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For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
cout<<'^';
}
cout<<'*'<<endl;
}
for(int i=0;i<=N;i++){
cout<<'*';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, 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 roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, 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 s=br.readLine().toString();
String[] str = s.split(" ");
float a=Float.parseFloat(str[0]);
float b=Float.parseFloat(str[1]);
float c=Float.parseFloat(str[2]);
float div = (float)(b*b-4*a*c);
if(div>0.0){
float alpha= (-b+(float)Math.sqrt(div))/(2*a);
float beta= (-b-(float)Math.sqrt(div))/(2*a);
System.out.printf("%.2f\n",alpha);
System.out.printf("%.2f",beta);
}
else{
float rp=-b/(2*a);
float ip=(float)Math.sqrt(-div)/(2*a);
System.out.printf("%.2f+i%.2f\n",rp,ip);
System.out.printf("%.2f-i%.2f\n",rp,ip);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: import math
a,b,c = map(int, input().split(' '))
disc = (b ** 2) - (4*a*c)
sq = disc ** 0.5
if disc > 0:
print("{:.2f}".format((-b + sq)/(2*a)))
print("{:.2f}".format((-b - sq)/(2*a)))
elif disc == 0:
print("{:.2f}".format(-b/(2*a)))
elif disc < 0:
r1 = complex((-b + sq)/(2*a))
r2 = complex((-b - sq)/(2*a))
print("{:.2f}+i{:.2f}".format(r1.real, abs(r1.imag)))
print("{:.2f}-i{:.2f}".format(r2.real, abs(r2.imag))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-19 02:44:22
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int main() {
float a, b, c;
float root1, root2, imaginary;
float discriminant;
scanf("%f%f%f", &a, &b, &c);
discriminant = (b * b) - (4 * a * c);
switch (discriminant > 0) {
case 1:
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("%.2f\n%.2f", root1, root2);
break;
case 0:
switch (discriminant < 0) {
case 1:
root1 = root2 = -b / (2 * a);
imaginary = sqrt(-discriminant) / (2 * a);
printf("%.2f + i%.2f\n%.2f - i%.2f", root1, imaginary, root2, imaginary);
break;
case 0:
root1 = root2 = -b / (2 * a);
printf("%.2f\n%.2f", root1, root2);
break;
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function <code>promiseMe</code>
Such that it takes a number as the first argument (time) and a string as the second argument(data).
It returns a promise which resolves after time milliseconds and data is returned.
Note:- You only have to implement the function, in the example it
shows your implemented question will be run.Function should take number as first argument and data to be returned as second.Resolves to the data given as inputpromiseMe(200, 'hi').then(data=>{
console.log(data) // prints hi
}), I have written this Solution Code: function promiseMe(number, dat) {
return new Promise((res,rej)=>{
setTimeout(()=>{
res(dat)
},number)
})
// return the output using return keyword
// do not console.log it
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: def findMin(arr, low, high):
if high < low:
return arr[0]
if high == low:
return arr[low]
mid = int((low + high)/2)
if mid < high and arr[mid+1] < arr[mid]:
return arr[mid+1]
if mid > low and arr[mid] < arr[mid - 1]:
return arr[mid]
if arr[high] > arr[mid]:
return findMin(arr, low, mid-1)
return findMin(arr, mid+1, high)
T =int(input())
for i in range(T):
N = int(input())
arr = list(input().split())
for k in range(len(arr)):
arr[k] = int(arr[k])
print(findMin(arr,0,N-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
if(a[1] < a[n]){
cout << a[1] << endl;
continue;
}
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] >= a[1])
l = m;
else
h = m;
}
cout << a[h] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main (String[] args) {
//code
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int j=0;j<t;j++){
int al = s.nextInt();
int a[] = new int[al];
for(int i=0;i<al;i++){
a[i] = s.nextInt();
}
binSearchSmallest(a);
}
}
public static void binSearchSmallest(int a[]) {
int s=0;
int e = a.length - 1;
int mid = 0;
while(s<=e){
mid = (s+e)/2;
if(a[s]<a[e]){
System.out.println(a[s]);
return;
}
if(a[mid]>=a[s]){
s=mid+1;
}
else{
e=mid;
}
if(s == e){
System.out.println(a[s]);
return;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: // arr is the input array
function findMin(arr,n) {
// write code here
// do not console.log
// return the number
let min = arr[0]
for(let i=1;i<arr.length;i++){
if(min > arr[i]){
min = arr[i]
}
}
return min
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: a, b = list(map(int, input().split(" ")))
print(str(a <= 10 and b >= 10).lower(), end=' ')
print(str(a % 2 == 0 or b % 2 == 0).lower(), end=' ')
print(str(not a == b).lower()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class Main {
static boolean Logical_AND(int a, int b){
if(a<=10 && b>=10){
return true;}
return false;}
static boolean Logical_OR(int a, int b){
if(a%2==0 || b%2==0){
return true;}
return false;}
static boolean Logical_NOT(int a, int b){
if(a!=b){
return true;}
return false;}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a=in.nextInt();
int b=in.nextInt();
System.out.print(Logical_AND(a, b)+" ");
System.out.print(Logical_OR(a,b)+" ");
System.out.print(Logical_NOT(a,b)+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: def lenOfLongSubarr(arr, N, K):
mydict = dict()
sum = 0
maxLen = 0
for i in range(N):
sum += arr[i]
if (sum == K):
maxLen = i + 1
elif (sum - K) in mydict:
maxLen = max(maxLen, i - mydict[sum - K])
if sum not in mydict:
mydict[sum] = i
return maxLen
if __name__ == '__main__':
T = int(input())
#N,K=list(map(int,input().split()))
for i in range(T):
N,k= [int(N)for N in input("").split()]
arr=list(map(int,input().split()))
N = len(arr)
print(lenOfLongSubarr(arr, N, k)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
unordered_map<long long,int> um;
int n,k;
cin>>n>>k;
long arr[n];
int maxLen=0;
for(int i=0;i<n;i++){cin>>arr[i];}
long long sum=0;
for(int i=0;i<n;i++){
sum += arr[i];
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (um.find(sum) == um.end())
um[sum] = i;
// check if 'sum-k' is present in 'um'
// or not
if (um.find(sum - k) != um.end()) {
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
cout<<maxLen<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, 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();
while(t-->0){
if(t%10==0){
System.gc();
}
int arrsize=sc.nextInt();
int k=sc.nextInt();
int[] arr=new int[arrsize];
for(int i=0;i<arrsize;i++){
arr[i]=sc.nextInt();
}
int subsize=0;
int sum=0;
HashMap<Integer, Integer> hash=new HashMap<>();
for(int i=0;i<arrsize;i++){
sum+=arr[i];
if(sum==k){
subsize=i+1;
}
if(!hash.containsKey(sum)){
hash.put(sum,i);
}
if(hash.containsKey(sum-k)){
if(subsize<(i-hash.get(sum-k))){
subsize=i-hash.get(sum-k);
}
}
}
System.out.println(subsize);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String[] input = in.readLine().split(" ");
int a = Integer.parseInt(input[0]);
int b = Integer.parseInt(input[1]);
int c = Integer.parseInt(input[2]);
if(a==b && b==c && c == a) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: a,b,c=map(int,input().split())
if a==b and b==c:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: #include <iostream>
using namespace std;
int main() {
int a, b, c;
cin >> a >> b >> c;
if (a == b && b == c) cout << "Yes";
else cout << "No";
}
, 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.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print 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: import java.util.*;
import java.io.*;
import java.lang.*;
public class Main
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int x=sc.nextInt();
int ans = 9 * (x-1);
System.out.print(ans);
}
}, In this Programming Language: Java, 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.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print 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:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int cnt[max1];
signed main(){
int n;
cin>>n;
out((n-1)*9);
}
, 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.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print 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: a=int(input())
print(9*(a-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],…, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between x−k and x+k.
The last line contains m integers b[1], b[2],…, b[m]: the size of each toy.
<b>Constraints</b>
1 ≤ n,m ≤ 200000
0 ≤ k ≤ 10<sup>9</sup>
1 ≤ a[i],b[i] ≤ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code: import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
import static java.lang.Math.pow;
class InputReader {
private boolean finished = false;
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 static InputReader getInputReader(boolean readFromTextFile) throws FileNotFoundException {
return ((readFromTextFile) ? new InputReader(new FileInputStream("src/input.txt"))
: new InputReader(System.in));
}
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 peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
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 long nextLong() {
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 String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
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 * pow(10, nextInt());
}
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 * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
}
class Main {
public static void main (String[] args) throws FileNotFoundException {
InputReader sc = InputReader.getInputReader(false);
int n = sc.nextInt();
int m = sc.nextInt();
long k = sc.nextLong();
long[] boyArray = new long[n];
long[] toyArray = new long[m];
for (int i = 0; i < n; i++) {
boyArray[i] = sc.nextLong();
}
for (int i = 0; i < m; i++) {
toyArray[i] = sc.nextLong();
}
Arrays.sort(boyArray);
Arrays.sort(toyArray);
int cnt = 0;
int l = 0 , h = n-1;
for (int i = 0; i < m; i++) {
while( h >= l){
if(toyArray[i] < boyArray[l]-k || toyArray[i] > boyArray[h]+k){
break;
}
else if(toyArray[i] >= boyArray[l]-k && toyArray[i] <= boyArray[l]+k){
l++;
cnt++;
break;
}
else if(toyArray[i] >= boyArray[h]-k && toyArray[i] <= boyArray[h]+k){
h--;
cnt++;
break;
}
else{
l++;
}
}
}
System.out.println(cnt);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],…, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between x−k and x+k.
The last line contains m integers b[1], b[2],…, b[m]: the size of each toy.
<b>Constraints</b>
1 ≤ n,m ≤ 200000
0 ≤ k ≤ 10<sup>9</sup>
1 ≤ a[i],b[i] ≤ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code: n,m,k=[int(x) for x in input().split()[:3]]
a=[int(x) for x in input().split()[:n]]
b=[int(x) for x in input().split()[:m]]
a.sort()
b.sort()
j = 0
sum1 = 0
for i in a:
while(j<len(b)):
if(i-k<=b[j] and b[j]<=i+k):
j = j+1
sum1 +=1
break
elif i+k<b[j]:
break
else:
j +=1
if(j>=len(b)):
break
print(sum1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],…, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between x−k and x+k.
The last line contains m integers b[1], b[2],…, b[m]: the size of each toy.
<b>Constraints</b>
1 ≤ n,m ≤ 200000
0 ≤ k ≤ 10<sup>9</sup>
1 ≤ a[i],b[i] ≤ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 200010
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
int c[max1];
int main()
{
int n,m;
cin>>n>>m;
ll k;
cin>>k;
ll a[n],b[m];
FOR(i,n){
cin>>a[i];
}
FOR(i,m){
cin>>b[i];}
sort(a,a+n);
sort(b,b+m);
int i=0,j=0;
int cnt=0;
while(i!=n && j!=m){
if(abs(b[j]-a[i])<=k){cnt++;i++;j++;}
else{
if(b[j]>a[i]){i++;}
else{j++;}
}
}
out(cnt);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==0:
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 a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, 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);
int x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(i,end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, 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. The array contains almost distinct elements with some duplicated. You have to print the elements in sorted order which appears more than once.The first line of input contains T, denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line of input contains size of array N. The second line contains N integers separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= A<sub>i</sub> <= 100000
Sum of N over all test cases do not exceed 100000For each test case, in a new line, print the required answer in separate line. If there are no duplicates print -1.Input:
1
9
3 4 5 7 8 1 2 1 3
Output:
1 3
Explanation : both 1,3 appeared 2 times
Input
1
5
1 2 3 4 5
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));
int t=Integer.parseInt(br.readLine());
String nums[];
int n;
TreeMap<Integer,Integer> tm=null;
int m,p;
boolean flag;
for(int a=0;a<t;++a)
{
n=Integer.parseInt(br.readLine());
nums=br.readLine().split("\\s");
tm=new TreeMap<>();
for(int i=0;i<n;++i)
{
p=Integer.parseInt(nums[i]);
tm.put(p,tm.getOrDefault(p,0)+1);
}
flag=false;
for(Integer i:tm.keySet())
{
m=tm.get(i);
if(m>1)
{
System.out.print(i+" ");
flag=true;
}
}
if(!flag)
{
System.out.println(-1);
}
else 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 an array A of size N. The array contains almost distinct elements with some duplicated. You have to print the elements in sorted order which appears more than once.The first line of input contains T, denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line of input contains size of array N. The second line contains N integers separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= A<sub>i</sub> <= 100000
Sum of N over all test cases do not exceed 100000For each test case, in a new line, print the required answer in separate line. If there are no duplicates print -1.Input:
1
9
3 4 5 7 8 1 2 1 3
Output:
1 3
Explanation : both 1,3 appeared 2 times
Input
1
5
1 2 3 4 5
Output
-1
, I have written this Solution Code: import numpy as np
from collections import defaultdict
t=int(input())
def solve():
d=defaultdict(int)
n=int(input())
a=np.array([input().strip().split()],int).flatten()
for i in a:
d[i]+=1
d=sorted(d.items())
c=0
for i in d:
if(i[1]>1):
c=1
print(i[0],end=" ")
if(c==0):
print(-1,end=" ")
while(t>0):
solve()
print()
t-=1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. The array contains almost distinct elements with some duplicated. You have to print the elements in sorted order which appears more than once.The first line of input contains T, denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line of input contains size of array N. The second line contains N integers separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= A<sub>i</sub> <= 100000
Sum of N over all test cases do not exceed 100000For each test case, in a new line, print the required answer in separate line. If there are no duplicates print -1.Input:
1
9
3 4 5 7 8 1 2 1 3
Output:
1 3
Explanation : both 1,3 appeared 2 times
Input
1
5
1 2 3 4 5
Output
-1
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#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 sz 200000
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
int n,m;
cin>>n;
map<int,int> A;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
A[a]++;
}
int ch=1;
int cnt=0;
for(auto it:A)
{
if(it.se>1){ cout<<it.fi<<" "; cnt++;}
}
if(cnt==0) cout<<-1;
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code: n = int(input())
arr = []
x = 4
prod = 1
a = input().split()
for i in range(0,n):
prod = prod*int(a[i])
if prod % 4 == 0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., 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);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int two=0,four=0;
for(int i=0;i<n;i++){
if(a[i]%2==0){
two++;
}
if(a[i]%4==0){
four++;
}
}
if(four>=1 || two>=2){
System.out.print("YES");
}
else{
System.out.print("NO");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 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(ll 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 int long long
#define ld 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
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 200005;
// 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
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n; cin>>n;
int ct = 0;
For(i, 0, n){
int a; cin>>a;
while(a%2==0){
ct++;
a/=2;
}
}
if(ct>=2){
cout<<"YES";
}
else{
cout<<"NO";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: def ispangram(str):
alphabet = "abcdefghijklmnopqrstuvwxyz"
for char in alphabet:
if char not in str.lower():
return False
return True
N = int(input())
arr = []
for i in range(N):
arr.append(input())
for i in range(N):
if(ispangram(arr[i]) == True):
print(1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: function pangrams(s) {
// write code here
// do not console.log it
// return 1 or 0
let alphabet = "abcdefghijklmnopqrstuvwxyz";
let regex = /\s/g;
let lowercase = s.toLowerCase().replace(regex, "");
for(let i = 0; i < alphabet.length; i++){
if(lowercase.indexOf(alphabet[i]) === -1){
return 0;
}
}
return 1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., 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);
int t = sc.nextInt();
while(t-->0){
String s = sc.next();
int check = 1;
int p =0;
for(char ch = 'a';ch<='z';ch++){
p=0;
for(int i = 0;i<s.length();i++){
if(s.charAt(i)==ch){p=1;}
}
if(p==0){check=0;}
}
System.out.println(check);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#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 sz 200000
int A[26],B[26];
signed main()
{
int t;
cin>>t;
string p;
getline(cin,p);
while(t>0)
{
t--;
string s;
getline(cin,s);
int n=s.size();
memset(A,0,sizeof(A));
int ch=1;
for(int i=0;i<n;i++)
{
int x=s[i]-'a';
if(x>=0 && x<26)
{
A[x]++;
}
x=s[i]-'A';
if(x>=0 && x<26)
{
A[x]++;
}
}
for(int i=0;i<26;i++)
{
if(A[i]==0) ch=0;
}
cout<<ch<<endl;
}
}, 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 two integers a and b, your task is to calculate values for each of the following operations:-
a + b
a - b
a * b
a/bSince this will be a functional problem, you don't have to take input. You have to complete the function
<b>operations()</b> that takes the integer a and b as parameters.
<b>Constraints:</b>
1 ≤ b ≤ a ≤1000
<b> It is guaranteed that a will be divisible by b.</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5, I have written this Solution Code: def operations(x, y):
print(x+y)
print(x-y)
print(x*y)
print(x//y), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a big number in form of a string A of characters from 0 to 9.
Check whether the given number is divisible by 30 .The first argument is the string A.
<b>Constraints</b>
1 ≤ |A| ≤ 10<sup>5</sup>Return "Yes" if it is divisible by 30 and "No" otherwise. Sample Input :
3033330
Sample Output:
Yes, 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 a= br.readLine();
int sum=0;
int n=a.length();
int s=0;
for(int i=0; i<n; i++){
sum=sum+ (a.charAt(i) - '0');
if(i == n-1){
s= (a.charAt(i) - '0');
}
}
if(sum%3==0 && s==0){
System.out.print("Yes");
}else{
System.out.print("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a big number in form of a string A of characters from 0 to 9.
Check whether the given number is divisible by 30 .The first argument is the string A.
<b>Constraints</b>
1 ≤ |A| ≤ 10<sup>5</sup>Return "Yes" if it is divisible by 30 and "No" otherwise. Sample Input :
3033330
Sample Output:
Yes, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-10 11:06:49
**/
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int solve(string a) {
int n = a.size();
int sum = 0;
if (a[n - 1] != '0') {
return 0;
}
for (auto &it : a) {
sum += (it - '0');
}
debug(sum % 3);
if (sum % 3 == 0) {
return 1;
}
return 0;
}
int32_t main() {
ios_base::sync_with_stdio(NULL);
cin.tie(0);
string str;
cin >> str;
int res = solve(str);
if (res) {
cout << "Yes\n";
} else {
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a big number in form of a string A of characters from 0 to 9.
Check whether the given number is divisible by 30 .The first argument is the string A.
<b>Constraints</b>
1 ≤ |A| ≤ 10<sup>5</sup>Return "Yes" if it is divisible by 30 and "No" otherwise. Sample Input :
3033330
Sample Output:
Yes, I have written this Solution Code: A=int(input())
if A%30==0:
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 a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, 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: 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 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: You are given two positive integers A and B.
Print "First", if A<sup>B</sup> > B<sup>A</sup>.
Print "Second", if A<sup>B</sup> < B<sup>A</sup>.
Otherwise, print "Equal".The only line contains two space-separated integers A and B.
<b>Constraints:</b>
1 ≤ A, B ≤ 4Print a single word – the answer to the problem.Sample Input 1:
1 3
Sample Output 1:
Second
Sample Explanation 1:
As 1<sup>3</sup> < 3<sup>1</sup>, we print "Second".
Sample Input 2:
2 2
Sample Output 2:
Equal
Sample Explanation 2:
As 2<sup>2</sup> = 2<sup>2</sup>, we print "Equal". , 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 s=br.readLine();
String[] arr=s.split(" ");
int a=Integer.parseInt(arr[0]);
int b=Integer.parseInt(arr[1]);
long ab=(long)Math.pow(a,b);
long ba=(long)Math.pow(b,a);
if(ab>ba) System.out.print("First");
else if(ab<ba) System.out.print("Second");
else System.out.print("Equal");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two positive integers A and B.
Print "First", if A<sup>B</sup> > B<sup>A</sup>.
Print "Second", if A<sup>B</sup> < B<sup>A</sup>.
Otherwise, print "Equal".The only line contains two space-separated integers A and B.
<b>Constraints:</b>
1 ≤ A, B ≤ 4Print a single word – the answer to the problem.Sample Input 1:
1 3
Sample Output 1:
Second
Sample Explanation 1:
As 1<sup>3</sup> < 3<sup>1</sup>, we print "Second".
Sample Input 2:
2 2
Sample Output 2:
Equal
Sample Explanation 2:
As 2<sup>2</sup> = 2<sup>2</sup>, we print "Equal". , I have written this Solution Code: a,b=map(int,input().split())
if a**b > b**a:
print("First")
elif a**b <b**a:
print("Second")
else :
print("Equal"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two positive integers A and B.
Print "First", if A<sup>B</sup> > B<sup>A</sup>.
Print "Second", if A<sup>B</sup> < B<sup>A</sup>.
Otherwise, print "Equal".The only line contains two space-separated integers A and B.
<b>Constraints:</b>
1 ≤ A, B ≤ 4Print a single word – the answer to the problem.Sample Input 1:
1 3
Sample Output 1:
Second
Sample Explanation 1:
As 1<sup>3</sup> < 3<sup>1</sup>, we print "Second".
Sample Input 2:
2 2
Sample Output 2:
Equal
Sample Explanation 2:
As 2<sup>2</sup> = 2<sup>2</sup>, we print "Equal". , 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 a,b;
cin>>a>>b;
int ans1=1;
for(int i=0;i<b;i++)
ans1*=a;
int ans2=1;
for(int i=0;i<a;i++)
ans2*=b;
if(ans1>ans2)
cout<<"First";
else if(ans1<ans2)
cout<<"Second";
else
cout<<"Equal";
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton has A cookies, and Einstein has B cookies. Newton will do the following action K times:
1) If Newton has one or more cookies, eat one of his cookies.
2) Otherwise, if Einstein has one or more cookies, eat one of Einstein's cookies.
3) If they both have no cookies, do nothing.
In the end, how many cookies will Newton and Einstein have, respectively?The first line of the input contains 3 integers, A, B and K
<b>Constraints:</b>
1) 0 ≤ A ≤ 10<sup>12</sup>
2) 0 ≤ B ≤ 10<sup>12</sup>
3) 0 ≤ K ≤ 10<sup>12</sup>Print the number of Newton's and Einstein's cookies after K actions.<b>Sample Input 1:</b>
2 3 3
<b>Sample Output 1:</b>
0 2
<b>Sample Explanation 1:</b>
Newton will do the following:
1) He has two cookies, so he eats one of them.
2) Now he has one cookie left, and he eats it.
3) Now he has no cookies left, but Einstein has three, so Newton eats one of them.
Thus, in the end, Newton will have 0 cookies, and Einstein will have 2.
, I have written this Solution Code: #include<iostream>
using namespace std;
int main(){
int64_t A, B, K; cin >> A >> B >>K;
B = A <= K ? max(B-(K-A), (int64_t)0) : B;
A = max(A-K, (int64_t)0);
cout << A << ' ' << B << endl;
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: def compound_interest(principle, rate, time):
Amount = principle * (pow((1 + rate / 100), time))
CI = Amount - principle
print( '%.2f'%CI)
principle,rate,time=map(int, input().split())
compound_interest(principle,rate,time), In this Programming Language: Python, 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: 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: 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: Stickler the thief wants to loot money from a society having n houses in a single line. He is a weird person and follows a certain rule when looting the houses. According to the rule, he will never loot two consecutive houses. At the same time, he wants to maximise the amount he loots. The thief knows which house has what amount of money but is unable to come up with an optimal looting strategy. He asks for your help to find the maximum money he can get if he strictly follows the rule. Each house has a[i] amount of money present in it.The first line of input contains an integer T denoting the number of test cases. T testcases follow. Each test case contains an integer n which denotes the number of houses. Next line contains space separated numbers denoting the amount of money in each house.
1 <= T <= 100
1 <= n <= 10^4
1 <= a[i] <= 10^4For each testcase, in a newline, print an integer which denotes the maximum amount he can take home.Sample Input:
2
6
5 5 10 100 10 5
3
1 2 3
Sample Output:
110
4
Explanation:
Testcase1:
5 + 100 + 5 = 110
Testcase2:
1 + 3 = 4, I have written this Solution Code: def maximize_loot(hval, n):
if n == 0:
return 0
value1 = hval[0]
if n == 1:
return value1
value2 = max(hval[0], hval[1])
if n == 2:
return value2
max_val = None
for i in range(2, n):
max_val = max(hval[i]+value1, value2)
value1 = value2
value2 = max_val
return max_val
t=int(input())
for l in range(0,t):
n=int(input())
arr=input().split()
one=0
two=0
for i in range(0,n):
arr[i]=int(arr[i])
print (maximize_loot(arr,n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Stickler the thief wants to loot money from a society having n houses in a single line. He is a weird person and follows a certain rule when looting the houses. According to the rule, he will never loot two consecutive houses. At the same time, he wants to maximise the amount he loots. The thief knows which house has what amount of money but is unable to come up with an optimal looting strategy. He asks for your help to find the maximum money he can get if he strictly follows the rule. Each house has a[i] amount of money present in it.The first line of input contains an integer T denoting the number of test cases. T testcases follow. Each test case contains an integer n which denotes the number of houses. Next line contains space separated numbers denoting the amount of money in each house.
1 <= T <= 100
1 <= n <= 10^4
1 <= a[i] <= 10^4For each testcase, in a newline, print an integer which denotes the maximum amount he can take home.Sample Input:
2
6
5 5 10 100 10 5
3
1 2 3
Sample Output:
110
4
Explanation:
Testcase1:
5 + 100 + 5 = 110
Testcase2:
1 + 3 = 4, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ld long double
#define ll long long
#define pb push_back
#define endl '\n'
#define pi pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define fi first
#define se second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e4 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N], dp[N][2];
void solve(){
int n; cin >> n;
for(int i = 1; i <= n; i++){
cin >> a[i];
dp[i][0] = max(dp[i-1][1], dp[i-1][0]);
dp[i][1] = a[i] + dp[i-1][0];
}
cout << max(dp[n][1], dp[n][0]) << endl;
}
void testcases(){
int tt = 1;
cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, 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 and an integer K, find and print the number of pairs of indices i, j (1 <= i < j <= N) such that A<sub>i</sub> * A<sub>j</sub> > K.First line of the input contains two integers N and K.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>12</sup>
1 <= A<sub>i</sub> <= 10<sup>6</sup>Print the number of pairs of indices i, j (1 <= i < j <= N) such that A<sub>i</sub> * A<sub>j</sub> > K.Sample Input:
7 20
5 7 2 3 2 9 1
Sample Output:
5
Explanation:
The following pairs of indices satisfy the condition (1-based indexing)
(1, 2) -> 5 * 7 = 35
(1, 6) -> 5 * 9 = 45
(2, 4) -> 7 * 3 = 21
(2, 6) -> 7 * 9 = 63
(4, 6) -> 3 * 9 = 27
All these products are greater than K (= 20)., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long long ll;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve(){
int n, k;
cin >> n >> k;
vector<int> a(n);
for(auto &i : a) cin >> i;
sort(all(a));
int ans = 0;
for(int i = 0; i < n; i++){
int x = k/a[i];
if(x * a[i] < k) x++;
int l = i + 1, r = n - 1, ind = n;
while(l <= r){
int m = (l + r)/2;
if(a[m] >= x){
r = m - 1;
ind = m;
}
else l = m + 1;
}
ans += n - ind;
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << 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 and an integer K, find and print the number of pairs of indices i, j (1 <= i < j <= N) such that A<sub>i</sub> * A<sub>j</sub> > K.First line of the input contains two integers N and K.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>12</sup>
1 <= A<sub>i</sub> <= 10<sup>6</sup>Print the number of pairs of indices i, j (1 <= i < j <= N) such that A<sub>i</sub> * A<sub>j</sub> > K.Sample Input:
7 20
5 7 2 3 2 9 1
Sample Output:
5
Explanation:
The following pairs of indices satisfy the condition (1-based indexing)
(1, 2) -> 5 * 7 = 35
(1, 6) -> 5 * 9 = 45
(2, 4) -> 7 * 3 = 21
(2, 6) -> 7 * 9 = 63
(4, 6) -> 3 * 9 = 27
All these products are greater than K (= 20)., 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 N = sc.nextInt();
long K = sc.nextLong();
long[] arr = new long[N];
for(int i=0; i<N; i++){
arr[i] = sc.nextLong();
}
Arrays.sort(arr);
int low = 0;
int high = N-1;
long count = 0;
while(low<high){
long num = arr[low]*arr[high];
if(num>K){
count += high-low;
high--;
} else {
low++;
}
}
System.out.println(count);
}
}, 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: 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: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, 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: #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: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<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>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: def MakeTheNumber(N) :
for i in range (1,10000):
cnt=0
for x in range (1,i+1):
if(i%x==0):
cnt=cnt+1
if(cnt==N):
return i
return -1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<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>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: int MakeTheNumber(int n){
for(int x=1;x<=10000;x++){
int cnt=0;
for(int i=1;i<=x;i++){
if(x%i==0){cnt++;}
}
if(cnt==n){
return x;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<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>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: public static int MakeTheNumber(int n){
for(int x=1;x<=10000;x++){
int cnt=0;
for(int i=1;i<=x;i++){
if(x%i==0){cnt++;}
}
if(cnt==n){
return x;}
}
return -1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<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>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: int MakeTheNumber(int n){
for(int x=1;x<=10000;x++){
int cnt=0;
for(int i=1;i<=x;i++){
if(x%i==0){cnt++;}
}
if(cnt==n){
return x;}
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
int a[]=new int[str.length];
int sum=0;
for(int i=0;i<str.length;i++)
{
a[i]=Integer.parseInt(str[i]);
sum=sum+a[i];
}
System.out.println(sum/4);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: a,b,c,d = map(int,input().split())
print((a+b+c+d)*100//400), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes, your task is to check if the given list is circular or not.
<b>Note: Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><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>check()</b> that takes the head node as a parameter.
<b>Note: 0 and 1 in sample input just show given LL is CLL or not. 1 denotes it is CLL, 0 denotes not</b>
Constraints:
1 <=N <= 1000
1 <= Node.data<= 1000
Return 1 if the given linked list is circular else return 0.Sample Input 1:-
3 0
1 2 3
Sample Output 1:-
0
Explanation:-
1->2->3
Sample Input 2:-
3 1
1 2 3
Sample Output 2:-
1
Explanation:-
1->2->3->1.......
, I have written this Solution Code: public static int check(Node head) {
if (head == null)
return 1;
Node node = head.next;
while (node != null && node != head)
node = node.next;
if(node==head){return 1;}
else {return 0;}
}, In this Programming Language: Java, 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 a positive integer n. The task is to find the sum of product of all possible pairs of (x, y) where x and y will be represented as such y = floor(n/x), x varies from 1 to n.The first line of input contains an integer T. Each test case contains an integer N.
Constraints:
1 <= T <= 100
1 <= N <= 10^6For each test case in a new line print the required output.Input:
2
5
10
Output:
21
87
Explanation:
Testcase 1:
Following are the possible pairs of (x, y): (1, 5), (2, 2), (3, 1), (4, 1), (5, 1).
So, 1*5 + 2*2 + 3*1 + 4*1 + 5*1
= 5 + 4 + 3 + 4 + 5
= 21., I have written this Solution Code: import math
def sumOfRange(a, b):
i = (a * (a + 1)) >> 1;
j = (b * (b + 1)) >> 1;
return (i - j);
def sumofproduct(n):
sum = 0;
root = int(math.sqrt(n));
for i in range(1, root + 1):
up = int(n / i);
low = max(int(n / (i + 1)), root);
sum += (i * sumOfRange(up, low));
sum += (i * int(n / i));
return sum;
T = int(input())
for i in range(T):
n = int(input())
print(sumofproduct(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer n. The task is to find the sum of product of all possible pairs of (x, y) where x and y will be represented as such y = floor(n/x), x varies from 1 to n.The first line of input contains an integer T. Each test case contains an integer N.
Constraints:
1 <= T <= 100
1 <= N <= 10^6For each test case in a new line print the required output.Input:
2
5
10
Output:
21
87
Explanation:
Testcase 1:
Following are the possible pairs of (x, y): (1, 5), (2, 2), (3, 1), (4, 1), (5, 1).
So, 1*5 + 2*2 + 3*1 + 4*1 + 5*1
= 5 + 4 + 3 + 4 + 5
= 21., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int p[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int ans = 0;
int n; cin >> n;
for(int i = 1; i <= n; i++){
ans += i*(n/i);
}
cout << ans << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer n. The task is to find the sum of product of all possible pairs of (x, y) where x and y will be represented as such y = floor(n/x), x varies from 1 to n.The first line of input contains an integer T. Each test case contains an integer N.
Constraints:
1 <= T <= 100
1 <= N <= 10^6For each test case in a new line print the required output.Input:
2
5
10
Output:
21
87
Explanation:
Testcase 1:
Following are the possible pairs of (x, y): (1, 5), (2, 2), (3, 1), (4, 1), (5, 1).
So, 1*5 + 2*2 + 3*1 + 4*1 + 5*1
= 5 + 4 + 3 + 4 + 5
= 21., I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t!=0){
int n=Integer.parseInt(br.readLine());
long res=0;
for(int i=1;i<=n;i++){
res=res+(i*(int)Math.floor(n/i));
}
System.out.println(res);
t--;
}
}
}, 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 |
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