HDU 5776 sum（思维题+前缀和）

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

大致题意：给你长度为n的序列，问你是否存在一个连续的子序列，使得该子序列的和能被m整除。

思路：假设sum[i]表示前i个数的前缀和，如果存在sum[i]%m==sum[j]%m，那么（sum[j]-sum[i]）%m==0。

代码如下

//#include<bits/stdc++.h>#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;  #define LL long long intconst int N=1e5+5;int sum[N];int f[5005];int main(){      int T;    scanf("%d",&T);    int n,m;    while(T--)    {        int flag=0;        scanf("%d%d",&n,&m);        sum[0]=0;        memset(f,0,sizeof f);        for(int i=1;i<=n;i++)        {            int x;            scanf("%d",&x);            sum[i]=sum[i-1]+x;            int wei=sum[i]%m;            f[wei]++;            if(f[wei]%2==0) flag=1;        }        if(flag==1||f[0])            printf("YES\n");        else             printf("NO\n");    }    return 0;}