hdu 5776 sum 尺取法

sum

   

 Accepts: 823

   

 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 ($1\le T\le 10$), 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 ($1\le n\le 100000$, $1\le m\le 5000$). 2.The second line contains n positive integers x ($1\le x\le 100$) according to the sequence.

Output

Output T lines, each line print a YES or NO.

Sample Input

23 31 2 35 76 6 6 6 6

Sample Output

YES
NO
题意：能不能找一个连续子区间的值等于m
代码：
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;int a[100005];int main (){    int t,n,m;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        int r=1,sum=0,flag=0;        for(int i=1;i<=n;i++){            while(r<=n&&sum<m)                sum+=a[r++];            if(sum==m){                flag=1;                break;            }            sum-=a[i];        }        if(flag)            printf("YES\n");        else printf("NO\n");    }}