Balls and Boxes codechef problem Problem Code: BALLBOX

                           Balls And Boxes 

Problem

You have $N$ balls and $K$ boxes. You want to divide the $N$ balls into $K$ boxes such that:

• Each box contains $\ge 1$ balls.
• No two boxes contain the same number of balls.

Determine if it is possible to do so.

Input Format

• The first line contains a single integer $T$ — the number of test cases. Then the test cases follow.
• The first and only line of each test case contains two space-separated integers $N$ and $K$ — the number of balls and the number of boxes respectively.

Output Format

For each test case, output YES if it is possible to divide the $N$ balls into $K$ boxes such that the conditions are satisfied. Otherwise, output NO.

You may print each character of YES and NO in uppercase or lowercase (for example, yes, yEs, Yes will be considered identical).

Sample 1:

Input

Output

4
3 4
30 3
2 2
1 1

NO
YES
NO
YES

Explanation:

Test Case 1: It is not possible to divide the $3$ balls into $4$ boxes such that each box contains $\ge 1$ balls.

Test Case 2: One way to divide the $30$ balls into $3$ boxes is the following: $[5, 9, 16]$.

Test Case 3: It is not possible to divide the $2$ balls into $2$ boxes such that no two boxes contain the same number of balls.

Test Case 4: We can divide $1$ ball into $1$ box

Solution

  
  #include <iostream>
using namespace std;

int main() {
	// your code goes here
	
	int t ;
	cin >> t;
	
	
	while(t--){
	    
	    int n,k, x;
	    cin >> n >> k ;
	    
	    x = (k * (k + 1)) / 2;
	    
	    if(n >= x){
	        cout << "YES" << endl;
	    }
	    else{
	        cout << "NO" << endl;
	    }
	    
	  
	}
	    
}

  

Code Explained

According to condition we have to put atleast single ball in a box and no two box contain same number of balls

First box contain atleast 1 ball and second box contain atleast 2 balls like this 3 number box contain atleast 3 balls , 

so the number of total balls must be more then or equals to 

                                        1 + 2 + 3 = 6

for fullfill all conditions.

   like that if we have k number of box then the least number of balls for fulfill all condition must be more then 

k * (k + 1) / 2 

balls ( which is sum of natural number from 1 to k )