Poj3061(尺取法)

Subsequence
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 17572 Accepted: 7481
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
Source

Southeastern Europe 2006

题目大意：
找到数组中最短的长度满足之和大于或等于s

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
long long int a[4353432];
int main()
{
    int t, n, m;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &n, &m);
        for(int i=0;i<n;i++)
            scanf("%lld", a+i);
        int l = 0, r = 0;//记录左边起点和右边终点
        int ans = INF, mik=0;//记录最小值
        while(1)
        {
            while(r<n&&mik<m)
                mik += a[r++];
            if(mik<m)//大于数组长度
                break;
            ans = min(ans, r-l);//取最短
            mik -= a[l++];//起始点左移
        }
        if(ans==INF)
            printf("0\n");
        else
            printf("%d\n", ans);
    }
    return 0;
}