很久以前就学啦尺取啦 但是一直都没来及整理  先贴一个简单题模板

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

Input

The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

Output

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

Sample Input

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[100010];
int min1(int a,int b)
{
    if(a<b)
        return a;
    else
        return b;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        long long sum=0;
        int st=0,en=0;
        int ans=0x3f3f3f3f;
        while(1)
        {
            while(en<n&&sum<m)           //主要就在这边每次确定好位置后把原来的sum减去
             sum+=a[en++];         
            if(sum<m)                    //初始位置的sum来减少数据量
             break;
            ans=min1(ans,en-st);
            sum-=a[st++];
        }
        if(ans==0x3f3f3f3f) ans=0;
        printf("%d\n",ans);
    }

    return 0;
 }