Instruction
stringlengths 261
35k
| Response
stringclasses 1
value |
|---|---|
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: function calculateCI(P, R, T)
{
let interest = P * (Math.pow(1.0 + R/100.0, T) - 1);
return interest.toFixed(2);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int p,r,t;
cin>>p>>r>>t;
double rate= (float)r/100;
double amt = (float)p*(pow(1+rate,t));
cout << fixed << setprecision(2) << (amt - p);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s= br.readLine().split(" ");
double[] darr = new double[s.length];
for(int i=0;i<s.length;i++){
darr[i] = Double.parseDouble(s[i]);
}
double ans = darr[0]*Math.pow(1+darr[1]/100,darr[2])-darr[0];
System.out.printf("%.2f",ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String[] st = bf.readLine().split(" ");
if(Integer.parseInt(st[1])==0)
System.out.print(-1);
else {
int f = (Integer.parseInt(st[0])/Integer.parseInt(st[1]));
System.out.print(f);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: D,Q = input().split()
D = int(D)
Q = int(Q)
if(0<=D and Q<=100 and Q >0):
print(int(D/Q))
else:
print('-1'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
if(m==0){cout<<-1;return 0;}
cout<<n/m;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: a, b, v = map(int, input().strip().split(" "))
c = abs(a-b)
t = c//v
print(t), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int k = sc.nextInt();
System.out.print(Time(n,m,k));
}
static int Time(int A, int B, int S){
if(B>A){
return (B-A)/S;
}
return (A-B)/S;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A Tribonacci number T(n) is the sum of the preceding three elements in a series.
Calculate the value of the nth Tribonacci number modulo 1000000007. The initial three numbers of this series will be a, b and c i.e., T(1)=a, T(2)=b, T(3)=c.The only line of input contains 4 integers n, a, b, c.
Constraints
4 ≤ n ≤ 10^5
1 ≤ a, b, c ≤ 1000000000Print the result containing one integer i.e., T(n) modulo 1000000007.Input1:
5 1 2 3
Output1:
11
Input2:
4 1 1 1
Output2:
3
Explanation(might now be the optimal solution):
Input 1:
Follow the below steps:-
T(1) = 1
T(2) = 2
T(3) = 3
T(4) = 6
T(5) = 11, I have written this Solution Code: n,a,b,c = map(int, input().split())
m = 1000000007
result = 0
while n > 3:
result = (a%m + b%m + c%m)%m
a = b
b = c
c = result
n -= 1
print(result), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A Tribonacci number T(n) is the sum of the preceding three elements in a series.
Calculate the value of the nth Tribonacci number modulo 1000000007. The initial three numbers of this series will be a, b and c i.e., T(1)=a, T(2)=b, T(3)=c.The only line of input contains 4 integers n, a, b, c.
Constraints
4 ≤ n ≤ 10^5
1 ≤ a, b, c ≤ 1000000000Print the result containing one integer i.e., T(n) modulo 1000000007.Input1:
5 1 2 3
Output1:
11
Input2:
4 1 1 1
Output2:
3
Explanation(might now be the optimal solution):
Input 1:
Follow the below steps:-
T(1) = 1
T(2) = 2
T(3) = 3
T(4) = 6
T(5) = 11, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main
{
static int mod = 1000000007;
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
//int t = Integer.parseInt(read.readLine());
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
long a =Long.parseLong(str[1]);
long b = Long.parseLong(str[2]);
long c = Long.parseLong(str[3]);
long dp[] = new long[n];
dp[0] = a;
dp[1] = b;
dp[2] = c;
for(int i = 3; i < n; i++)
{
dp[i] = (dp[i-1]%mod + dp[i-2]%mod + dp[i-3]%mod)%mod;
}
System.out.println(dp[n-1]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A Tribonacci number T(n) is the sum of the preceding three elements in a series.
Calculate the value of the nth Tribonacci number modulo 1000000007. The initial three numbers of this series will be a, b and c i.e., T(1)=a, T(2)=b, T(3)=c.The only line of input contains 4 integers n, a, b, c.
Constraints
4 ≤ n ≤ 10^5
1 ≤ a, b, c ≤ 1000000000Print the result containing one integer i.e., T(n) modulo 1000000007.Input1:
5 1 2 3
Output1:
11
Input2:
4 1 1 1
Output2:
3
Explanation(might now be the optimal solution):
Input 1:
Follow the below steps:-
T(1) = 1
T(2) = 2
T(3) = 3
T(4) = 6
T(5) = 11, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e3+ 5;
const int mod = 1e9 + 7;
const int inf = 1e18 + 9;
void solve(){
int n, a, b, c;
cin >> n >> a >> b >> c;
for(int i = 4; i <= n; i++){
int x = (a + b + c) % mod;
a = b;
b = c;
c = x;
}
cout << c << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Saloni has recently distributed N chocolates to some kids. She does not remember the number of kids she distributed the chocolates to. But she knows for sure that she did not break any chocolate while distributing. Can you tell her the number of possible values for the number of children?The first and the only line of input contains an integer N.
Constraints
1 <= N <= 10<sup>12</sup>Output a single integer, the number of possible values for the number of children.Sample Input
6
Sample Output
4
Explanation: The possible values for the number of children are 1, 2, 3, and 6.
Sample Input
10
Sample Output
4, I have written this Solution Code: import math
n=int(input())
count=0
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
count=count+1
else:
count=count+2
i = i + 1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Saloni has recently distributed N chocolates to some kids. She does not remember the number of kids she distributed the chocolates to. But she knows for sure that she did not break any chocolate while distributing. Can you tell her the number of possible values for the number of children?The first and the only line of input contains an integer N.
Constraints
1 <= N <= 10<sup>12</sup>Output a single integer, the number of possible values for the number of children.Sample Input
6
Sample Output
4
Explanation: The possible values for the number of children are 1, 2, 3, and 6.
Sample Input
10
Sample Output
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
int ans = 0;
for(int i=1; i*i<n; i++){
if(n%i == 0){
ans += 2;
}
}
int nn = sqrt(n);
if(nn*nn == n) {
ans++;
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Saloni has recently distributed N chocolates to some kids. She does not remember the number of kids she distributed the chocolates to. But she knows for sure that she did not break any chocolate while distributing. Can you tell her the number of possible values for the number of children?The first and the only line of input contains an integer N.
Constraints
1 <= N <= 10<sup>12</sup>Output a single integer, the number of possible values for the number of children.Sample Input
6
Sample Output
4
Explanation: The possible values for the number of children are 1, 2, 3, and 6.
Sample Input
10
Sample Output
4, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
long num = sc.nextLong();
System.out.println(possibleValues(num));
}
static int possibleValues(long num)
{
int ans = 0;
for(int i = 1; (long)i*i < num; i++)
{
if(num%i == 0)
ans += 2;
}
int nn = (int)Math.sqrt(num);
if((long)(nn*nn) == num)
ans++;
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a stack containing some integers, your task is to reverse the given stack.
Note:- Try to do this question using recursion, do not use any loop.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>Reverse_stack()</b> that takes no parameter.
</b>Constraints:</b>
1 <= Elements in stack <= 100
<b> Custom Input: </b>
First line of input should contain the number of elements N of the stack, the next line of input should contain N space separated integers depicting the elements of the stack.
You don't need to return or print anything just complete the given function.
Note:- For the custom input if your code is correct then the elements will be printed in the same order.Sample Input:-
Stack = {1, 2, 3, 4, 5}, where top = 5
Sample Output:-
Stack = {5, 4, 3, 2, 1} where top = 1, I have written this Solution Code: static Stack <Integer> St = new Stack();
static void Reverse_Stack(){
if(St.size()==0){
return;
}
int x=St.peek();
St.pop();
Reverse_Stack();
bottom_insert(x);
}
static void bottom_insert(int x){
if(St.isEmpty()){
St.push(x);
}
else{
int a = St.peek();
St.pop();
bottom_insert(x);
St.push(a);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a stack containing some integers, your task is to reverse the given stack.
Note:- Try to do this question using recursion, do not use any loop.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>Reverse_stack()</b> that takes no parameter.
</b>Constraints:</b>
1 <= Elements in stack <= 100
<b> Custom Input: </b>
First line of input should contain the number of elements N of the stack, the next line of input should contain N space separated integers depicting the elements of the stack.
You don't need to return or print anything just complete the given function.
Note:- For the custom input if your code is correct then the elements will be printed in the same order.Sample Input:-
Stack = {1, 2, 3, 4, 5}, where top = 5
Sample Output:-
Stack = {5, 4, 3, 2, 1} where top = 1, I have written this Solution Code: n=int(input())
lst=[int(x) for x in input().split()]
for i in lst:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N positive integers and a number K. The task is to find the kth largest element in the array.
Note: DO NOT USE sort() stl.First line of input contains number of testcases. For each testcase, there will be a single line of input containing number of elements in the array and K. Next line contains N elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^5
1 <= K <= NFor each testcase, print a single line of output containing the kth largest element in the array.Sample Input:
2
5 3
3 5 4 2 9
5 5
4 3 7 6 5
Sample Output:
4
3
Explanation:
Testcase 1: Third largest element in the array is 4.
Testcase 2: Fifth largest element in the array is 3., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
Reader sc = new Reader();
int t = sc.nextInt();
while(t!=0)
{
int n= sc.nextInt();
int k=sc.nextInt();
int arr[]= new int[n];
for(int i=0;i<n;i++)
arr[i]=sc.nextInt();
System.out.println(findKthLargest(arr,k));
t--;
}
}
public static int largestEle(int []arr,int k)
{
PriorityQueue<Integer> pq= new PriorityQueue<>(Collections.reverseOrder ());
for(int i=0;i<arr.length;i++)
{
pq.offer(arr[i]);
}
for(int i=0;i<k-1;i++)
{
pq.poll();
}
return pq.peek();
}
public static int findKthLargest(int[] nums, int k) {
int start = 0, end = nums.length - 1, index = nums.length - k;
while (start < end) {
int pivot = partion(nums, start, end);
if (pivot < index) start = pivot + 1;
else if (pivot > index) end = pivot - 1;
else return nums[pivot];
}
return nums[start];
}
private static int partion(int[] nums, int start, int end) {
int pivot = start, temp;
while (start <= end) {
while (start <= end && nums[start] <= nums[pivot]) start++;
while (start <= end && nums[end] > nums[pivot]) end--;
if (start > end) break;
temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
}
temp = nums[end];
nums[end] = nums[pivot];
nums[pivot] = temp;
return end;
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N positive integers and a number K. The task is to find the kth largest element in the array.
Note: DO NOT USE sort() stl.First line of input contains number of testcases. For each testcase, there will be a single line of input containing number of elements in the array and K. Next line contains N elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^5
1 <= K <= NFor each testcase, print a single line of output containing the kth largest element in the array.Sample Input:
2
5 3
3 5 4 2 9
5 5
4 3 7 6 5
Sample Output:
4
3
Explanation:
Testcase 1: Third largest element in the array is 4.
Testcase 2: Fifth largest element in the array is 3., I have written this Solution Code:
t=int(input())
while t>0:
t-=1
n,k=map(int,input().split())
arr=list(map(int,input().split()))
arr.sort(reverse=True)
print(arr[k-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N positive integers and a number K. The task is to find the kth largest element in the array.
Note: DO NOT USE sort() stl.First line of input contains number of testcases. For each testcase, there will be a single line of input containing number of elements in the array and K. Next line contains N elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^5
1 <= K <= NFor each testcase, print a single line of output containing the kth largest element in the array.Sample Input:
2
5 3
3 5 4 2 9
5 5
4 3 7 6 5
Sample Output:
4
3
Explanation:
Testcase 1: Third largest element in the array is 4.
Testcase 2: Fifth largest element in the array is 3., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];}
sort(a,a+n);
cout<<a[n-k]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2[] = str1.split(" ");
int n = Integer.parseInt(str2[0]);
int k = Integer.parseInt(str2[1]);
String str3 = br.readLine();
String str4[] = str3.split(" ");
long[] arr = new long[n];
for(int i = 0; i < n; ++i) {
arr[i] = Long.parseLong(str4[i]);
}
Arrays.sort(arr);
int i=0,j=2;
long ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((arr[j]-arr[i])>k){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: n, p = input().split(' ')
n = int(n)
p = int(p)
arr = input().split(' ')[:n]
for i in range(n):
arr[i] = int(arr[i])
arr.sort()
i = 0
k = 2
count = 0
while k!=n:
if i == k-1:
k = k + 1
continue
if (arr[k] - arr[i]) <= p:
count = count + (int)(((k-i)*(k-i-1))/2)
k = k + 1
else:
i = i + 1
print(count)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,p;
cin>>n>>p;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
int i=0,j=2;
int ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((a[j]-a[i])>p){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Determine the total salary of each department.
Output the salary along with the corresponding department.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'worker', 'columns': [{'name': 'worker_id', 'type': 'int64'}, {'name': 'first_name', 'type': 'object'}, {'name': 'last_name', 'type': 'object'}, {'name': 'salary', 'type': 'int64'}, {'name': 'joining_date', 'type': 'datetime64[ns]'}, {'name': 'department', 'type': 'object'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: df=worker.groupby(by='department').sum()
for i,row in df.iterrows():
print(f"{i}|{row['salary']}"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Determine the total salary of each department.
Output the salary along with the corresponding department.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'worker', 'columns': [{'name': 'worker_id', 'type': 'int64'}, {'name': 'first_name', 'type': 'object'}, {'name': 'last_name', 'type': 'object'}, {'name': 'salary', 'type': 'int64'}, {'name': 'joining_date', 'type': 'datetime64[ns]'}, {'name': 'department', 'type': 'object'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: SELECT
department,
sum(salary)
FROM worker
GROUP BY
department, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Your task is to change the given number into into Binary form.The input contains a single integer N.
Constraints:-
1 <= N <= 1000Print the Binary form of the given number N.Sample Input:-
8
Sample Output:-
1000
Sample Input:-
15
Sample Output:-
1111, I have written this Solution Code: def decimalToBinary(n):
return bin(n).replace("0b", "")
n=int(input())
print(decimalToBinary(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Your task is to change the given number into into Binary form.The input contains a single integer N.
Constraints:-
1 <= N <= 1000Print the Binary form of the given number N.Sample Input:-
8
Sample Output:-
1000
Sample Input:-
15
Sample Output:-
1111, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
long n=sc.nextLong();
long ans=0;
long p=1;
while(n>0){
ans+=(n%2)*p;
p*=10;
n/=2;
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Phoebe is a big GCD fan. Being bored, she starts counting number of pairs of integers (A, B) such that following conditions are satisfied:
<li> GCD(A, B) = X (As X is Phoebe's favourite integer)
<li> A <= B <= L
As Phoebe's performance is coming up, she needs your help to find the number of such pairs possible.
Note: GCD refers to the <a href="https://en.wikipedia.org/wiki/Greatest_common_divisor">Greatest common divisor</a>.Input contains two integers L and X.
Constraints:
1 <= L, X <= 1000000000Print a single integer denoting number of pairs possible.Sample Input
5 2
Sample Output
2
Explanation: Pairs satisfying all conditions are: (2, 2), (2, 4)
Sample Input
5 3
Sample Output
1
Explanation: Pairs satisfying all conditions are: (3, 3), I have written this Solution Code: import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve()
{
out.println(sumTotient(ni()/ni()));
}
public static int[] enumTotientByLpf(int n, int[] lpf)
{
int[] ret = new int[n+1];
ret[1] = 1;
for(int i = 2;i <= n;i++){
int j = i/lpf[i];
if(lpf[j] != lpf[i]){
ret[i] = ret[j] * (lpf[i]-1);
}else{
ret[i] = ret[j] * lpf[i];
}
}
return ret;
}
public static int[] enumLowestPrimeFactors(int n)
{
int tot = 0;
int[] lpf = new int[n+1];
int u = n+32;
double lu = Math.log(u);
int[] primes = new int[(int)(u/lu+u/lu/lu*1.5)];
for(int i = 2;i <= n;i++)lpf[i] = i;
for(int p = 2;p <= n;p++){
if(lpf[p] == p)primes[tot++] = p;
int tmp;
for(int i = 0;i < tot && primes[i] <= lpf[p] && (tmp = primes[i]*p) <= n;i++){
lpf[tmp] = primes[i];
}
}
return lpf;
}
public static long sumTotient(int n)
{
if(n == 0)return 0L;
if(n == 1)return 1L;
int s = (int)Math.sqrt(n);
long[] cacheu = new long[n/s];
long[] cachel = new long[s+1];
int X = (int)Math.pow(n, 0.66);
int[] lpf = enumLowestPrimeFactors(X);
int[] tot = enumTotientByLpf(X, lpf);
long sum = 0;
int p = cacheu.length-1;
for(int i = 1;i <= X;i++){
sum += tot[i];
if(i <= s){
cachel[i] = sum;
}else if(p > 0 && i == n/p){
cacheu[p] = sum;
p--;
}
}
for(int i = p;i >= 1;i--){
int x = n/i;
long all = (long)x*(x+1)/2;
int ls = (int)Math.sqrt(x);
for(int j = 2;x/j > ls;j++){
long lval = i*j < cacheu.length ? cacheu[i*j] : cachel[x/j];
all -= lval;
}
for(int v = ls;v >= 1;v--){
long w = x/v-x/(v+1);
all -= cachel[v]*w;
}
cacheu[(int)i] = all;
}
return cacheu[1];
}
void run() throws Exception
{
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
public static void main(String[] args) throws Exception { new Main().run(); }
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
private int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private double nd() { return Double.parseDouble(ns()); }
private char nc() { return (char)skip(); }
private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Phoebe is a big GCD fan. Being bored, she starts counting number of pairs of integers (A, B) such that following conditions are satisfied:
<li> GCD(A, B) = X (As X is Phoebe's favourite integer)
<li> A <= B <= L
As Phoebe's performance is coming up, she needs your help to find the number of such pairs possible.
Note: GCD refers to the <a href="https://en.wikipedia.org/wiki/Greatest_common_divisor">Greatest common divisor</a>.Input contains two integers L and X.
Constraints:
1 <= L, X <= 1000000000Print a single integer denoting number of pairs possible.Sample Input
5 2
Sample Output
2
Explanation: Pairs satisfying all conditions are: (2, 2), (2, 4)
Sample Input
5 3
Sample Output
1
Explanation: Pairs satisfying all conditions are: (3, 3), I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
ull X[20000001];
ull cmp(ull N){
return N*(N+1)/2;
}
ull solve(ull N){
if(N==1)
return 1;
if(N < 20000001 && X[N] != 0)
return X[N];
ull res = 0;
ull q = floor(sqrt(N));
for(int k=2;k<N/q+1;++k){
res += solve(N/k);
}
for(int m=1;m<q;++m){
res += (N/m - N/(m+1)) * solve(m);
}
res = cmp(N) - res;
if(N < 20000001)
X[N] = res;
return res;
}
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int l,x;
cin>>l>>x;
if(l<x)
cout<<0;
else
cout<<solve(l/x);
#ifdef ANIKET_GOYAL
cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a function called <code>createCounter</code> that takes in <code>a number</code> (initialValue) as an argument and returns an object with two methods: <code>increment</code> and <code>decrement</code>.
<ul>
<li>The <code>increment</code> method should <code>increment the number by 1</code> and return the updated value. </li>
<li>The <code>decrement</code> method should <code>decrement the number by 1</code> and return the updated value. </li>
</ul>
On calling the <code>increment</code> and <code>decrement</code> methods at various times, the operations should be on the <code>initialValue</code> and the initialValue should be updated and stored so that the next time any of these methods is called the operation is on the updated value and that updated value is stored as well. Use <code>closure</code> to complete this task.The <code>createCounter</code> function takes a <code>number</code> as an argumentThe <code>createCounter</code> function should return an <code>object</code>.
Inside the object, both <code>increment</code> and <code>decrement</code> methods should return a <code>number</code>.const counter = createCounter(5);
console.log(counter.increment()); // prints 6
console.log(counter.increment()); // prints 7
console.log(counter.decrement()); // prints 6, I have written this Solution Code: function createCounter(initialValue) {
let count = initialValue;
return {
increment: function() {
count++;
return count;
},
decrement: function() {
count--;
return count;
}
};
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N elements. Find the maximum value of GCD(Arr[i], Arr[j]) where i != j.First line of input contains a single integer N.
Second line of input contains N space separated integers, denoting array Arr.
Constraints:
2 <= N <= 100000
1 <= Arr[i] <= 100000Print the maximum value of GCD(Arr[i], Arr[j]) where i != j.Sample Input 1
5
2 4 5 2 2
Sample Output 1
2
Explanation: We can select index 1 and index 4, GCD(2, 2) = 2
Sample Input 2
6
4 3 4 1 6 5
Sample Output 2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int result(int a[],int n)
{
int maxe =0;
for(int i=0;i<n;i++)
maxe = Math.max(maxe,a[i]);
int count[]=new int[maxe+1];
for(int i=0;i<n;i++)
{
for(int j=1;j<Math.sqrt(a[i]);j++)
{
if(a[i]%j==0)
{
count[j]++;
if (j != a[i] / j)
count[a[i] / j]++;
}
}
}
for(int i=maxe;i>0;i--)
{
if(count[i]>1)
return i;
}
return 1;
}
public static void main (String[] args) throws IOException{
InputStreamReader i = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(i);
int n = Integer.parseInt(br.readLine());
int a[] = new int[n];
String str = br.readLine();
String[] strs = str.trim().split(" ");
for (int j = 0; j < n; j++)
{
a[j] = Integer.parseInt(strs[j]);
}
System.out.println(result(a,n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N elements. Find the maximum value of GCD(Arr[i], Arr[j]) where i != j.First line of input contains a single integer N.
Second line of input contains N space separated integers, denoting array Arr.
Constraints:
2 <= N <= 100000
1 <= Arr[i] <= 100000Print the maximum value of GCD(Arr[i], Arr[j]) where i != j.Sample Input 1
5
2 4 5 2 2
Sample Output 1
2
Explanation: We can select index 1 and index 4, GCD(2, 2) = 2
Sample Input 2
6
4 3 4 1 6 5
Sample Output 2
4, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int v[100001]={};
int n;
cin>>n;
for(int i=1;i<=n;++i){
int d;
cin>>d;
for(int j=1;j*j<=d;++j){
if(d%j==0){
v[j]++;
if(j!=d/j)
v[d/j]++;
}
}
}
for(int i=100000;i>=1;--i)
if(v[i]>1){
cout<<i;
return 0;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given Q queries. In each query, determine whether there exists an array A of size N such that:
1. All the elements are positive integers.
2. The number of subarrays such that their XOR-sum is 0 is exactly K. In other words, there are exactly K pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ N and A<sub>l</sub> ⊕ A<sub>l+1</sub> ⊕ ... A<sub>r</sub> = 0.
If there exists such an array, print "YES", otherwise print "NO".The first line of the input contains a single integer Q — the number of queries (1 ≤ Q ≤ 10<sup>5</sup>).
Q lines follow, each line containing two space separated integers N (1 ≤ N ≤ 1000) and K (0 ≤ K ≤ N(N+1)/2).For each test case, print "YES", if there exists such an array, otherwise print "NO" (without the quotes).
Note that the output is case sensitive.Sample Input
3
2 2
3 2
2 1
Sample Output
NO
YES
YES, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
FastScanner in;
PrintWriter out;
boolean systemIO = true;
int MAXSUM = 250000;
int[] dp = new int[MAXSUM];
int[] minn = new int[MAXSUM];
public void precalc() {
for (int i = 1; i < minn.length; i++) {
minn[i] = 1002;
dp[i] = 1002;
}
for (int step = 2; step <= 501; step++) {
int delta = step * (step - 1) / 2;
for (int j = 0; j + delta < MAXSUM; j++) {
if (dp[j] + step < dp[j + delta]) {
dp[j + delta] = dp[j] + step;
minn[j + delta] = Math.min(minn[j + delta], Math.max(dp[j + delta] - 1, 2 * step - 2));
}
}
}
}
public boolean clever(int n, int k) {
if (k >= MAXSUM) {
return false;
}
return n >= minn[k];
}
public void solve() {
precalc();
for (int qwerty = in.nextInt(); qwerty > 0; --qwerty) {
int n = in.nextInt();
int k = in.nextInt();
if (clever(n, k)) {
out.println("YES");
} else {
out.println("NO");
}
}
}
public void run() {
try {
if (systemIO) {
in = new FastScanner(System.in);
out = new PrintWriter(System.out);
} else {
String fileName = "60-huge-inexact";
in = new FastScanner(new File(fileName + ".txt"));
out = new PrintWriter(new File(fileName + ".out"));
}
solve();
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f));
}
String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
return null;
}
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
public static void main(String[] arg) {
long time = System.currentTimeMillis();
new Main().run();
System.err.println(System.currentTimeMillis() - time);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given Q queries. In each query, determine whether there exists an array A of size N such that:
1. All the elements are positive integers.
2. The number of subarrays such that their XOR-sum is 0 is exactly K. In other words, there are exactly K pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ N and A<sub>l</sub> ⊕ A<sub>l+1</sub> ⊕ ... A<sub>r</sub> = 0.
If there exists such an array, print "YES", otherwise print "NO".The first line of the input contains a single integer Q — the number of queries (1 ≤ Q ≤ 10<sup>5</sup>).
Q lines follow, each line containing two space separated integers N (1 ≤ N ≤ 1000) and K (0 ≤ K ≤ N(N+1)/2).For each test case, print "YES", if there exists such an array, otherwise print "NO" (without the quotes).
Note that the output is case sensitive.Sample Input
3
2 2
3 2
2 1
Sample Output
NO
YES
YES, I have written this Solution Code: #include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e9;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
struct info {int n, k, idx;};
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
read(t);
vector<vector <info>> q(1001);
FOR (i, 1, t)
{
readb(n, k);
q[n/2 + 1].pb({n, k, i});
}
vi dp(1001*500 + 5, inf);
dp[0] = 0;
bool ans[t + 1] = {};
FOR (i, 1, 502)
{
FOR (j, i*(i - 1)/2, 1001*500)
dp[j] = min(dp[j], dp[j - i*(i - 1)/2] + i);
for (auto x: q[i])
if (dp[x.k] <= x.n + 1) ans[x.idx] = 1;
}
FOR (i, 1, t)
if (ans[i]) cout << "YES" << endl;
else cout << "NO" << endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: def MaximumProduct(N):
ans=N//4
if N %4 !=0:
ans=ans+1
return ans, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: static int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where N is even. You have to pair up the elements into N/2 pairs such that each element is in exactly 1 pair. You need to find minimum possible X such that there exists a way to pair the N elements and for no pair sum of its elements is greater than X.First line contains N.
Second line contains N space separated integers, denoting array.
Constraints:
1 <= N <= 100000
1 <= elements of the array <= 1000000000Print a single integer, minimum possible X.Sample Input
4
3 1 1 4
Sample Output
5
Explanation: we can pair (1, 3) and (1, 4) so all pairs have sum less than or equal to 5., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int[]sort(int n, int a[]){
int i,key;
for(int j=1;j<n;j++){
key=a[j];
i=j-1;
while(i>=0 && a[i]>key){
a[i+1]=a[i];
i=i-1;
}
a[i+1]=key;
}
return a;
}
public static void main (String[] args) throws IOException {
InputStreamReader io = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(io);
int n = Integer.parseInt(br.readLine());
String str = br.readLine();
String stra[] = str.trim().split(" ");
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = Integer.parseInt(stra[i]);
}
a=sort(n,a);
int max=0;
for(int i=0;i<n;i++)
{
if(a[i]+a[n-i-1]>max)
{
max=a[i]+a[n-i-1];
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where N is even. You have to pair up the elements into N/2 pairs such that each element is in exactly 1 pair. You need to find minimum possible X such that there exists a way to pair the N elements and for no pair sum of its elements is greater than X.First line contains N.
Second line contains N space separated integers, denoting array.
Constraints:
1 <= N <= 100000
1 <= elements of the array <= 1000000000Print a single integer, minimum possible X.Sample Input
4
3 1 1 4
Sample Output
5
Explanation: we can pair (1, 3) and (1, 4) so all pairs have sum less than or equal to 5., I have written this Solution Code: n=int(input())
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
arr=sorted(arr,key=int)
start=0
end=n-1
ans=0
while(start<end):
ans=max(ans,arr[end]+arr[start])
start+=1
end-=1
print (ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where N is even. You have to pair up the elements into N/2 pairs such that each element is in exactly 1 pair. You need to find minimum possible X such that there exists a way to pair the N elements and for no pair sum of its elements is greater than X.First line contains N.
Second line contains N space separated integers, denoting array.
Constraints:
1 <= N <= 100000
1 <= elements of the array <= 1000000000Print a single integer, minimum possible X.Sample Input
4
3 1 1 4
Sample Output
5
Explanation: we can pair (1, 3) and (1, 4) so all pairs have sum less than or equal to 5., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
sort(a,a+n);
int ma=0;
for(int i=0;i<n;++i){
ma=max(ma,a[i]+a[n-i-1]);
}
cout<<ma;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N, with all values as non-negative integers. Bodega asked you whether it is possible to partition the array into exactly K non-empty subarrays such that the sum of values of each subarray is equal. If it is possible print "Yes", otherwise print "No".
Note:
Every element of the array should belong to exactly one subarray in the partition.The first line consists of two space-separated integers N and K respectively.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub>.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 20
0 ≤ A<sub>i</sub> ≤ 10<sup>6</sup>
There exists at least one non-zero element in array A.Print a single word "Yes" if the array can be partitioned into K equal sum parts, otherwise print "No".
Note:
The quotation marks are only for clarity, you should not print them in output.
The output is case-sensitive.Sample Input 1:
5 2
2 2 1 2 2
Sample Output 1:
No
Sample Input 2:
10 3
3 0 1 2 0 0 6 0 1 5
Sample Output 2:
Yes
Explanation:
The array can be partitioned as {3, 0, 1, 2, 0, 0}, {6}, {0, 1, 5}., I have written this Solution Code: import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve(int TC) {
int n = ni(), k = ni();
long sum = 0L, tempSum = 0;
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = ni();
sum += a[i];
}
if (sum % k != 0) {
pn("No");
return;
}
long req = sum / (long) k;
for (int i : a) {
tempSum += i;
if (tempSum == req) {
tempSum = 0;
} else if (tempSum > req) {
pn("No");
return;
}
}
pn(tempSum == 0 ? "Yes" : "No");
}
boolean TestCases = false;
public static void main(String[] args) throws Exception {
new Main().run();
}
void hold(boolean b) throws Exception {
if (!b) throw new Exception("Hold right there, Sparky!");
}
static void dbg(Object... o) {
System.err.println(Arrays.deepToString(o));
}
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for (int t = 1; t <= T; t++) solve(t);
out.flush();
if (!INPUT.isEmpty()) tr(System.currentTimeMillis() - s + "ms");
}
void p(Object o) {
out.print(o);
}
void pn(Object o) {
out.println(o);
}
void pni(Object o) {
out.println(o);
out.flush();
}
int ni() {
int num = 0, b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() {
return Double.parseDouble(ns());
}
char nc() {
return (char) skip();
}
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if (lenbuf == -1) throw new InputMismatchException();
if (ptrbuf >= lenbuf) {
ptrbuf = 0;
try {
lenbuf = is.read(inbuf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (lenbuf <= 0) return -1;
}
return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b)) ;
return b;
}
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while (p < n && !(isSpaceChar(b))) {
buf[p++] = (char) b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) {
if (INPUT.length() > 0) System.out.println(Arrays.deepToString(o));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N, with all values as non-negative integers. Bodega asked you whether it is possible to partition the array into exactly K non-empty subarrays such that the sum of values of each subarray is equal. If it is possible print "Yes", otherwise print "No".
Note:
Every element of the array should belong to exactly one subarray in the partition.The first line consists of two space-separated integers N and K respectively.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub>.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 20
0 ≤ A<sub>i</sub> ≤ 10<sup>6</sup>
There exists at least one non-zero element in array A.Print a single word "Yes" if the array can be partitioned into K equal sum parts, otherwise print "No".
Note:
The quotation marks are only for clarity, you should not print them in output.
The output is case-sensitive.Sample Input 1:
5 2
2 2 1 2 2
Sample Output 1:
No
Sample Input 2:
10 3
3 0 1 2 0 0 6 0 1 5
Sample Output 2:
Yes
Explanation:
The array can be partitioned as {3, 0, 1, 2, 0, 0}, {6}, {0, 1, 5}., I have written this Solution Code:
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define pi 3.141592653589793238
#define int long long
#define ll long long
#define ld long double
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long powm(long long a, long long b,long long mod) {
long long res = 1;
while (b > 0) {
if (b & 1)
res = res * a %mod;
a = a * a %mod;
b >>= 1;
}
return res;
}
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
#ifndef ONLINE_JUDGE
if(fopen("input.txt","r"))
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
}
#endif
int n,k;
cin>>n>>k;
int a[n];
int sum=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
sum+=a[i];
}
if(sum%k)
cout<<"No";
else
{
int tot=0;
int cur=0;
sum/=k;
for(int i=0;i<n;i++)
{
cur+=a[i];
if(cur==sum)
{
tot++;
cur=0;
}
}
if(cur==0 && tot==k)
cout<<"Yes";
else
cout<<"No";
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N, with all values as non-negative integers. Bodega asked you whether it is possible to partition the array into exactly K non-empty subarrays such that the sum of values of each subarray is equal. If it is possible print "Yes", otherwise print "No".
Note:
Every element of the array should belong to exactly one subarray in the partition.The first line consists of two space-separated integers N and K respectively.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub>.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 20
0 ≤ A<sub>i</sub> ≤ 10<sup>6</sup>
There exists at least one non-zero element in array A.Print a single word "Yes" if the array can be partitioned into K equal sum parts, otherwise print "No".
Note:
The quotation marks are only for clarity, you should not print them in output.
The output is case-sensitive.Sample Input 1:
5 2
2 2 1 2 2
Sample Output 1:
No
Sample Input 2:
10 3
3 0 1 2 0 0 6 0 1 5
Sample Output 2:
Yes
Explanation:
The array can be partitioned as {3, 0, 1, 2, 0, 0}, {6}, {0, 1, 5}., I have written this Solution Code: N,K=map(int,input().split())
S=0
a=input().split()
for i in a:
S+=int(i)
if S%K!=0:
print('No')
else:
su=0
count=0
asd=S/K
for i in range(N):
su+=int(a[i])
if su==asd:
count+=1
su=0
if count==K:
print('Yes')
else:
print('No'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N. You need to print the squares of all numbers from 1 to N in different lines.The only input line contains an integer N.
Constraints:
1 <= N <= 1000Print N lines where the ith line (1-based indexing) contains an integer i<sup>2</sup>.Sample Input 1:
4
Output:
1
4
9
16
Explanation:
1*1 = 1
2*2 = 4
3*3 = 9
4*4 = 16
Sample Input 2:
2
Output:
1
4
Explanation:
1*1 = 1
2*2 = 4, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
print(i**2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N. You need to print the squares of all numbers from 1 to N in different lines.The only input line contains an integer N.
Constraints:
1 <= N <= 1000Print N lines where the ith line (1-based indexing) contains an integer i<sup>2</sup>.Sample Input 1:
4
Output:
1
4
9
16
Explanation:
1*1 = 1
2*2 = 4
3*3 = 9
4*4 = 16
Sample Input 2:
2
Output:
1
4
Explanation:
1*1 = 1
2*2 = 4, I have written this Solution Code: import java.io.*;
import java.util.Scanner;
class Main {
public static void main (String[] args) {
int n;
Scanner s = new Scanner(System.in);
n = s.nextInt();
for(int i=1;i<=n;i++){
System.out.println(i*i);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
long z=0,x=0,y=0;
int choice;
Scanner in = new Scanner(System.in);
choice = in.nextInt();
String s="";
int f = 1;
while(f<=choice){
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
System.out.println((long)(Math.max((z-x),(z-y))));
f++;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: n = int(input())
for i in range(n):
l = list(map(int,input().split()))
print(l[2]-min(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll mod2 = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
read(t);
assert(1 <= t && t <= ll(1e4));
while (t--)
{
readc(x, y, z);
assert(1 <= x && x <= ll(1e15));
assert(1 <= y && y <= ll(1e15));
assert(max(x, y) < z && z <= ll(1e15));
int r = 2*z - x - y - 1;
int l = z - max(x, y);
print(r - l + 1);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int t;
cin>>t;
while(t--)
{
int x, y, z;
cin>>x>>y>>z;
cout<<max(z - y, z- x)<<endl;
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args)throws IOException {
Reader sc = new Reader();
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;i++){
arr[i] = sc.nextInt();
}
int max=0;
if(arr[0]<arr[N-1])
System.out.print(N-1);
else{
for(int i=0;i<N-1;i++){
int j = N-1;
while(j>i){
if(arr[i]<arr[j]){
if(max<j-i){
max = j-i;
} break;
}
j--;
}
if(i==j)
break;
if(j==N-1)
break;
}
if(max==0)
System.out.print("-1");
else
System.out.print(max);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
/* For a given array arr[],
returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = new int[(sizeof(int) * n)];
int *RMax = new int[(sizeof(int) * n)];
/* Construct LMin[] such that
LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i - 1]);
/* Construct RMax[] such that
RMax[j] stores the maximum value from
(arr[j], arr[j+1], ..arr[n-1]) */
RMax[n - 1] = arr[n - 1];
for (j = n - 2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j + 1]);
/* Traverse both arrays from left to right
to find optimum j - i. This process is similar to
merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j - i);
j = j + 1;
}
else
i = i + 1;
}
return maxDiff;
}
// Driver Code
signed main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int maxDiff = maxIndexDiff(a, n);
cout << maxDiff;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
rightMax = [0] * n
rightMax[n - 1] = arr[n - 1]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], arr[i])
maxDist = -2**31
i = 0
j = 0
while (i < n and j < n):
if (rightMax[j] >= arr[i]):
maxDist = max(maxDist, j - i)
j += 1
else:
i += 1
if maxDist==0:
maxDist=-1
print(maxDist), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: def StrangeNumber(N):
return 9*(N-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: static int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String arr[][]=new String[n][n];
String transpose[][]=new String[n][n];
int row;
int cols;
for(row=0;row<n;row++)
{
String rowNum=br.readLine();
String rowVals[]=rowNum.split(" ");
for(cols=0; cols<n;cols++)
{
arr[row][cols]=rowVals[cols];
}
}
for(row=0;row<n;row++)
{
for(cols=0; cols<n;cols++)
{
transpose[row][cols]=arr[cols][row];
System.out.print(transpose[row][cols]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: x=int(input())
l1=[]
for i in range(x):
a1=list(map(int,input().split()))
l1.append(a1)
l4=[]
for j in range(x):
l3=[]
for i in range(x):
l3.append(l1[i][j])
l4.append(l3)
for i in range(x):
for j in range(x):
print(l4[i][j], end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>a[j][i];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Cf cf = new Cf();
cf.solve();
}
static class Cf {
InputReader in = new InputReader(System.in);
OutputWriter out = new OutputWriter(System.out);
int mod = (int)1e9+7;
public void solve() {
int t = in.readInt();
if(t>=6) {
out.printLine("No");
}else {
out.printLine("Yes");
}
}
public long findPower(long x,long n) {
long ans = 1;
long nn = n;
while(nn>0) {
if(nn%2==1) {
ans = (ans*x) % mod;
nn-=1;
}else {
x = (x*x)%mod;
nn/=2;
}
}
return ans%mod;
}
public static int log2(int x) {
return (int) (Math.log(x) / Math.log(2));
}
private static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
private static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
writer.flush();
}
public void printLine(Object... objects) {
print(objects);
writer.println();
writer.flush();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n<6)
cout<<"Yes";
else
cout<<"No";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: n=int(input())
if(n>=6):
print("No")
else:
print("Yes"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: N = int(input())
if N > 5:
print("Greater than 5")
elif(N == 1):
print("one")
elif(N == 2):
print("two")
elif(N == 3):
print("three")
elif(N == 4):
print("four")
elif(N == 5):
print("five"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.