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本文介绍了一个编程问题：给定一系列正整数及一个目标值S，如何找到该序列中连续元素子序列的最短长度，使得这些元素的总和大于或等于S。文章提供了完整的C语言代码实现，并详细解释了其工作原理。

描述
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

Sample Output
2
3

#include<stdio.h>
int n,s,a[100005];
bool check(int ans)
{
    int sum=0;
    for(int j=0;j<ans;j++)
        sum+=a[j];
    for(int j=ans;j<=n;j++)
    {
        if(sum>=s)
            return true;//ans一定成立 让top=ans 
        sum+=a[j]-a[j-ans];
    }
    return false;//ans一定不成立 让low=ans 
}
int main()
{
    int T;
    while(~scanf("%d",&T))
    {
        while(T--)
        {
            scanf("%d%d",&n,&s);int sum=0,max=-100000; 
            for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
                if((sum+=a[i])<0) sum=0;
                if(sum>max) max=sum;//记录最大连续片段和 
            }
            if(max<s)//如果 大连续片段和小于 s 显然答案不存在 输出0 
            {
                printf("0\n");
                continue;
            }
            int low=0,top=n;
            while(top-low>1)//寻找满足条件的最小值 
            {
                int m=(top+low)>>1;
                if(check(m))
                    top=m;
                else
                    low=m;
            }
            printf("%d\n",top);
        }
    }
    return 0;
}