POJ 3061 Subsequence 尺取法附基础模板

最新推荐文章于 2021-01-06 02:16:56 发布

原创最新推荐文章于 2021-01-06 02:16:56 发布 · 363 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法 #acm #poj

ACM-区间-尺取法专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种寻找连续子序列最小长度的算法，使得该子序列的元素之和大于或等于给定数值S。通过双指针技术实现高效求解，并提供了一个完整的C++代码示例。

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15909		Accepted: 6728

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

[Submit] [Go Back] [Status] [Discuss]

传送门：http://poj.org/problem?id=3061

AC代码&&模板：

#include <cstdio>
#include <algorithm>
#include <cstring>

const int INF =1000000;


using namespace std;
int a[100010];
int n, t, ans = INF;
int sum, s;

int main()
{
    scanf("%d", &t);
    while (t--){
        scanf("%d %d", &n, &s);
        for (int i = 0; i < n; i++) scanf("%d", a+i);
        int st = 0, en = 0;
        ans =INF ; sum = 0;
        while (1){
            while (en<n && sum<s) sum += a[en++];
            if (sum < s) break;
            ans = min(ans, en-st);
            sum -= a[st++];
        }
        if (ans == INF) ans = 0;
        printf("%d\n", ans);
    }
    return 0;
}