Alternate Additions Solution Codechef

Alternate Additions Solution Codechef

Chef has 2 numbers A and B (A<B). Chef will perform some operations on A.

In the i^th operation:

• Chef will add 1 to A if i is odd.
• Chef will add 2 to A if i is even.

Chef can stop at any instant. Can Chef make A equal to B?

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 A and B.

Output Format

• For each test case, output YES if Chef can make A and B equal, NO otherwise.
• Note that the checker is case-insensitive. So, YES, Yes, yEs are all considered same.

Constraints

• 1≤T≤1000
• 1≤A<B≤10⁹

Sample 1:

Input

4
1 2
3 6
4 9
10 20

Output

YES
YES
NO
YES

Explanation:

SOLUTION

Program: Alternate Additions Solution in Python

for _ in  range(int(input())):
    a,b = map(int,input().split(' '))
    N = b-a
    if(N%3==0 or N%3==1):
        print('yes')
    else:
        print('No')

Program: Alternate Additions Solution in C++

#include <iostream>
using namespace std;
int main() 
{
    int t;
    cin>>t;
    while(t--)
    {
         int a,b;
         cin>>a>>b; 
          if(b-a<0)
          {
              cout<<"NO"<<endl;
          }
          else
          {
             if((b-a)%3==0||(b-a-1)%3==0)
             {
              cout<<"YES"<<endl;
             }
             else
             {
               cout<<"NO"<<endl;
             }
          }
    } 
	return 0;
}

Program: Alternate Additions Solution in Java

import java.util.*;
public class Main
{
	public static void main(String[] args) 
	{
	    Scanner sc=new Scanner(System.in);
	    int t=sc.nextInt();
	    while(t--!=0)
	    {
	        int a=sc.nextInt();
	        int b=sc.nextInt();
	        if(Math.abs(b-a)%3!=1 && Math.abs(b-a)%3!=0)
	        System.out.println("NO");
	        else
	        System.out.println("YES");
	    }
		
	}
}

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