HDU 5776 sum（抽屉原理）

题目传送：http://acm.hdu.edu.cn/showproblem.php?pid=5776

Problem Description

Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO

Input

The first line of the input has an integer T (1≤T≤10), which represents the number of test cases. 
For each test case, there are two lines:
1.The first line contains two positive integers n, m (1≤n≤100000, 1≤m≤5000).
2.The second line contains n positive integers x (1≤x≤100) according to the sequence.

Output

Output T lines, each line print a YES or NO.

Sample Input

2

3 3

1 2 3

5 7

6 6 6 6 6

Sample Output

YES

NO

 

直接拿这道代码改了一下，两道题一个意思http://www.cnblogs.com/Annetree/p/7095096.html

 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string.h>
#include<cmath>
using namespace std;
#define N 100005


struct node
{
    int num,r;
}k[N];

bool cmp(node a,node b)
{
    if(a.r==b.r)
        return a.num<b.num;
    return a.r<b.r;
}

int main()
{
    long long sum,a;
    int n,m;
    bool flag;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        flag=false;
        if(n>m)
            flag=true;
        sum=0;
        k[0].num=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a);
            sum+=a;
            k[i].r=sum%m;
            k[i].num=i;
            if(k[i].r==0&&flag==false)
            {
                flag=true;
            }
        }
        if(!flag)
        {
            sort(k+1,k+1+n,cmp);
            for(int i=2;i<=n;i++)
            {
                if(k[i].r==k[i-1].r)
                {
                    flag=true;
                    break;
                }
            }
        }
        if(!flag)
            printf("NO
");
        else
            printf("YES
");
    }
    return 0;
}