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 int
const int N=1e5+5;
int sum
;
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;
}