hdu 5776 sum 前缀和

sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1126    Accepted Submission(s): 494

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

 

Source

BestCoder Round #85

  思路：当有两个前缀和相等，可以得到这段区间的和整除m；

　　　余m==0特判；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 0.00000000001
const int N=5e3+10,M=1e6+10,inf=1e9+10,mod=1000000007;
int flag[N];
int main()
{
    int x,y,z,i,t;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(flag,0,sizeof(flag));
        scanf("%d%d",&x,&z);
        int ans=0,sum=0;
        for(i=0;i<x;i++)
        {
            scanf("%d",&y);
            sum+=y;
            sum%=z;
            if(flag[sum])
            ans=1;
            flag[sum]=1;
        }
        if(ans||flag[0])
        printf("YES
");
        else
        printf("NO
");
    }
    return 0;
}