[POJ](3061)Subsequence ---- 尺取法

Description

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

尺取法讲解： 尺取法（一）

这个题一年之前做过，没想到当时自己写出了尺取法，还不知道当时是神马情况，蒟蒻！变捞了！

注释： POJ还是加速外挂和万用头文件用不了呀！

AC代码：

#include
#include
#include
using namespace std;
const int maxn = 1e5+10;
int a[maxn];
int n,s;
int flag;
int inchworm()
{
    int i=0,j=0;
    int sum = a[0];
    int ans = 0x3f3f3f3f;
    while(i<=j && jif(sum>=s)
        {
            ans = min(ans,j-i+1);
            sum-=a[i];
            i++;
            flag = 1;
        }
        else
        {
            j++;
            sum+=a[j];
        }
    }
    return ans;
}
int main()
{
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    #endif //LOCAL
//    ios_base::sync_with_stdio(false);
//    cin.tie(NULL),cout.tie(NULL);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        flag = 0;
        scanf("%d %d",&n,&s);
        for(int i=0;iscanf("%d",&a[i]);
        int ans = inchworm();
        if(flag) printf("%d\n",ans);
        else
            printf("0\n");
    }
    return 0;
}