BestCoder Round #85 sum

大晚上的更一道下午的水题吧。(虽然WA了好多次= =，但真实情况是我比较水)

描述

Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO.

输入

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.

输出

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

样例输入

2
3 3
1 2 3
5 7
6 6 6 6 6

样例输出

YES
NO

代码如下：

 #include <cstdio>
 #define N 100100 

 int main()
 {
     int t,n,m,a[N];
     int flag=;
     scanf("%d",&t);
     while(t--){
         scanf("%d %d",&n,&m);
         int sum=;
         int k=;
         for(int j=;j<=n;j++){
            scanf("%d",&a[j]);
         }
         for(int i=;i<=n;i++){
                while(k<=n&&sum<m){
                   sum+=a[k++];
                }
                if(sum==m){
                      flag=;
                      break;
                }
                sum-=a[i];

         }
         if(flag==){
             printf("YES\n");
         }else
             printf("NO\n");
         flag=;
     }
     return ;
 } 

题意：

题意：能不能找一个连续子区间的和是m的倍数。

思路总结：

简单尺取法。既然说找的是m的倍数，那就从头开始加起来。加到数小于m且加起来得到的和是最大的小于m的数。开始判断。如果失败就从头开始减一个，在接着从后加一个。一点点枚举。像尺子一样~~所以叫尺取法~~HOHO~