hdu 5776 sum（前缀和取模）

sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 323 Accepted Submission(s): 156



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

题意：有n个数，问存不存在连续子序列之和是m的倍数

思路：比赛的时候看错题意，以为是求等于m的，故用尺取法秒解，居然还AC了，真是费解 ...

现在重新做一遍，此题应该用前缀和来处理问题。

如果前缀和存在%m后==0的不用考虑，YES

如果存在余数为1~m-1的个数大于一个时我们也能保证为YES 为什么呢？ 我们假设sum[1~i]%m为k,sum[1~j]%m为k（j>i）

那么容易想到sum[i+1,j]%m==0

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 100050
int a
,sum
;
int vis
;
int main()
{
int T,n,m,t;
scanf("%d",&T);
while(T--)
{
memset(vis,0,sizeof(vis));
scanf("%d %d",&n,&m);
sum[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=sum[i-1]+a[i];
vis[sum[i]%m]++;
}
int flag=0;
if(vis[0]) flag=1;
else
{
for(int i=1;i<m;i++)
if(vis[i]>=2)
flag=1;
}
if(flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}