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?
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 ≤ 10^1000), which are the amount of gentlemen and ladies.
For each test case, output “YES” if it is possible to find a way, output “NO” if not.
“浪潮杯”山东省第六届ACM大学生程序设计竞赛
题解：
判断是否被11整除。根据11的倍数的特征 奇数位数字之和与偶数位数字之和的差是11的倍数
或者通过高精度取余计算
代码：
/** 11的倍数的特征 */ #include<bits/stdc++.h> using namespace std; string a,b; int main() { int caset;scanf("%d",&caset); while(caset--) { cin>>a>>b; if(a == b) { int ji = 0,ou = 0; for(int i=0;i<a.length();i++) { if(i & 1) ji += a[i] - 48; else ou += a[i] - 48; } if(abs(ji-ou) % 11 == 0) printf("YES\n"); else printf("NO\n"); } else printf("NO\n"); } return 0; }