Subsequence POJ - 3061 （尺取模板）

题意：给定长度n的整数数组，以及整数s，求出总和不小于s的子数组长度的最小值。

 

Input

第一行输入n和s，第二行输入n个正整数。

 

Output

找到满足要求的子数组的最小长度，如果没有，输出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

 

代码；

 

#include <iostream>
#include <algorithm>
#include <string.h>
#include <cstdio>
#include <string>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
//#include <unordered_map>
#define Fbo friend bool operator < (node a, node b)
#define mem(a, b) memset(a, b, sizeof(a))
#define FOR(a, b, c) for (int a = b; a <= c; a++)
#define RFOR(a, b, c) for (int a = b; a >= c; a--)
#define off ios::sync_with_stdio(0)
#define sc(a) scanf("%d",&a) 
bool check1(int a) { return (a & (a - 1)) == 0 ? true : false; }

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
const int INF = 0x3f3f3f3f;//1e10
const int mod = 1e9 + 7;
const int Maxn = 2e5+9;
const double pi = acos(-1.0);
const double eps = 1e-8;

int a[Maxn];

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, s;
        cin >> n >> s;
        FOR(i, 1, n)sc(a[i]);
        int i = 1, j = 1, sum = 0, ans = INF;
        while (1) {
            while (j <= n && sum < s) {
                sum += a[j++];
            }
            if (sum < s) break;
            ans = min(ans, j - i);
            sum -= a[i++];
        }
        if (ans == INF)printf("0\n");
        else printf("%d\n", ans);
    }
}