Poj 3061Subsequence (How to take the ruler at first sight)

 The time complexity of the ruler method is O(n), because the two pointers can move up to 2*n times in total. The specific movement is in the comment.

code：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <stack>
#include <map>
#include <sstream>
#include <cstring>
#include <set>
#include <cctype>
#include <bitset>
#define IO                       \
    ios::sync_with_stdio(false); \
    // cin.tie(0);                  \
    // cout.tie(0);
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const LL mod = 11092019;
int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
int a[maxn];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif
    IO;
    int T;
    int n, s;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d %d", &n, &s);
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        int L = 1;
        int R = 1;
        int t = a[1];
        int ans = inf;
        while (L <= n && R <= n) // 区间和 >= s
        {
            if (t >= s)
            {
                ans = min(ans, R - L + 1); // 更新答案
                t -= a[L]; // 左端点左移
                ++L; 
            }
            else // 区间和小于 s 时 
            {
                ++R; // 向右拓
                t += a[R];
            }
        }
        if (ans == inf)
            cout << 0 << endl;
        else
            cout << ans << endl;
    }
    return 0;
}