poj3061(Subsequence)尺取法

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

2
3

很简单的一道题，尺取法就是对数组保存起点和终点，通过根据实际情况交替推进两个端点得到答案。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;

int a[100010];

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n, S;
        cin >> n >> S;
        for (int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        int ans = n + 1;
        int s = 0, t = 0, sum = 0;
        while (1)
        {
            while (t < n && sum < S)
                sum += a[t++];
            if (sum < S)
                break;
            ans = min(ans, t-s);
            sum -= a[s++];
        }
        if (ans > n)
            ans = 0;
        cout << ans << endl;
    }
    return 0;
}

poj3061(Subsequence)尺取法

原文：http://blog.csdn.net/god_weiyang/article/details/47683155