Single Round Match

最新推荐文章于 2022-12-10 16:39:26 发布

梧桐下的四叶草

最新推荐文章于 2022-12-10 16:39:26 发布

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分类专栏：水题文章标签： c++

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本文介绍了一个基于特定条件的情侣匹配问题。该问题要求在派对上将N位男士与M位女士分成11组，每组包含相同数量的情侣。通过分析输入数据，并利用能否被11整除的特性来判断是否可以实现这一目标。本文提供了具体的输入输出样例及C++实现代码。

Single Round Match

(Input File: single.in / Standard Output)

Association for Couples Match (ACM) is a non-profit organization which is engaged in helping single people to find his/her other half. As November 11th is “Singles 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 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

与能否被11整除的题结合起来的题。。。
AC代码：

#include <iostream>
#include <cstring>
using namespace std;
int fun(char* p)
{
    int i=0;
    long long sum=0;
    while (p[i]!='\0')
    {
        if(i&1)
        sum+=-(p[i]-'0');
        else
            sum+=p[i]-'0';
        i++;
    }
    if (sum%11==0)return 1;
    return 0;
}
int main()
{
    int N;
    char n[1001],m[1001];
    cin >> N;

    while (N--)
    {
        cin >> n >> m;
        if (strcmp(n,m)==0&&fun(m))
        {
            cout << "YES" << endl;
        }
        else
        {
            cout << "NO" << endl;
        }
    }
    return 0;
}