补写：Best Coder #85 1001 Sum(前缀和)

sum

Accepts: 640

Submissions: 1744

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 131072/131072 K (Java/Others)

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≤101 \leq T \leq 101≤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≤1000001 \leq n \leq 1000001≤n≤100000, 1≤m≤50001 \leq m \leq 50001≤m≤5000). 2.The second line contains n positive integers x (1≤x≤1001 \leq x \leq 1001≤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

/*第一次来打BC，作为大一的大水B，水出一道，已经很高兴了*/
/*用前缀和来遍历每一个连续数列的和*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 100010
using namespace std;
long long a[N],sum[N];
int main()
{
    //freopen("in.txt", "r", stdin);
    int t;
    int n,m;
    scanf("%d",&t);
    while(t--)
    {
        memset(sum,,sizeof sum);
        memset(a,,sizeof a);
        scanf("%d%d",&n,&m);
        for(int i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-]+a[i];
        }
        int flag=;
        for(int i=;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                if((sum[j]-sum[i-])%m==)
                {
                    puts("YES");
                    flag=;
                    break;
                }
            }
            if(!flag)
                break;
        }
        if(flag)
            printf("NO\n");
    }
    return ;
}

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