input in the following format:
X
Output
If Takahashi's growth will be celebrated, print `YES`; if it will not, print `NO`.
Examples
Input
5
Output
YES
Input
6
Output
NO | ```python
a=int(input())
print('YES' if a in [3,5,7] else 'NO')
``` |
p03359 AtCoder Beginner Contest 096 - Day of Takahashi | In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year.
For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year.
In this country, a date is called Takahashi when the month and the day are equal as numbers. For example, 5-5 is Takahashi.
How many days from 2018-1-1 through 2018-a-b are Takahashi?
Constraints
* a is an integer between 1 and 12 (inclusive).
* b is an integer between 1 and 31 (inclusive).
* 2018-a-b is a valid date in Gregorian calendar.
Input
Input is given from Standard Input in the following format:
a b
Output
Print the number of days from 2018-1-1 through 2018-a-b that are Takahashi.
Examples
Input
5 5
Output
5
Input
2 1
Output
1
Input
11 30
Output
11 | ```python
a, b = map(int, input().split())
print(a if (a <= b) else a - 1)
``` |
p03517 CODE FESTIVAL 2017 Elimination Tournament Round 2 (Parallel) - Colorful MST | Ringo has an undirected graph G with N vertices numbered 1,2,...,N and M edges numbered 1,2,...,M. Edge i connects Vertex a_{i} and b_{i} and has a length of w_i.
Now, he is in the middle of painting these N vertices in K colors numbered 1,2,...,K. Vertex i is already painted in Color c_i, except when c_i = 0, in which case Vertex i is not yet painted.
After he paints each vertex that is not yet painted in one of the K colors, he will give G to Snuke.
Based on G, Snuke will make another undirected graph G' with K vertices numbered 1,2,...,K and M edges. Initially, there is no edge in G'. The i-th edge will be added as follows:
* Let x and y be the colors of the two vertices connected by Edge i in G.
* Add an edge of length w_i connecting Vertex x and y in G'.
What is the minimum possible sum of the lengths of the edges in the minimum spanning tree of G'? If G' will not be connected regardless of how Ringo paints the vertices, print -1.
Constraints
* 1 \leq N,M \leq 10^{5}
* 1 \leq K \leq N
* 0 \leq c_i \leq K
* 1 \leq a_i,b_i \leq N
* 1 \leq w_i \leq 10^{9}
* The given graph may NOT be simple or connected.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M K
c_1 c_2 ... c_{N}
a_1 b_1 w_1
:
a_M b_M w_M
Output
Print the answer.
Examples
Input
4 3 3
1 0 1 2
1 2 10
2 3 20
2 4 50
Output
60
Input
5 2 4
0 0 0 0 0
1 2 10
2 3 10
Output
-1
Input
9 12 9
1 2 3 4 5 6 7 8 9
6 9 9
8 9 6
6 7 85
9 5 545631016
2 1 321545
1 6 33562944
7 3 84946329
9 7 15926167
4 7 53386480
5 8 70476
4 6 4549
4 8 8
Output
118901402
Input
18 37 12
5 0 4 10 8 7 2 10 6 0 9 12 12 11 11 11 0 1
17 1 1
11 16 7575
11 15 9
10 10 289938980
5 10 17376
18 4 1866625
8 11 959154208
18 13 200
16 13 2
2 7 982223
12 12 9331
13 12 8861390
14 13 743
2 10 162440
2 4 981849
7 9 1
14 17 2800
2 7 7225452
3 7 85
5 17 4
2 13 1
10 3 45
1 15 973
14 7 56553306
16 17 70476
7 18 9
9 13 27911
18 14 7788322
11 11 8925
9 13 654295
2 10 9
10 1 545631016
3 4 5
17 12 1929
2 11 57
1 5 4
1 17 7807368
Output
171 | ```python
class Unionfindtree:
def __init__(self, number):
self.par = [i for i in range(number)]
self.rank = [0] * (number)
def find(self, x): # 親を探す
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y): # x,yを繋げる
px = self.find(x)
py = self.find(y)
if px == py:
return
if self.rank[px] < self.rank[py]:
self.par[px] = py
else:
self.par[py] = px
if self.rank[px] == self.rank[py]:
self.rank[px] += 1
def connect(self, x, y): # 親が同じかみる
return self.find(x) == self.find(y)
N, M, K = map(int, input().split())
C = [int(i) for i in input().split()]
table = []
t = K
for i in range(N):
if C[i]==0:
t+=1
C[i]=t
for i in range(M):
s, u, c = map(int, input().split())
a, b = C[s - 1], C[u - 1]
table.append((c, a, b))
tree = Unionfindtree(t+1)
table.sort()
ct = K - 1
ans = 0
for c, a, b in table:
if not tree.connect(a, b) and ct > 0:
tree.union(a, b)
ans += c
ct -= 1
if ct > 0:
print(-1)
else:
print(ans)
``` |
p03680 AtCoder Beginner Contest 065 - Trained? | Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | ```python
n=int(input())
l=[int(input()) for i in range(n)]
a=1
ans=0
while a!=2:
a=l[a-1]
ans+=1
if ans==n+1:
ans=-1
break
print(ans)
``` |
p03835 AtCoder Beginner Contest 051 - Sum of Three Integers | You are given two integers K and S.
Three variable X, Y and Z takes integer values satisfying 0≤X,Y,Z≤K.
How many different assignments of values to X, Y and Z are there such that X + Y + Z = S?
Constraints
* 2≤K≤2500
* 0≤S≤3K
* K and S are integers.
Input
The input is given from Standard Input in the following format:
K S
Output
Print the number of the triples of X, Y and Z that satisfy the condition.
Examples
Input
2 2
Output
6
Input
5 15
Output
1 | ```python
k,s=map(int,input().split());r=range;print(sum(1for x in r(k+1)for y in r(k+1)if 0<=s-x-y<=k))
``` |
p04000 AtCoder Beginner Contest 045 - Snuke's Coloring | We have a grid with H rows and W columns. At first, all cells were painted white.
Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column.
Compute the following:
* For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
Constraints
* 3 \leq H \leq 10^9
* 3 \leq W \leq 10^9
* 0 \leq N \leq min(10^5,H×W)
* 1 \leq a_i \leq H (1 \leq i \leq N)
* 1 \leq b_i \leq W (1 \leq i \leq N)
* (a_i, b_i) \neq (a_j, b_j) (i \neq j)
Input
The input is given from Standard Input in the following format:
H W N
a_1 b_1
:
a_N b_N
Output
Print 10 lines. The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.
Examples
Input
4 5 8
1 1
1 4
1 5
2 3
3 1
3 2
3 4
4 4
Output
0
0
0
2
4
0
0
0
0
0
Input
10 10 20
1 1
1 4
1 9
2 5
3 10
4 2
4 7
5 9
6 4
6 6
6 7
7 1
7 3
7 7
8 1
8 5
8 10
9 2
10 4
10 9
Output
4
26
22
10
2
0
0
0
0
0
Input
1000000000 1000000000 0
Output
999999996000000004
0
0
0
0
0
0
0
0
0 | ```python
from collections import defaultdict
H,W,N=map(int,input().split())
ans=[(H-2)*(W-2)]+[0]*9
blacks=defaultdict(int)
for i in range(N):
a, b = map(int,input().split())
for dx in [-1,0,1]:
for dy in [-1,0,1]:
if 2<=a+dx<H and 2<=b+dy<W:
ans[blacks[(a+dx,b+dy)]]-=1
ans[blacks[(a+dx,b+dy)]+1]+=1
blacks[(a+dx,b+dy)]+=1
print(*ans,sep="\n")
``` |
p00088 The Code A Doctor Loved | Dr .: Peter, I've finally done it.
Peter: What's wrong, Dr. David? Is it a silly invention again?
Dr .: This table, this table.
| Character | Sign
--- | ---
(Blank) | 101
'| 000000
, | 000011
-| 10010001
. | 010001
? | 000001
A | 100101
B | 10011010
| Character | Sign
--- | ---
C | 0101
D | 0001
E | 110
F | 01001
G | 10011011
H | 010000
I | 0111
J | 10011000
| Character | Sign
--- | ---
K | 0110
L | 00100
M | 10011001
N | 10011110
O | 00101
P | 111
Q | 10011111
R | 1000
| Character | Sign
--- | ---
S | 00110
T | 00111
U | 10011100
V | 10011101
W | 000010
X | 10010010
Y | 10010011
Z | 10010000
Peter: What? This table is.
Dr .: Okay, just do what you say. First, write your name on a piece of paper.
Peter: Yes, "PETER POTTER".
Dr .: Then, replace each character with the "code" in this table.
Peter: Well, change "P" to "111" and "E" to "110" ... It's quite annoying.
111 110 00111 110 1000 101 111 00101 00111 00111 110 1000
became. It looks like a barcode.
Dr .: All right. Then connect all the replaced strings and separate them by 5 characters.
Peter: Yes, if you connect and separate.
11111 00011 11101 00010 11110 01010 01110 01111 10100 0
It turned out to be something like this. But what about the last "0" guy?
Dr .: Add 0 to make it 5 letters.
Peter: Well, there is only one 0 at the end, so I should add four more 0s. I was able to do it.
11111 00011 11101 00010 11110 01010 01110 01111 10100 00000
Dr .: Next, use this table.
| Sign | Character
--- | ---
00000 | A
00001 | B
00010 | C
00011 | D
00100 | E
00101 | F
00110 | G
00111 | H
| Sign | Character
--- | ---
01000 | I
01001 | J
01010 | K
01011 | L
01100 | M
01101 | N
01110 | O
01111 | P
| Sign | Character
--- | ---
10000 | Q
10001 | R
10010 | S
10011 | T
10100 | U
10101 | V
10110 | W
10111 | X
| Sign | Character
--- | ---
11000 | Y
11001 | Z
11010 | (blank)
11011 | ..
11100 |,
11101 |-
11110 |'
11111 |?
Peter: How do you use this ... yeah! Now you're going to replace the code with a letter!
Dr .: That's right. If it is "11111", go to "?", If it is "00011", go to "D".
Peter: This is simple ... well, it became "? D-C'KOPUA". But it doesn't make sense.
Dr .: Count the number of characters.
Peter: It's 10 characters. Oh, "PETER POTTER" was 12 characters, but 2 characters are reduced.
Dr .: Yes, you can use this table to reduce the number of letters. Now do the same thing in this sentence.
PETER PIPER PICKED A PECK OF PICKLED PEPPERS. A PECK OF PICKLED PEPPERS
PETER PIPER PICKED. IF PETER PIPER PICKED A PECK OF PICKLED PEPPERS, WHERE'S
THE PECK OF PICKLED PEPPERS PETER PIPER PICKED?
Peter: Gyogyo, the lines are separated, what do you do?
Dr .: Due to space limitations, there are only 3 lines, but think of it as a long line with a space instead of a newline character.
Peter: Yes, yes. There is a blank space between the lines. But it's a hassle ...
Dr .: Then why not let the program do it?
So, instead of Peter, create a program that converts the read character string into a code and outputs it.
input
Given multiple datasets. Each dataset is given a line of strings (consisting of the characters contained in the table). The string contains at least 1 and no more than 100 characters.
The number of datasets does not exceed 200.
output
For each data set, output the converted character string on one line.
Example
Input
PETER POTTER
Output
?D-C'KOPUA | ```python
# Aizu Problem 0088: The Code A Doctor Loved
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def bin5(k):
b = bin(k)[2:]
while len(b) < 5:
b = '0' + b
return b
CODE1 = {' ': '101', "'": '000000', ",": '000011', '-': '10010001',
'.': '010001', '?': '000001',
'A': '100101', 'B': '10011010', 'C': '0101', 'D': '0001',
'E': '110', 'F': '01001', 'G': '10011011', 'H': '010000',
'I': '0111', 'J': '10011000', 'K': '0110', 'L': '00100',
'M': '10011001', 'N': '10011110', 'O': '00101', 'P': '111',
'Q': '10011111', 'R': '1000', 'S': '00110', 'T': '00111',
'U': '10011100', 'V': '10011101', 'W': '000010',
'X': '10010010', 'Y': '10010011', 'Z': '10010000'}
CODE2 = {bin5(k): chr(65+k) for k in range(26)}
symbols = [' ', '.', ',', '-', "'", '?']
for k in range(len(symbols)):
CODE2[bin5(26+k)] = symbols[k]
def encode(string):
enc = ''.join([CODE1[char] for char in string])
while len(enc) % 5 != 0:
enc += '0'
return ''.join([CODE2[enc[5*k:5*k+5]] for k in range(len(enc) // 5)])
for line in sys.stdin:
print(encode(line.replace('\n', '')))
``` |
p00220 Binary Digit A Doctor Loved | "What are your shoe sizes?"
Suddenly, the doctor asked me when I met him for the first time.
"It's 23.5"
"Oh, that's a really nice number. It's 2 to the 4th power plus 2 to the 2nd power, 2 to the 1st power, 2 to the 0th power, and 2 to the 1st power."
Then the doctor asked.
"You, how tall are you?"
"Yes, it's 158.1."
He folded his arms and closed his eyes. After a while of silence, I opened my mouth.
"Nah ~"
After that, during the time I spent together, I gradually became able to understand the behavior of the doctor.
First, I say the real number at the request of the doctor. If the real number is represented by a binary number with no more than 8 digits for the integer part and no more than 4 digits for the decimal part, he is happy to say the result of the conversion to binary. If not, it will sadly yell "Nah ~". This repeats until I say a negative real number.
By the way, as he got older, it became more and more difficult to make long calculations. Therefore, please make a program for you to input real numbers and convert / output them to binary numbers on your behalf. However, if the binary representation does not fit within the limit number of digits (integer part within 8 digits + decimal part within 4 digits), output NA (half-width alphabetic characters). The input real number shall fit within 8 digits of the integer part and within 4 digits of the decimal part, and the binary representation to be output should be output with 8 digits of the integer part and 4 digits of the decimal part.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single negative real line. One real number n is given to one row for each dataset.
The number of datasets does not exceed 1200.
Output
Outputs the conversion result to binary number for each input data set.
Example
Input
23.5
158.1
-1.0
Output
00010111.1000
NA | ```python
def ans(N): # N:foat
n=int(N)
d=N-n
ans_l=bin(n)[2:]
ans_r=''
for _ in range(4):
ans_r+=str(int(d*2))
d=d*2-int(d*2)
if n>=256 or d!=0:
return 'NA'
else:
return '0'*(8-len(ans_l))+ans_l+'.'+ans_r
while True:
INP=float(input())
if INP<0: break
print(ans(INP))
``` |
p00382 Taxi | PCK Taxi in Aizu city, owned by PCK company, has adopted a unique billing system: the user can decide the taxi fare. Today as usual, many people are waiting in a queue at the taxi stand in front of the station.
In front of the station, there are $N$ parking spaces in row for PCK taxis, each with an index running from $1$ to $N$. Each of the parking areas is occupied by a taxi, and a queue of potential passengers is waiting for the ride. Each one in the queue has his/her own plan for how much to pay for the ride.
To increase the company’s gain, the taxi driver is given the right to select the passenger who offers the highest taxi fare, rejecting others.
The driver in the $i$-th parking space can perform the following actions any number of times in any sequence before he finally selects a passenger and starts driving.
1. Offer a ride to the passenger who is at the head of the $i$-th parking space’s queue.
2. Reject to offer a ride to the passenger who is at the head of the $i$-th parking space’s queue. The passenger is removed from the queue.
3. Move to the $i + 1$-th parking area if it is empty. If he is in the $N$-th parking area, he leaves the taxi stand to cruise the open road.
A preliminary listening is made as to the fare the users offer. Your task is to maximize the sales volume of PCK Taxi in reference to the table of offered fares. A taxi cannot accommodate more than one passenger.
Given the number of taxi parking spaces and information regarding the persons waiting in the parking areas, calculate the maximum possible volume of sales.
Input
The input is given in the following format.
$N$
$s_1$
$s_2$
$...$
$s_N$
The first line provides the number of taxi parking areas $N$ ($1 \leq N \leq 300,000$). Each of the subsequent $N$ lines provides information on the customers queueing in the $i$-th taxi parking area in the following format:
$M$ $c_1$ $c_2$ ... $c_M$
The first integer $M$ ($1 \leq M \leq 300,000$) indicates the number of customers in the queue, and the subsequent array of integers $c_j$ ($1 \leq c_j \leq 10,000$) indicates the fare the $j$-th customer in the queue is willing to pay. The total number of customers in the taxi stand is equal to or less than $300,000$.
Output
Output the maximum volume of sales.
Example
Input
3
3 8 10 1
4 7 1 2 15
3 11 8 19
Output
45 | ```python
from heapq import heappush, heappop
def main():
n = int(input())
que = []
for i in range(n):
lst = sorted(list(map(int, input().split()))[1:], reverse=True)[:i+1]
heappush(que, lst[-1])
for v in lst[:-1]:
heappush(que, v)
heappop(que)
print(sum(que))
main()
``` |
p00598 Operations with Finite Sets | Let A, B, C, D, E be sets of integers and let U is a universal set that includes all sets under consideration. All elements in any set are different (no repetitions).
u - union of two sets, AuB = {x ∈ U : x ∈ A or x ∈ B} is the set of all elements which belong to A or B.
i - intersection of two sets, AiB = {x ∈ U : x ∈ A and x ∈ B} is the set of all elements which belong to both A and B.
d - difference of two sets, AdB = {x ∈ U : x ∈ A, x ∉ B} is the set of those elements of A which do not belong to B.
s - symmetric difference of two sets, AsB = (AdB)u(BdA) consists of those elements which belong to A or B but not to both.
c - complement of a set, cA = {x ∈ U : x ∉ A}, is set of elements which belong to U but do not belong to A. Unary operator c has higest precedence.
The universal set U is defined as a union of all sets specified in data.
Your task is to determine the result of an expression, which includes sets, set operations and parenthesis (any number of parenthesis and any correct enclosure of parenthesis may take place).
Input
Input consists of several pairs of lines difining sets and one pair of lines defining an expression. Each pair of lines for set definition includes the following.
Line 1: Set name (A, B, C, D, E), number of elements in a set.
Line 2: Set elements separated by blanks.
Pair of lines for expression definition:
Line 1: R 0
Line 2: Expression consisting of set names, operators and parenthesis (no blanks).
Number of sets can vary from 1 to 5. Set names can be specified in any order. Each set consists of 1-100 elements. Pair of lines for expression definition signals the end of data set. Input file includes several data sets. The number of datasets is less than 20.
Output
For each data set, the output should contain one line with resulting set elements sorted in ascending order separated by blanks. If the result contains no set elements then the line should contain the text NULL.
Example
Input
A 3
1 3 -1
B 4
3 1 5 7
D 1
5
R 0
cAiBdD
C 3
1 2 3
A 4
2 10 8 3
B 3
2 4 8
R 0
(As(AiB))uC
Output
7
1 2 3 10 | ```python
import sys
def rpn(str):
r = []
stack = []
for i in range(0, len(str)):
c = str[i]
if c in "idsu":
while len(stack) > 0:
if stack[-1] in "idsuc":
a = stack.pop()
r.extend(a)
else:
break
stack.extend(c)
elif c == "c":
stack.extend(c)
elif c == "(":
stack.extend(c)
elif c == ")":
while len(stack) > 0:
a = stack.pop()
if a == "(":
break
r.extend(a)
else:
r.extend(c)
while len(stack) > 0:
a = stack.pop()
r.extend(a)
return r
def intersect(a, b):
r = []
for e in a:
if e in b:
r.extend([e])
return r
def union(a, b):
r = list(set(a + b))
return r
def diff(a, b):
r = []
for e in a:
if e not in b:
r.extend([e])
return r
def universal(sets):
r = []
for v in sets.values():
r.extend(v)
r = list(set(r))
return r
def calc(rpn, sets):
stack = []
U = universal(sets)
for c in rpn:
if c in "iuds":
op2 = stack.pop()
op1 = stack.pop()
if c == "i":
x = intersect(op1, op2)
stack.append(x)
elif c == "u":
x = union(op1, op2)
stack.append(x)
elif c == "d":
x = diff(op1, op2)
stack.append(x)
elif c == "s":
x = diff(op1, op2)
y = diff(op2, op1)
z = union(x, y)
stack.append(z)
elif c == "c":
op1 = stack.pop()
x = diff(U, op1)
stack.append(x)
else:
stack.append(sets[c])
return stack.pop()
lno = 0
sets = {}
name = ""
for line in sys.stdin:
lno += 1
if lno % 2 == 1:
name = line.strip().split()[0]
elif name != "R":
elem = list(map(int, line.strip().split()))
sets[name] = elem
else:
e = rpn(line.strip())
result = calc(e, sets)
result.sort()
if len(result) > 0:
print(" ".join([str(n) for n in result]))
else:
print("NULL")
sets = {}
``` |
p00734 Equal Total Scores | Taro and Hanako have numbers of cards in their hands. Each of the cards has a score on it. Taro and Hanako wish to make the total scores of their cards equal by exchanging one card in one's hand with one card in the other's hand. Which of the cards should be exchanged with which?
Note that they have to exchange their cards even if they already have cards of the same total score.
Input
The input consists of a number of datasets. Each dataset is formatted as follows.
> n m
> s1
> s2
> ...
> sn
> sn+1
> sn+2
> ...
> sn+m
>
The first line of a dataset contains two numbers n and m delimited by a space, where n is the number of cards that Taro has and m is the number of cards that Hanako has. The subsequent n+m lines list the score for each of the cards, one score per line. The first n scores (from s1 up to sn) are the scores of Taro's cards and the remaining m scores (from sn+1 up to sn+m) are Hanako's.
The numbers n and m are positive integers no greater than 100. Each score is a non-negative integer no greater than 100.
The end of the input is indicated by a line containing two zeros delimited by a single space.
Output
For each dataset, output a single line containing two numbers delimited by a single space, where the first number is the score of the card Taro gives to Hanako and the second number is the score of the card Hanako gives to Taro. If there is more than one way to exchange a pair of cards that makes the total scores equal, output a pair of scores whose sum is the smallest.
In case no exchange can make the total scores equal, output a single line containing solely -1. The output must not contain any superfluous characters that do not conform to the format.
Sample Input
2 2
1
5
3
7
6 5
3
9
5
2
3
3
12
2
7
3
5
4 5
10
0
3
8
1
9
6
0
6
7 4
1
1
2
1
2
1
4
2
3
4
3
2 3
1
1
2
2
2
0 0
Output for the Sample Input
1 3
3 5
-1
2 2
-1
Example
Input
2 2
1
5
3
7
6 5
3
9
5
2
3
3
12
2
7
3
5
4 5
10
0
3
8
1
9
6
0
6
7 4
1
1
2
1
2
1
4
2
3
4
3
2 3
1
1
2
2
2
0 0
Output
1 3
3 5
-1
2 2
-1 | ```python
while True:
n,m = map(int,input().split())
if n == 0:
exit()
t = []
h = []
ans = []
for i in range(n):
t.append(int(input()))
for i in range(m):
h.append(int(input()))
for i in range(n):
for j in range(m):
if (sum(t) - t[i] + h[j]) == (sum(h) - h[j] + t[i]):
ans.append([t[i] + h[j], t[i], h[j]])
ans = sorted(ans)
if len(ans) != 0:
print(str(ans[0][1]) + " " + str(ans[0][2]))
else:
print(-1)
``` |
p00874 Cubist Artwork | International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | ```python
while True:
w,d = map(int,input().split())
if w == 0 and d == 0:
break
h = list(map(int,input().split()))
m = list(map(int,input().split()))
hl = [0] * 30
ml = [0] * 30
for i in h:
hl[i] += 1
for i in m:
ml[i] += 1
ans = 0
for i in range(30):
ans += i * max(hl[i],ml[i])
print (ans)
``` |
p01137 Space Coconut Grab | Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | ```python
import math
def solve(e):
k = 2 ** 32
for z in range(100, -1, -1):
z3 = z * z * z
if z3 > e:
continue
e2 = e - z3
ylm = int(math.sqrt(e2))
xzlm = 3 * z * z + 3 * z + 1
for y in range(ylm, -1, -1):
y2 = y * y
if e2 > (y + 1) * (y + 1):
break
e3 = e2 - y2
xylm = 2 * y + 1
x = e3
if x > xylm or x > xzlm:
continue
k = min(k, x + y + z)
return k
def main():
while True:
a = int(input())
if a == 0:
break
print(solve(a))
if __name__ == '__main__':
main()
``` |
p01752 Prowler | Mr. A and Mr. B live in an N × M rectangular grid area. Each square is either a road, a wall, or a house. Since this area is famous for the frequent occurrence of molestation damage due to the complicated and intricate roads, the boundary between this area and the outside is completely surrounded by walls and isolated.
Mr. B somehow felt like it, so he decided to go to see Mr. A's house. However, unfortunately, Mr. B has an apparently suspicious face, so if he takes any action that seems to be suspected of being a molester on his way to Mr. A's house, he will be caught immediately. In particular, never walk with your right hand off the wall. Can Mr. B reach Mr. A's house without letting go of his right hand for a moment?
Mr. B is always facing up, down, left, or right, and the range that Mr. B's right hand can reach is only four squares, front, diagonally right front, right, and diagonally right back with respect to the direction that Mr. B is facing. Mr. B repeats one of the following actions. However, these cannot be done at the same time.
* If there is no wall in front, go one square.
* Change the direction you are facing to the right by 90 degrees.
* Change the direction you are facing to the left by 90 degrees.
* Change the square that your right hand touches. However, at this time, Mr. B cannot release his right hand, so he must have something in common between the square before the change and the square after the change.
Constraints
* 1 ≤ N ≤ 50
* 1 ≤ M ≤ 50
* Only one of the characters "^", "v", "<", ">" always appears in the input.
* Similarly, only one letter "G" always appears in the input.
* In the initial state, Mr. B's right hand touches the wall on the right from the direction in which Mr. B is facing.
Input
The input is given in the following format.
N M
S1
S2
::
SN
Si (1 ≤ i ≤ N) is a string of M characters, each character representing the following.
* "^", "V", "<", ">" indicate the first position of Mr. B and the direction (up, down, left, right).
* "." Is an empty cell. Mr. B can move on this square.
* "#" Represents a wall. You cannot move on the wall.
* "G" represents the location of Mr. A's house. Mr. B repeats the movement until he reaches Mr. A's house.
Output
If you can reach Mr. A's house without releasing your right hand, output the minimum number of different squares that you have passed to reach Mr. A's house. If you cannot reach Mr. A's house, output -1.
Examples
Input
3 3
G##
.#.
.
Output
4
Input
3 3
G##
.#.
.<.
Output
4
Input
3 3
G##
.#.
.>.
Output
6
Input
3 3
...
.G.
.>.
Output
-1
Input
4 4
....
.#.G
...#
^#..
Output
8 | ```python
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m = LI()
a = ['#'*(m+2)] + ['#' + S() + '#' for _ in range(n)] + ['#'*(m+2)]
st = (-1,-1,-1)
for i in range(1,n+1):
for j in range(m+1):
c = a[i][j]
if c == '^':
st = (i,j,0,1)
elif c == 'v':
st = (i,j,2,1)
elif c == '>':
st = (i,j,1,1)
elif c == '<':
st = (i,j,3,1)
v = collections.defaultdict(bool)
vs = set()
r = -1
kb = [[(-1,1), (0,1), (1,1)], [(1,1),(1,0),(1,-1)], [(1,-1),(0,-1),(-1,-1)], [(-1,-1),(-1,0),(-1,1)]]
while True:
if v[st]:
return -1
v[st] = True
i,j,di,ki = st
vs.add((i,j))
if a[i][j] == 'G':
return len(vs)
if ki < 2:
if a[i+dd[di][0]][j+dd[di][1]] != '#':
r += 1
st = (i+dd[di][0], j+dd[di][1], di, ki+1)
else:
st = (i, j, (di-1)%4, 1)
else:
dk = (di+1) % 4
if a[i+dd[dk][0]][j+dd[dk][1]] == '#':
st = (i, j, di, 1)
else:
st = (i, j, dk, 0)
return -1
print(main())
``` |
p01892 SNS | Problem statement
AOR Ika-chan is in a bad mood lately. Apparently, I don't like the ratio of the number of followers to the number of followers of "Ikatta". Currently, AOR Ika-chan has $ A $ followers, $ B $ followers, and a ratio of $ A: B $.
Therefore, AOR Ika decided to increase or decrease the number of followers so that the ratio of the number of followers to the number of followers would be an integer ratio that she liked. The integer ratio I like is a ratio that can be expressed so that both values included in the ratio are integers of $ 1 $ or more and $ N $ or less.
However, AOR Ika doesn't want to change the number of followers as much as possible, so I want to make the absolute value of the difference from before the change as small as possible. Create a program that asks at least how many followers you need to change to make AOR Ika feel good.
Input constraints
$ 1 \ le A, \ B \ le 10 ^ {12} $
$ 1 \ leq N \ leq 100 $
sample
Sample input 1
19 30 3
Sample output 1
1
Sample input 2
3 7 7
Sample output 2
0
Sample input 3
3 7 1
Sample output 3
Four
Sample input 4
102 30 3
Sample output 4
12
By reducing the number of followers by $ 12 $, it becomes $ 90: 30 \ (= 3: 1) $, and the higher ratio number is $ 3 $ or less.
At this time, the amount of change is $ 12 $.
Sample input 5
3 4 2
Sample output 5
1
If you unfollow one person, it will be $ 2: 4 \ (= 1: 2) $, and if you follow it, it will be $ 4: 4 \ (= 1: 1) $. In both cases, the absolute value of increase / decrease is $ 1 $, which is the answer.
Sample input 6
1 100 2
Sample output 6
49
Please note that at least $ 1 $ people must be following.
input
$ A \ B \ N $
output
Output the minimum absolute value of the amount of change in $ A $, which can be an integer ratio you like.
Example
Input
19 30 3
Output
1 | ```python
a, b, n = map(int, input().split())
score = abs(a - b)
for x in range(1, n + 1):
for y in range(1, n + 1):
if b * x % y == 0:
new_a = b * x // y
score = min(score, abs(a - new_a))
print(score)
``` |
p02312 Intersection of a Circle and a Polygon | Find the area of intersection between a circle $c$ and a polygon $g$. The center coordinate of the circle is ($0, 0$).
The polygon $g$ is represented by a sequence of points $p_1$, $p_2$,..., $p_n$ where line segments connecting $p_i$ and $p_{i+1}$ ($1 \leq i \leq n−1$) are sides of the polygon. The line segment connecting $p_n$ and $p_1$ is also a side of the polygon.
Note that the polygon is not necessarily convex.
Constraints
* $3 \leq n \leq 100$
* $1 \leq r \leq 100$
* $-100 \leq x_i, y_i \leq 100$
Input
The input is given in the following format.
$n$ $r$
$x_1$ $y_1$
$x_2$ $y_2$
:
$x_n$ $y_n$
In the first line, an integer n representing the number of points in the polygon is given. The coordinate of a point $p_i$ is given by two integers $x_i$ and $y_i$. The coordinates of the points are given in the order of counter-clockwise visit of them. All input values are given in integers.
Output
Print the area of intersection in a line. The output values should be in a decimal fraction with an error less than 0.00001.
Examples
Input
3 5
1 1
4 1
5 5
Output
4.639858417607
Input
4 5
0 0
-3 -6
1 -3
5 -4
Output
11.787686807576 | ```python
import cmath
import itertools
import math
from collections import defaultdict
INF = float("inf")
PI = cmath.pi
TAU = cmath.pi * 2
EPS = 1e-10
class Point:
"""
2次元空間上の点
"""
# 反時計回り側にある
CCW_COUNTER_CLOCKWISE = 1
# 時計回り側にある
CCW_CLOCKWISE = -1
# 線分の後ろにある
CCW_ONLINE_BACK = 2
# 線分の前にある
CCW_ONLINE_FRONT = -2
# 線分上にある
CCW_ON_SEGMENT = 0
def __init__(self, x: float, y: float):
self.c = complex(x, y)
@property
def x(self):
return self.c.real
@property
def y(self):
return self.c.imag
@staticmethod
def from_complex(c: complex):
return Point(c.real, c.imag)
@staticmethod
def from_polar(r: float, phi: float):
c = cmath.rect(r, phi)
return Point(c.real, c.imag)
def __add__(self, p):
"""
:param Point p:
"""
c = self.c + p.c
return Point(c.real, c.imag)
def __iadd__(self, p):
"""
:param Point p:
"""
self.c += p.c
return self
def __sub__(self, p):
"""
:param Point p:
"""
c = self.c - p.c
return Point(c.real, c.imag)
def __isub__(self, p):
"""
:param Point p:
"""
self.c -= p.c
return self
def __mul__(self, f: float):
c = self.c * f
return Point(c.real, c.imag)
def __imul__(self, f: float):
self.c *= f
return self
def __truediv__(self, f: float):
c = self.c / f
return Point(c.real, c.imag)
def __itruediv__(self, f: float):
self.c /= f
return self
def __repr__(self):
return "({}, {})".format(round(self.x, 10), round(self.y, 10))
def __neg__(self):
c = -self.c
return Point(c.real, c.imag)
def __eq__(self, p):
return abs(self.c - p.c) < EPS
def __abs__(self):
return abs(self.c)
@staticmethod
def ccw(a, b, c):
"""
線分 ab に対する c の位置
線分上にあるか判定するだけなら on_segment とかのが速い
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_C&lang=ja
:param Point a:
:param Point b:
:param Point c:
"""
b = b - a
c = c - a
det = b.det(c)
if det > EPS:
return Point.CCW_COUNTER_CLOCKWISE
if det < -EPS:
return Point.CCW_CLOCKWISE
if b.dot(c) < -EPS:
return Point.CCW_ONLINE_BACK
if c.norm() - b.norm() > EPS:
return Point.CCW_ONLINE_FRONT
return Point.CCW_ON_SEGMENT
def dot(self, p):
"""
内積
:param Point p:
:rtype: float
"""
return self.x * p.x + self.y * p.y
def det(self, p):
"""
外積
:param Point p:
:rtype: float
"""
return self.x * p.y - self.y * p.x
def dist(self, p):
"""
距離
:param Point p:
:rtype: float
"""
return abs(self.c - p.c)
def norm(self):
"""
原点からの距離
:rtype: float
"""
return abs(self.c)
def phase(self):
"""
原点からの角度
:rtype: float
"""
return cmath.phase(self.c)
def angle(self, p, q):
"""
p に向いてる状態から q まで反時計回りに回転するときの角度
-pi <= ret <= pi
:param Point p:
:param Point q:
:rtype: float
"""
return (cmath.phase(q.c - self.c) - cmath.phase(p.c - self.c) + PI) % TAU - PI
def area(self, p, q):
"""
p, q となす三角形の面積
:param Point p:
:param Point q:
:rtype: float
"""
return abs((p - self).det(q - self) / 2)
def projection_point(self, p, q, allow_outer=False):
"""
線分 pq を通る直線上に垂線をおろしたときの足の座標
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_A&lang=ja
:param Point p:
:param Point q:
:param allow_outer: 答えが線分の間になくても OK
:rtype: Point|None
"""
diff_q = q - p
# 答えの p からの距離
r = (self - p).dot(diff_q) / abs(diff_q)
# 線分の角度
phase = diff_q.phase()
ret = Point.from_polar(r, phase) + p
if allow_outer or (p - ret).dot(q - ret) < EPS:
return ret
return None
def reflection_point(self, p, q):
"""
直線 pq を挟んで反対にある点
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_B&lang=ja
:param Point p:
:param Point q:
:rtype: Point
"""
# 距離
r = abs(self - p)
# pq と p-self の角度
angle = p.angle(q, self)
# 直線を挟んで角度を反対にする
angle = (q - p).phase() - angle
return Point.from_polar(r, angle) + p
def on_segment(self, p, q, allow_side=True):
"""
点が線分 pq の上に乗っているか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_C&lang=ja
:param Point p:
:param Point q:
:param allow_side: 端っこでギリギリ触れているのを許容するか
:rtype: bool
"""
if not allow_side and (self == p or self == q):
return False
# 外積がゼロ: 面積がゼロ == 一直線
# 内積がマイナス: p - self - q の順に並んでる
return abs((p - self).det(q - self)) < EPS and (p - self).dot(q - self) < EPS
class Line:
"""
2次元空間上の直線
"""
def __init__(self, a: float, b: float, c: float):
"""
直線 ax + by + c = 0
"""
self.a = a
self.b = b
self.c = c
@staticmethod
def from_gradient(grad: float, intercept: float):
"""
直線 y = ax + b
:param grad: 傾き
:param intercept: 切片
:return:
"""
return Line(grad, -1, intercept)
@staticmethod
def from_segment(p1, p2):
"""
:param Point p1:
:param Point p2:
"""
a = p2.y - p1.y
b = p1.x - p2.x
c = p2.y * (p2.x - p1.x) - p2.x * (p2.y - p1.y)
return Line(a, b, c)
@property
def gradient(self):
"""
傾き
"""
return INF if self.b == 0 else -self.a / self.b
@property
def intercept(self):
"""
切片
"""
return INF if self.b == 0 else -self.c / self.b
def is_parallel_to(self, l):
"""
平行かどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Line l:
"""
# 法線ベクトル同士の外積がゼロ
return abs(Point.from_rect(self.a, self.b).det(Point.from_rect(l.a, l.b))) < EPS
def is_orthogonal_to(self, l):
"""
直行しているかどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Line l:
"""
# 法線ベクトル同士の内積がゼロ
return abs(Point.from_rect(self.a, self.b).dot(Point.from_rect(l.a, l.b))) < EPS
def intersection_point(self, l):
"""
交差する点
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_C&lang=ja
:param Line l:
:rtype: Point|None
"""
a1, b1, c1 = self.a, self.b, self.c
a2, b2, c2 = l.a, l.b, l.c
det = a1 * b2 - a2 * b1
if abs(det) < EPS:
# 並行
return None
x = (b1 * c2 - b2 * c1) / det
y = (a2 * c1 - a1 * c2) / det
return Point.from_rect(x, y)
def dist(self, p):
"""
他の点との最短距離
:param Point p:
"""
raise NotImplementedError()
def has_point(self, p):
"""
p が直線上に乗っているかどうか
:param Point p:
"""
return abs(self.a * p.x + self.b * p.y + self.c) < EPS
class Segment:
"""
2次元空間上の線分
"""
def __init__(self, p1, p2):
"""
:param Point p1:
:param Point p2:
"""
self.p1 = p1
self.p2 = p2
def norm(self):
"""
線分の長さ
"""
return abs(self.p1 - self.p2)
def phase(self):
"""
p1 を原点としたときの p2 の角度
"""
return (self.p2 - self.p1).phase()
def is_parallel_to(self, s):
"""
平行かどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Segment s:
:return:
"""
# 外積がゼロ
return abs((self.p1 - self.p2).det(s.p1 - s.p2)) < EPS
def is_orthogonal_to(self, s):
"""
直行しているかどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Segment s:
:return:
"""
# 内積がゼロ
return abs((self.p1 - self.p2).dot(s.p1 - s.p2)) < EPS
def intersects_with(self, s, allow_side=True):
"""
交差するかどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_B&lang=ja
:param Segment s:
:param allow_side: 端っこでギリギリ触れているのを許容するか
"""
if self.is_parallel_to(s):
# 並行なら線分の端点がもう片方の線分の上にあるかどうか
return (s.p1.on_segment(self.p1, self.p2, allow_side) or
s.p2.on_segment(self.p1, self.p2, allow_side) or
self.p1.on_segment(s.p1, s.p2, allow_side) or
self.p2.on_segment(s.p1, s.p2, allow_side))
else:
# allow_side ならゼロを許容する
det_upper = EPS if allow_side else -EPS
ok = True
# self の両側に s.p1 と s.p2 があるか
ok &= (self.p2 - self.p1).det(s.p1 - self.p1) * (self.p2 - self.p1).det(s.p2 - self.p1) < det_upper
# s の両側に self.p1 と self.p2 があるか
ok &= (s.p2 - s.p1).det(self.p1 - s.p1) * (s.p2 - s.p1).det(self.p2 - s.p1) < det_upper
return ok
def closest_point(self, p):
"""
線分上の、p に最も近い点
:param Point p:
"""
# p からおろした垂線までの距離
d = (p - self.p1).dot(self.p2 - self.p1) / self.norm()
# p1 より前
if d < EPS:
return self.p1
# p2 より後
if -EPS < d - self.norm():
return self.p2
# 線分上
return Point.from_polar(d, (self.p2 - self.p1).phase()) + self.p1
def dist(self, p):
"""
他の点との最短距離
:param Point p:
"""
return abs(p - self.closest_point(p))
def dist_segment(self, s):
"""
他の線分との最短距離
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_D&lang=ja
:param Segment s:
"""
if self.intersects_with(s):
return 0.0
return min(
self.dist(s.p1),
self.dist(s.p2),
s.dist(self.p1),
s.dist(self.p2),
)
def has_point(self, p, allow_side=True):
"""
p が線分上に乗っているかどうか
:param Point p:
:param allow_side: 端っこでギリギリ触れているのを許容するか
"""
return p.on_segment(self.p1, self.p2, allow_side=allow_side)
class Polygon:
"""
2次元空間上の多角形
"""
def __init__(self, points):
"""
:param list of Point points:
"""
self.points = points
def iter2(self):
"""
隣り合う2点を順に返すイテレータ
:rtype: typing.Iterator[(Point, Point)]
"""
return zip(self.points, self.points[1:] + self.points[:1])
def iter3(self):
"""
隣り合う3点を順に返すイテレータ
:rtype: typing.Iterator[(Point, Point, Point)]
"""
return zip(self.points,
self.points[1:] + self.points[:1],
self.points[2:] + self.points[:2])
def area(self):
"""
面積
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_3_A&lang=ja
"""
# 外積の和 / 2
dets = []
for p, q in self.iter2():
dets.append(p.det(q))
return abs(math.fsum(dets)) / 2
def is_convex(self, allow_straight=False, allow_collapsed=False):
"""
凸多角形かどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_3_B&lang=ja
:param allow_straight: 3点がまっすぐ並んでるのを許容するかどうか
:param allow_collapsed: 面積がゼロの場合を許容するか
"""
ccw = []
for a, b, c in self.iter3():
ccw.append(Point.ccw(a, b, c))
ccw = set(ccw)
if len(ccw) == 1:
if ccw == {Point.CCW_CLOCKWISE}:
return True
if ccw == {Point.CCW_COUNTER_CLOCKWISE}:
return True
if allow_straight and len(ccw) == 2:
if ccw == {Point.CCW_ONLINE_FRONT, Point.CCW_CLOCKWISE}:
return True
if ccw == {Point.CCW_ONLINE_FRONT, Point.CCW_COUNTER_CLOCKWISE}:
return True
if allow_collapsed and len(ccw) == 3:
return ccw == {Point.CCW_ONLINE_FRONT, Point.CCW_ONLINE_BACK, Point.CCW_ON_SEGMENT}
return False
def has_point_on_edge(self, p):
"""
指定した点が辺上にあるか
:param Point p:
:rtype: bool
"""
for a, b in self.iter2():
if p.on_segment(a, b):
return True
return False
def contains(self, p, allow_on_edge=True):
"""
指定した点を含むか
Winding Number Algorithm
https://www.nttpc.co.jp/technology/number_algorithm.html
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_3_C&lang=ja
:param Point p:
:param bool allow_on_edge: 辺上の点を許容するか
"""
angles = []
for a, b in self.iter2():
if p.on_segment(a, b):
return allow_on_edge
angles.append(p.angle(a, b))
# 一周以上するなら含む
return abs(math.fsum(angles)) > EPS
@staticmethod
def convex_hull(points, allow_straight=False):
"""
凸包。x が最も小さい点のうち y が最も小さい点から反時計回り。
Graham Scan O(N log N)
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_4_A&lang=ja
:param list of Point points:
:param allow_straight: 3点がまっすぐ並んでるのを許容するかどうか
:rtype: list of Point
"""
points = points[:]
points.sort(key=lambda p: (p.x, p.y))
# allow_straight なら 0 を許容する
det_lower = -EPS if allow_straight else EPS
sz = 0
#: :type: list of (Point|None)
ret = [None] * (len(points) * 2)
for p in points:
while sz > 1 and (ret[sz - 1] - ret[sz - 2]).det(p - ret[sz - 1]) < det_lower:
sz -= 1
ret[sz] = p
sz += 1
floor = sz
for p in reversed(points[:-1]):
while sz > floor and (ret[sz - 1] - ret[sz - 2]).det(p - ret[sz - 1]) < det_lower:
sz -= 1
ret[sz] = p
sz += 1
ret = ret[:sz - 1]
if allow_straight and len(ret) > len(points):
# allow_straight かつ全部一直線のときに二重にカウントしちゃう
ret = points
return ret
@staticmethod
def diameter(points):
"""
直径
凸包構築 O(N log N) + カリパー法 O(N)
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_4_B&lang=ja
:param list of Point points:
"""
# 反時計回り
points = Polygon.convex_hull(points, allow_straight=False)
if len(points) == 1:
return 0.0
if len(points) == 2:
return abs(points[0] - points[1])
# x軸方向に最も遠い点対
si = points.index(min(points, key=lambda p: (p.x, p.y)))
sj = points.index(max(points, key=lambda p: (p.x, p.y)))
n = len(points)
ret = 0.0
# 半周回転
i, j = si, sj
while i != sj or j != si:
ret = max(ret, abs(points[i] - points[j]))
ni = (i + 1) % n
nj = (j + 1) % n
# 2つの辺が並行になる方向にずらす
if (points[ni] - points[i]).det(points[nj] - points[j]) > 0:
j = nj
else:
i = ni
return ret
def convex_cut_by_line(self, line_p1, line_p2):
"""
凸多角形を直線 line_p1-line_p2 でカットする。
凸じゃないといけません
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_4_C&lang=ja
:param line_p1:
:param line_p2:
:return: (line_p1-line_p2 の左側の多角形, line_p1-line_p2 の右側の多角形)
:rtype: (Polygon|None, Polygon|None)
"""
n = len(self.points)
line = Line.from_segment(line_p1, line_p2)
# 直線と重なる点
on_line_points = []
for i, p in enumerate(self.points):
if line.has_point(p):
on_line_points.append(i)
# 辺が直線上にある
has_on_line_edge = False
if len(on_line_points) >= 3:
has_on_line_edge = True
elif len(on_line_points) == 2:
# 直線上にある点が隣り合ってる
has_on_line_edge = abs(on_line_points[0] - on_line_points[1]) in [1, n - 1]
# 辺が直線上にある場合、どっちか片方に全部ある
if has_on_line_edge:
for p in self.points:
ccw = Point.ccw(line_p1, line_p2, p)
if ccw == Point.CCW_COUNTER_CLOCKWISE:
return Polygon(self.points[:]), None
if ccw == Point.CCW_CLOCKWISE:
return None, Polygon(self.points[:])
ret_lefts = []
ret_rights = []
d = line_p2 - line_p1
for p, q in self.iter2():
det_p = d.det(p - line_p1)
det_q = d.det(q - line_p1)
if det_p > -EPS:
ret_lefts.append(p)
if det_p < EPS:
ret_rights.append(p)
# 外積の符号が違う == 直線の反対側にある場合は交点を追加
if det_p * det_q < -EPS:
intersection = line.intersection_point(Line.from_segment(p, q))
ret_lefts.append(intersection)
ret_rights.append(intersection)
# 点のみの場合を除いて返す
l = Polygon(ret_lefts) if len(ret_lefts) > 1 else None
r = Polygon(ret_rights) if len(ret_rights) > 1 else None
return l, r
def closest_pair(points):
"""
最近点対 O(N log N)
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_5_A&lang=ja
:param list of Point points:
:rtype: (float, (Point, Point))
:return: (距離, 点対)
"""
assert len(points) >= 2
def _rec(xsorted):
"""
:param list of Point xsorted:
:rtype: (float, (Point, Point))
"""
n = len(xsorted)
if n <= 2:
return xsorted[0].dist(xsorted[1]), (xsorted[0], xsorted[1])
if n <= 3:
# 全探索
d = INF
pair = None
for p, q in itertools.combinations(xsorted, r=2):
if p.dist(q) < d:
d = p.dist(q)
pair = p, q
return d, pair
# 分割統治
# 両側の最近点対
ld, lp = _rec(xsorted[:n // 2])
rd, rp = _rec(xsorted[n // 2:])
if ld <= rd:
d = ld
ret_pair = lp
else:
d = rd
ret_pair = rp
mid_x = xsorted[n // 2].x
# 中央から d 以内のやつを集める
mid_points = []
for p in xsorted:
# if abs(p.x - mid_x) < d:
if abs(p.x - mid_x) - d < -EPS:
mid_points.append(p)
# この中で距離が d 以内のペアがあれば更新
mid_points.sort(key=lambda p: p.y)
mid_n = len(mid_points)
for i in range(mid_n - 1):
j = i + 1
p = mid_points[i]
q = mid_points[j]
# while q.y - p.y < d
while (q.y - p.y) - d < -EPS:
pq_d = p.dist(q)
if pq_d < d:
d = pq_d
ret_pair = p, q
j += 1
if j >= mid_n:
break
q = mid_points[j]
return d, ret_pair
return _rec(list(sorted(points, key=lambda p: p.x)))
def closest_pair_randomized(points):
"""
最近点対 乱択版 O(N)
http://ir5.hatenablog.com/entry/20131221/1387557630
グリッドの管理が dict だから定数倍気になる
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_5_A&lang=ja
:param list of Point points:
:rtype: (float, (Point, Point))
:return: (距離, 点対)
"""
n = len(points)
assert n >= 2
if n == 2:
return points[0].dist(points[1]), (points[0], points[1])
# 逐次構成法
import random
points = points[:]
random.shuffle(points)
DELTA_XY = list(itertools.product([-1, 0, 1], repeat=2))
grid = defaultdict(list)
delta = INF
dist = points[0].dist(points[1])
ret_pair = points[0], points[1]
for i in range(2, n):
if delta < EPS:
return 0.0, ret_pair
# i 番目より前までを含む grid を構築
# if dist < delta:
if dist - delta < -EPS:
delta = dist
grid = defaultdict(list)
for a in points[:i]:
grid[a.x // delta, a.y // delta].append(a)
else:
p = points[i - 1]
grid[p.x // delta, p.y // delta].append(p)
p = points[i]
dist = delta
grid_x = p.x // delta
grid_y = p.y // delta
# 周り 9 箇所だけ調べれば OK
for dx, dy in DELTA_XY:
for q in grid[grid_x + dx, grid_y + dy]:
d = p.dist(q)
# if d < dist:
if d - dist < -EPS:
dist = d
ret_pair = p, q
return min(delta, dist), ret_pair
class Circle:
def __init__(self, o, r):
"""
:param Point o:
:param float r:
"""
self.o = o
self.r = r
def __eq__(self, other):
return self.o == other.o and abs(self.r - other.r) < EPS
def ctc(self, c):
"""
共通接線 common tangent の数
4: 離れてる
3: 外接
2: 交わってる
1: 内接
0: 内包
inf: 同一
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_7_A&lang=ja
:param Circle c:
:rtype: int
"""
if self.o == c.o:
return INF if abs(self.r - c.r) < EPS else 0
# 円同士の距離
d = self.o.dist(c.o) - self.r - c.r
if d > EPS:
return 4
elif d > -EPS:
return 3
# elif d > -min(self.r, c.r) * 2:
elif d + min(self.r, c.r) * 2 > EPS:
return 2
elif d + min(self.r, c.r) * 2 > -EPS:
return 1
return 0
def has_point_on_edge(self, p):
"""
指定した点が円周上にあるか
:param Point p:
:rtype: bool
"""
return abs(self.o.dist(p) - self.r) < EPS
def contains(self, p, allow_on_edge=True):
"""
指定した点を含むか
:param Point p:
:param bool allow_on_edge: 辺上の点を許容するか
"""
if allow_on_edge:
# return self.o.dist(p) <= self.r
return self.o.dist(p) - self.r < EPS
else:
# return self.o.dist(p) < self.r
return self.o.dist(p) - self.r < -EPS
def area(self):
"""
面積
"""
return self.r ** 2 * PI
def circular_segment_area(self, angle):
"""
弓形⌓の面積
:param float angle: 角度ラジアン
"""
# 扇形の面積
sector_area = self.area() * angle / TAU
# 三角形部分を引く
return sector_area - self.r ** 2 * math.sin(angle) / 2
def intersection_points(self, other, allow_outer=False):
"""
:param Segment|Circle other:
:param bool allow_outer:
"""
if isinstance(other, Segment):
return self.intersection_points_with_segment(other, allow_outer=allow_outer)
if isinstance(other, Circle):
return self.intersection_points_with_circle(other)
raise NotImplementedError()
def intersection_points_with_segment(self, s, allow_outer=False):
"""
線分と交差する点のリスト
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_7_D&lang=ja
:param Segment s:
:param bool allow_outer: 線分の間にない点を含む
:rtype: list of Point
"""
# 垂線の足
projection_point = self.o.projection_point(s.p1, s.p2, allow_outer=True)
# 線分との距離
dist = self.o.dist(projection_point)
# if dist > self.r:
if dist - self.r > EPS:
return []
if dist - self.r > -EPS:
if allow_outer or s.has_point(projection_point):
return [projection_point]
else:
return []
# 足から左右に diff だけ動かした座標が答え
diff = Point.from_polar(math.sqrt(self.r ** 2 - dist ** 2), s.phase())
ret1 = projection_point + diff
ret2 = projection_point - diff
ret = []
if allow_outer or s.has_point(ret1):
ret.append(ret1)
if allow_outer or s.has_point(ret2):
ret.append(ret2)
return ret
def intersection_points_with_circle(self, other):
"""
円と交差する点のリスト
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_7_E&langja
:param circle other:
:rtype: list of Point
"""
ctc = self.ctc(other)
if not 1 <= ctc <= 3:
return []
if ctc == 3:
# 外接
return [Point.from_polar(self.r, (other.o - self.o).phase()) + self.o]
if ctc == 1:
# 内接
if self.r > other.r:
return [Point.from_polar(self.r, (other.o - self.o).phase()) + self.o]
else:
return [Point.from_polar(self.r, (self.o - other.o).phase()) + self.o]
# 2つ交点がある
assert ctc == 2
a = other.r
b = self.r
c = self.o.dist(other.o)
# 余弦定理で cos(a) を求めます
cos_a = (b ** 2 + c ** 2 - a ** 2) / (2 * b * c)
angle = math.acos(cos_a)
phi = (other.o - self.o).phase()
return [
self.o + Point.from_polar(self.r, phi + angle),
self.o + Point.from_polar(self.r, phi - angle),
]
def tangent_points_with_point(self, p):
"""
p を通る接線との接点
:param Point p:
:rtype: list of Point
"""
dist = self.o.dist(p)
# if dist < self.r:
if dist - self.r < -EPS:
# p が円の内部にある
return []
if dist - self.r < EPS:
# p が円周上にある
return [Point(p.x, p.y)]
a = math.sqrt(dist ** 2 - self.r ** 2)
b = self.r
c = dist
# 余弦定理で cos(a) を求めます
cos_a = (b ** 2 + c ** 2 - a ** 2) / (2 * b * c)
angle = math.acos(cos_a)
phi = (p - self.o).phase()
return [
self.o + Point.from_polar(self.r, phi + angle),
self.o + Point.from_polar(self.r, phi - angle),
]
def tangent_points_with_circle(self, other):
"""
other との共通接線との接点
:param Circle other:
:rtype: list of Point
"""
ctc = self.ctc(other)
if ctc > 4:
raise ValueError('2つの円が同一です')
if ctc == 0:
return []
if ctc == 1:
return self.intersection_points_with_circle(other)
assert ctc in (2, 3, 4)
ret = []
# 共通外接線を求める
# if self.r == other.r:
if abs(self.r - other.r) < EPS:
# 半径が同じ == 2つの共通外接線が並行
phi = (other.o - self.o).phase()
ret.append(self.o + Point.from_polar(self.r, phi + PI / 2))
ret.append(self.o + Point.from_polar(self.r, phi - PI / 2))
else:
# 2つの共通外接線の交点から接線を引く
intersection = self.o + (other.o - self.o) / (self.r - other.r) * self.r
ret += self.tangent_points_with_point(intersection)
# 共通内接線を求める
# 2つの共通内接線の交点から接線を引く
intersection = self.o + (other.o - self.o) / (self.r + other.r) * self.r
ret += self.tangent_points_with_point(intersection)
return ret
def round_point(p):
"""
8桁で四捨五入して、-0を0に変換する
:param p:
:return:
"""
return round(p.x, 10) + 0, round(p.y, 10) + 0
def solve():
import sys
N, R = list(map(int, sys.stdin.buffer.readline().split()))
XY = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N)]
circle = Circle(Point(0, 0), R)
points = []
for x, y in XY:
points.append(Point(x, y))
polygon = Polygon(points)
# 円との交点を含む点リスト
points_with_on_circle = []
for p, q in polygon.iter2():
points_with_on_circle.append(p)
s = Segment(p, q)
intersections = circle.intersection_points_with_segment(s, allow_outer=False)
# p に近い順にする
intersections.sort(key=lambda ip: p.dist(ip))
for ip in intersections:
if ip != p and ip != q:
points_with_on_circle.append(ip)
outer_index = None
for i, p in enumerate(points_with_on_circle):
if not circle.contains(p, allow_on_edge=True):
outer_index = i
break
# 全部円の内側
if outer_index is None:
return polygon.area()
inner_index = None
i = outer_index + 1
while i != outer_index:
j = (i + 1) % len(points_with_on_circle)
p = points_with_on_circle[i]
q = points_with_on_circle[j]
if circle.contains(p, allow_on_edge=True) and circle.contains(q, allow_on_edge=True):
inner_index = i
break
i = j
# 全部円の外側
if inner_index is None:
return 0.0
points_with_on_circle = points_with_on_circle[inner_index:] + points_with_on_circle[:inner_index]
points_on_circle = []
# 次外に出るか中に入るか
for pl, p, pr in Polygon(points_with_on_circle).iter3():
if circle.has_point_on_edge(p) and (circle.contains(pl) or circle.contains(pr)):
points_on_circle.append((p, not circle.contains(pr, allow_on_edge=True)))
# 偏角ソート
points_on_circle.sort(key=lambda pf: (pf[0] - circle.o).phase())
seen = [False] * len(points_on_circle)
ret = 0
# O(N) にしたい
while not all(seen):
partial_points = []
i = seen.index(False)
p, out = points_on_circle[i]
seen[i] = True
partial_points.append(p)
j = points_with_on_circle.index(p)
end_p = p
first = True
while first or p != end_p:
# 円周上かつ out なら次の円周上の点に飛ばす
if out:
i = (i + 1) % len(points_on_circle)
q, out = points_on_circle[i]
ret += circle.circular_segment_area(circle.o.angle(p, q) % TAU)
p = q
partial_points.append(p)
j = points_with_on_circle.index(p)
else:
j = (j + 1) % len(points_with_on_circle)
p = points_with_on_circle[j]
partial_points.append(p)
if circle.has_point_on_edge(p):
i = [i for i, (q, _) in enumerate(points_on_circle) if p == q][0]
p, out = points_on_circle[i]
if seen[i]:
pp = [p]
partial_points.pop()
while partial_points and partial_points[-1] != p:
pp.append(partial_points.pop())
ret += Polygon(pp).area()
out = True
seen[i] = True
first = False
return ret
print(solve())
``` |
p02460 Map: Delete | For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print the value with the specified $key$. Print 0 if there is no such element.
* delete($key$): Delete the element with the specified $key$.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consits of lower case letters
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
where the first digits 0, 1 and 2 represent insert, get and delete operations respectively.
Output
For each get operation, print an integer in a line.
Example
Input
8
0 blue 4
0 red 1
0 white 5
1 red
1 blue
2 red
1 black
1 red
Output
1
4
0
0 | ```python
import sys
d = {}
input()
for q in sys.stdin:
q = q.split()
if q[0] == '0':
d[q[1]] = q[2]
elif q[0] == '1':
print(d.get(q[1], 0))
else:
if q[1] in d:
del d[q[1]]
``` |
1004_D. Sonya and Matrix | Since Sonya has just learned the basics of matrices, she decided to play with them a little bit.
Sonya imagined a new type of matrices that she called rhombic matrices. These matrices have exactly one zero, while all other cells have the Manhattan distance to the cell containing the zero. The cells with equal numbers have the form of a rhombus, that is why Sonya called this type so.
The Manhattan distance between two cells (x_1, y_1) and (x_2, y_2) is defined as |x_1 - x_2| + |y_1 - y_2|. For example, the Manhattan distance between the cells (5, 2) and (7, 1) equals to |5-7|+|2-1|=3.
<image> Example of a rhombic matrix.
Note that rhombic matrices are uniquely defined by n, m, and the coordinates of the cell containing the zero.
She drew a n× m rhombic matrix. She believes that you can not recreate the matrix if she gives you only the elements of this matrix in some arbitrary order (i.e., the sequence of n⋅ m numbers). Note that Sonya will not give you n and m, so only the sequence of numbers in this matrix will be at your disposal.
Write a program that finds such an n× m rhombic matrix whose elements are the same as the elements in the sequence in some order.
Input
The first line contains a single integer t (1≤ t≤ 10^6) — the number of cells in the matrix.
The second line contains t integers a_1, a_2, …, a_t (0≤ a_i< t) — the values in the cells in arbitrary order.
Output
In the first line, print two positive integers n and m (n × m = t) — the size of the matrix.
In the second line, print two integers x and y (1≤ x≤ n, 1≤ y≤ m) — the row number and the column number where the cell with 0 is located.
If there are multiple possible answers, print any of them. If there is no solution, print the single integer -1.
Examples
Input
20
1 0 2 3 5 3 2 1 3 2 3 1 4 2 1 4 2 3 2 4
Output
4 5
2 2
Input
18
2 2 3 2 4 3 3 3 0 2 4 2 1 3 2 1 1 1
Output
3 6
2 3
Input
6
2 1 0 2 1 2
Output
-1
Note
You can see the solution to the first example in the legend. You also can choose the cell (2, 2) for the cell where 0 is located. You also can choose a 5× 4 matrix with zero at (4, 2).
In the second example, there is a 3× 6 matrix, where the zero is located at (2, 3) there.
In the third example, a solution does not exist. | ```python
def get(n,m,a,b,t):
freq=[0]*(t+1)
for i in range(n):
for j in range(m):
val=abs(i-a)+abs(j-b)
freq[val]+=1
return freq
t=int(input())
a=list(map(int,input().split()))
mx=max(a)
f=[0]*(t+1)
for i in a:
f[i]+=1
b=1
for i in range(1,mx+1):
if f[i]!=4*i:
b=i
break
n=1
a=-1
x=0
y=0
mila=False
while n*n<=t:
if t%n==0:
m=t//n
a=n+m-mx-b
x,y=n,m
if a>0 and a<=n and b>0 and b<=m and f==get(n,m,a-1,b-1,t):
mila=True
break
if a>0 and a<=m and b>0 and b<=n and f==get(n,m,b-1,a-1,t):
mila=True
a,b=b,a
break
n+=1
if not mila:
print(-1)
else:
print(x,y)
print(a,b)
``` |
1028_D. Order book | Let's consider a simplified version of order book of some stock. The order book is a list of orders (offers) from people that want to buy or sell one unit of the stock, each order is described by direction (BUY or SELL) and price.
At every moment of time, every SELL offer has higher price than every BUY offer.
In this problem no two ever existed orders will have the same price.
The lowest-price SELL order and the highest-price BUY order are called the best offers, marked with black frames on the picture below.
<image> The presented order book says that someone wants to sell the product at price 12 and it's the best SELL offer because the other two have higher prices. The best BUY offer has price 10.
There are two possible actions in this orderbook:
1. Somebody adds a new order of some direction with some price.
2. Somebody accepts the best possible SELL or BUY offer (makes a deal). It's impossible to accept not the best SELL or BUY offer (to make a deal at worse price). After someone accepts the offer, it is removed from the orderbook forever.
It is allowed to add new BUY order only with prices less than the best SELL offer (if you want to buy stock for higher price, then instead of adding an order you should accept the best SELL offer). Similarly, one couldn't add a new SELL order with price less or equal to the best BUY offer. For example, you can't add a new offer "SELL 20" if there is already an offer "BUY 20" or "BUY 25" — in this case you just accept the best BUY offer.
You have a damaged order book log (in the beginning the are no orders in book). Every action has one of the two types:
1. "ADD p" denotes adding a new order with price p and unknown direction. The order must not contradict with orders still not removed from the order book.
2. "ACCEPT p" denotes accepting an existing best offer with price p and unknown direction.
The directions of all actions are lost. Information from the log isn't always enough to determine these directions. Count the number of ways to correctly restore all ADD action directions so that all the described conditions are satisfied at any moment. Since the answer could be large, output it modulo 10^9 + 7. If it is impossible to correctly restore directions, then output 0.
Input
The first line contains an integer n (1 ≤ n ≤ 363 304) — the number of actions in the log.
Each of the next n lines contains a string "ACCEPT" or "ADD" and an integer p (1 ≤ p ≤ 308 983 066), describing an action type and price.
All ADD actions have different prices. For ACCEPT action it is guaranteed that the order with the same price has already been added but has not been accepted yet.
Output
Output the number of ways to restore directions of ADD actions modulo 10^9 + 7.
Examples
Input
6
ADD 1
ACCEPT 1
ADD 2
ACCEPT 2
ADD 3
ACCEPT 3
Output
8
Input
4
ADD 1
ADD 2
ADD 3
ACCEPT 2
Output
2
Input
7
ADD 1
ADD 2
ADD 3
ADD 4
ADD 5
ACCEPT 3
ACCEPT 5
Output
0
Note
In the first example each of orders may be BUY or SELL.
In the second example the order with price 1 has to be BUY order, the order with the price 3 has to be SELL order. | ```python
from sys import stdin
import heapq
MOD = pow(10, 9) + 7
n=int(stdin.readline())
a=[]
for i in range(n):
x=stdin.readline().split()
if x[0]=='ADD':
a.append((0,int(x[1])))
else:
a.append((1,int(x[1])))
next_accept=[-1]*n
accept = -1
for i in range(n-1, -1, -1):
if a[i][0]== 1:
accept=i
next_accept[i] = accept
top = []
bottom = []
buysell_n = 0
last_n=0
invalid = False
for i in range(n):
if a[i][0] == 0:
if next_accept[i] != -1:
if a[i][1] > a[next_accept[i]][1]:
heapq.heappush(top, a[i][1])
elif a[i][1] < a[next_accept[i]][1]:
heapq.heappush(bottom, -a[i][1])
elif (len(top) == 0 or a[i][1] < top[0]) and (len(bottom) == 0 or a[i][1] > -bottom[0]):
last_n += 1
else:
if len(top) > 0 and a[i][1] == top[0]:
heapq.heappop(top)
elif len(bottom) > 0 and a[i][1] == -bottom[0]:
heapq.heappop(bottom)
else:
if len(top) > 0 and a[i][1] > top[0] or len(bottom) > 0 and a[i][1] < -bottom[0]:
invalid = True
break
buysell_n += 1
if invalid:
ans = 0
else:
ans = (pow(2, buysell_n, MOD)*(last_n+1))%MOD
print(ans)
``` |
1092_D2. Great Vova Wall (Version 2) | Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches.
The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall.
Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however).
Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it).
Note that Vova can't put bricks vertically.
Vova is a perfectionist, so he considers the wall completed when:
* all parts of the wall has the same height;
* the wall has no empty spaces inside it.
Can Vova complete the wall using any amount of bricks (possibly zero)?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall.
Output
Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero).
Print "NO" otherwise.
Examples
Input
5
2 1 1 2 5
Output
YES
Input
3
4 5 3
Output
NO
Input
2
10 10
Output
YES
Note
In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5].
In the second example Vova can put no bricks in the wall.
In the third example the wall is already complete. | ```python
n=int(input())
a=list(map(int,input().split()))
down=[]
good=True
for guy in a:
if len(down)==0:
down.append(guy)
elif down[-1]>guy:
down.append(guy)
elif down[-1]==guy:
down.pop()
else:
good=False
break
if not good:
print("NO")
elif len(down)>1:
print("NO")
elif len(down)==1 and down[0]!=max(a):
print("NO")
else:
print("YES")
``` |
1111_C. Creative Snap | Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B ⋅ n_a ⋅ l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 ≤ n ≤ 30, 1 ≤ k ≤ 10^5, 1 ≤ A,B ≤ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, …, a_{k} (1 ≤ a_{i} ≤ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer — the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 ⋅ 2 ⋅ 4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 ⋅ 1 ⋅ 2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 ⋅ 1 ⋅ 1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | ```python
def solve (l,r,y) :
if (len(y) == 0) :
return a
if (l == r) :
return b*len(y)
c = []
d = []
m = (l+r)//2
for i in y :
if i <= m :
c += [i]
else :
d += [i]
return min(b*(r-l+1)*len(y) , solve(l,m,c) + solve(m+1,r,d))
n,k,a,b = list(map(int, input().split()))
x = list(map(int, input().split()))
print(solve(1,1<<n,x))
``` |
1141_A. Game 23 | Polycarp plays "Game 23". Initially he has a number n and his goal is to transform it to m. In one move, he can multiply n by 2 or multiply n by 3. He can perform any number of moves.
Print the number of moves needed to transform n to m. Print -1 if it is impossible to do so.
It is easy to prove that any way to transform n to m contains the same number of moves (i.e. number of moves doesn't depend on the way of transformation).
Input
The only line of the input contains two integers n and m (1 ≤ n ≤ m ≤ 5⋅10^8).
Output
Print the number of moves to transform n to m, or -1 if there is no solution.
Examples
Input
120 51840
Output
7
Input
42 42
Output
0
Input
48 72
Output
-1
Note
In the first example, the possible sequence of moves is: 120 → 240 → 720 → 1440 → 4320 → 12960 → 25920 → 51840. The are 7 steps in total.
In the second example, no moves are needed. Thus, the answer is 0.
In the third example, it is impossible to transform 48 to 72. | ```python
def fun(a,b):
if a % b != 0:
return -1
temp = 0
m = a/b
while m % 2 == 0:
m = m / 2
temp += 1
while m % 3 == 0:
m = m/3
temp += 1
if m == 1:
return temp
else :
return -1
a,b = map(int,input().split())
print(fun(b,a))
``` |
1217_C. The Number Of Good Substrings | You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The only line of each query contains string s (1 ≤ |s| ≤ 2 ⋅ 10^5), consisting of only digits 0 and 1.
It is guaranteed that ∑_{i=1}^{t} |s_i| ≤ 2 ⋅ 10^5.
Output
For each query print one integer — the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3 | ```python
#!/usr/bin/env python3
#Educational Codeforces Round 72 C
import sys
import math
import bisect
sys.setrecursionlimit(1000000000)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
t = I()
for _ in range(t):
s = input()
ans = 0
for l in range(1, min(20, len(s))+1):
p = 0
for i in range(len(s)-l+1):
if s[i]=='0':
p += 1
continue
x = int(s[i:i+l], 2)
if x >= l and (x-l)<=p:
ans+=1
p = 0
print(ans)
``` |
1260_A. Heating | Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. | ```python
n=int(input())
for i in range(n):
x = list(map(int, input().split(" ")))
print(((x[1]%x[0])*((x[1]//x[0]+1)**2))+(((x[1]//x[0])**2)*(x[0]-(x[1]%x[0]))))
``` |
1282_B1. K for the Price of One (Easy Version) | This is the easy version of this problem. The only difference is the constraint on k — the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky — today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i — the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p ≥ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p ≥ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ 2⋅10^9, k=2) — the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 ≤ a_i ≤ 10^4) — the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m — the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3 | ```python
"""
Author : thekushalghosh
Team : CodeDiggers
"""
import sys,math
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(s[:len(s) - 1])
def invr():
return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
t = 1
t = inp()
for tt in range(t):
n,p,k = invr()
a = inlt()
a.sort()
b = [0] * (len(a) + k + 3)
q = -1
for i in range(len(a)):
if i < k - 1:
b[i] = b[i - 1] + a[i]
else:
b[i] = b[i - k] + a[i]
for i in range(len(a)):
if p >= b[i]:
q = i
print(q + 1)
``` |
1326_B. Maximums | Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9.
Output
Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. | ```python
input()
b = list(map(int,input().split()))
a = list()
sumnum = 0
negnum = 0
for i in range(len(b)):
sumnum += b[i]
a.append(sumnum+abs(negnum))
if b[i]<0:
negnum += b[i]
print(*a)
``` |
1345_A. Puzzle Pieces | You are given a special jigsaw puzzle consisting of n⋅ m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 ≤ n,m ≤ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image> | ```python
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict
from bisect import bisect_left as bl, bisect_right as br
from collections import Counter
from collections import deque
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
t = ip()
for _ in range(t):
n,m = inp()
if n == 1 or m == 1:
print("YES")
elif n == 2 and m == 2:
print("YES")
else:
print("NO")
``` |
1408_G. Clusterization Counting | There are n computers in the company network. They are numbered from 1 to n.
For each pair of two computers 1 ≤ i < j ≤ n you know the value a_{i,j}: the difficulty of sending data between computers i and j. All values a_{i,j} for i<j are different.
You want to separate all computers into k sets A_1, A_2, …, A_k, such that the following conditions are satisfied:
* for each computer 1 ≤ i ≤ n there is exactly one set A_j, such that i ∈ A_j;
* for each two pairs of computers (s, f) and (x, y) (s ≠ f, x ≠ y), such that s, f, x are from the same set but x and y are from different sets, a_{s,f} < a_{x,y}.
For each 1 ≤ k ≤ n find the number of ways to divide computers into k groups, such that all required conditions are satisfied. These values can be large, so you need to find them by modulo 998 244 353.
Input
The first line contains a single integer n (1 ≤ n ≤ 1500): the number of computers.
The i-th of the next n lines contains n integers a_{i,1}, a_{i,2}, …, a_{i,n}(0 ≤ a_{i,j} ≤ (n (n-1))/(2)).
It is guaranteed that:
* for all 1 ≤ i ≤ n a_{i,i} = 0;
* for all 1 ≤ i < j ≤ n a_{i,j} > 0;
* for all 1 ≤ i < j ≤ n a_{i,j} = a_{j,i};
* all a_{i,j} for i <j are different.
Output
Print n integers: the k-th of them should be equal to the number of possible ways to divide computers into k groups, such that all required conditions are satisfied, modulo 998 244 353.
Examples
Input
4
0 3 4 6
3 0 2 1
4 2 0 5
6 1 5 0
Output
1 0 1 1
Input
7
0 1 18 15 19 12 21
1 0 16 13 17 20 14
18 16 0 2 7 10 9
15 13 2 0 6 8 11
19 17 7 6 0 4 5
12 20 10 8 4 0 3
21 14 9 11 5 3 0
Output
1 1 2 3 4 3 1
Note
Here are all possible ways to separate all computers into 4 groups in the second example:
* \{1, 2\}, \{3, 4\}, \{5\}, \{6, 7\};
* \{1\}, \{2\}, \{3, 4\}, \{5, 6, 7\};
* \{1, 2\}, \{3\}, \{4\}, \{5, 6, 7\}. | ```python
# import itertools as it
# import functools as ft
import math
teststring = """4
0 3 4 6
3 0 2 1
4 2 0 5
6 1 5 0
"""
online = __file__ != "/home/jhli/py/Grakn/Problem_G2.py"
true, false = True, False
if True:
def spitout():
for c in teststring.splitlines():
yield c
_ito = spitout()
if not online:
def input():
return next(_ito)
def build_enum(*a):
built = dict()
for i, c in enumerate(a):
built[c] = i
return lambda x: built[x]
# T = 1
# T = int(input())
##-----------------start coding-----------------
n = int(input())
E = [(0, 0)] * (int(n*(n-1)/2))
P = 998244353
for i in range(n):
L = list(map(int, input().split(" ")))
for j in range(i+1, n):
E[L[j]-1] = (i, j)
R = list(range(n))
C = [[0, 1] for _ in range(n)]
Nv = [1] * n
Ne = [0] * n
def root(x):
if x == R[x]:
return x
else:
R[x] = y = root(R[x])
return y
def prod(A, B, da, db):
C = [0] * (min(da+db, n) + 1)
for i in range(da+1):
for j in range(db+1):
if i + j <= n:
C[i+j] += A[i] * B[j]
C[i+j] %= P
return C
# print(E)
# print("")
for (x, y) in E:
r = rx = root(x)
ry = root(y)
# print((x, y, w), (rx, ry))
if rx != ry:
if r > ry: r = ry
R[rx] = R[ry] = r
C[r] = prod(C[rx], C[ry], Nv[rx], Nv[ry])
Nv[r] = Nv[rx] + Nv[ry]
Ne[r] = Ne[rx] + Ne[ry] + 1
else:
Ne[r] += 1
if Ne[r]*2 == Nv[r] * (Nv[r] - 1):
C[r][1] = 1
# print("R", R)
# print("Nv", Nv)
# print("Ne", Ne)
# print("C", C)
# print("")
print(" ".join(map(str, C[0][1:n+1])))
# print('Case #{}: {}'.format(ti, '...'))
``` |
142_A. Help Farmer | Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | ```python
import math
n = int(input())
mx = 0
mn = 4 * 10**18
for A in range(1, n+1):
if A * A * A > n:
break
if n % A:
continue
nn = n // A
for B in range(1, nn+1):
if B * B > nn:
break
if nn % B:
continue
C = nn // B
mn = min(mn, (A + 1) * (B + 2) * (C + 2) - n)
mx = max(mx, (A + 1) * (B + 2) * (C + 2) - n)
mn = min(mn, (A + 1) * (C + 2) * (B + 2) - n)
mx = max(mx, (A + 1) * (C + 2) * (B + 2) - n)
mn = min(mn, (B + 1) * (A + 2) * (C + 2) - n)
mx = max(mx, (B + 1) * (A + 2) * (C + 2) - n)
mn = min(mn, (B + 1) * (C + 2) * (A + 2) - n)
mx = max(mx, (B + 1) * (C + 2) * (A + 2) - n)
mn = min(mn, (C + 1) * (A + 2) * (B + 2) - n)
mx = max(mx, (C + 1) * (A + 2) * (B + 2) - n)
mn = min(mn, (C + 1) * (B + 2) * (A + 2) - n)
mx = max(mx, (C + 1) * (B + 2) * (A + 2) - n)
print(str(mn) + " " + str(mx))
``` |
1452_E. Two Editorials | Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well. | ```python
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
###########################
# Sorted list
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
#
# # to find factorial and ncr
# tot = 400005
# mod = 10 ** 9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, tot + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
def solve():
n,m,k=sep()
a=[lis() for _ in range(m)]
def solv(lst):
res = sum(lst[:k]);
s = res
for i in range(n - k): s += lst[i + k] - lst[i];res = max(res, s)
return res
a.sort(key=lambda x: sum(x) / 2);
apr = [0] * n
for el in a:
for i in range(el[0] - 1, el[1]): apr[i] += 1
res = solv(apr);
bpr = [0] * n
for r, l in a:
for i in range(r - 1, l): apr[i] -= 1;bpr[i] += 1
nres = solv(apr) + solv(bpr);
res = max(res, nres)
print(res)
solve()
#testcase(int(inp()))
# 5
# 3 6
# 5 10
# 4 3
# 2 1
# 1 3
``` |
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