Janmansh and Assignments CodeChef Solution

Problem – Janmansh and Assignments CodeChef Solution

Problem

Janmansh has to submit 33 assignments for Chingari before 1010 pm and he starts to do the assignments at XX pm. Each assignment takes him 11 hour to complete. Can you tell whether he’ll be able to complete all assignments on time or not?

Input Format

  • The first line will contain TT – the number of test cases. Then the test cases follow.
  • The first and only line of each test case contains one integer XX – the time when Janmansh starts doing the assignemnts.

Output Format

For each test case, output Yes if he can complete the assignments on time. Otherwise, output No.

You may print each character of Yes and No in uppercase or lowercase (for example, yesyEsYES will be considered identical).

Constraints

  • 1 \le T \le 101≤T≤10
  • 1 \le X \le 91≤X≤9

Sample 1:

Input: 2
7
9
Output: Yes
No

Explanation:

Test case-1: He can start at 77pm and finish by 1010 pm. Therefore he can complete the assignments.

Test case-2: He can not complete all the 33 assignments if he starts at 99 pm.

Janmansh and Assignments CodeChef Solution in Java

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		while (t-- > 0) {
		  int x = Integer.parseInt(br.readLine());
		  System.out.println(10 - x >= 3 ? "Yes" : "No");
		}
	}
}

Janmansh and Assignments CodeChef Solution in Pyth 3

# cook your dish here
t=int(input())
for i in range(t):
    x=int(input())
    if x+3<=10:
        print("yes")
    else:
        print("no")
    
    

Janmansh and Assignments CodeChef Solution in C++17

#include <iostream>
using namespace std;

int main() {
	int t;
	cin>> t;
	while(t--){
	    int x;
	    cin>>x;
	    if(10-x>=3){
	        cout<<"Yes";
	    }
	    else{
	        cout<<"No";
	    }
	    cout<<endl;
	}
	return 0;
}
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