ZCMU—J

J - Single Round Math

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

Association for Couples Math (ACM) is a non-profit organization which is engaged in helping single people to find his/her other half. As November 11th is “Single Day”, on this day, ACM invites a large group of singles to the party. People round together, chatting with others, and matching partners.

There are N gentlemen and M ladies in the party, each gentleman should only match with a lady and vice versa. To memorize the Singles Day, ACM decides to divides to divide people into 11 groups, each group should have the same amount of couples and no people are left without the groups.

Can ACM achieve the goal?

Input

The first line of the input is a positive integer T. T is the number of test cases followed. Each test case contains two integer N and M (0 ≤ N, M ≤ 101000), which are the amount of gentlemen and ladies.

Output

For each test case, output “YES” if it is possible to find a way, output “NO” if not.

Sample Input

3
1 1
11 11
22 11

Sample Output

NO
YES
NO

【分析】

题意：给你一群单身狗，n个男性单身狗和m个女性单身狗，问你他们能不能完全成对并且均分成11组..所以首先要满足n=m

然后满足n%11=0就可以了...10^1000次所以大数除法...这里要提一下讲道理稍微学一下java会做个大数四则运算还是有用的....

大数除法是比较麻烦的所以直接java了，大数加减乘法最好还是要熟练的会写

【代码】

import java.math.BigInteger;  
import java.util.Scanner;   
public class Main 
{  
    public static void main(String[] args)
	{        
        BigInteger MOD=new BigInteger("11");  
        BigInteger zero=new BigInteger("0");  
        Scanner sc=new Scanner(System.in);  
        int pp;
    	pp=sc.nextInt();  
        while((pp--)>0)
		{        
            BigInteger n=sc.nextBigInteger();  
            BigInteger m=sc.nextBigInteger();  
            if(n.compareTo(m)==0 && n.mod(MOD).compareTo(zero)==0)  
                System.out.println("YES");  
            else
                System.out.println("NO");  
        }  
          
    }  
}