2678 - Subsequence

最新推荐文章于 2018-08-20 16:50:07 发布

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本文介绍了一个编程问题：给定一系列正整数及一个目标值S，如何找到连续子序列中最小长度使得其元素之和大于等于S。通过滑动窗口算法解决此问题，并提供了完整的C++代码实现。

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

Many test cases will be given. 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 there isn't such a subsequence, print 0 on a line by itself.

Sample Input

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

Sample Output

#include <iostream>
#include <stdio.h>
using namespace std;
const int MAXN = 100010;
int g[MAXN];
int n, m;
int main()
{
    int i, sum, ans, l;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        ans = n+1;
        l = 0;
        sum = 0;
        for(i=0; i<n; i++)
        {
            scanf("%d",&g[i]);
        }
        for(i=0; i<n; i++)
        {
            sum+=g[i];
            while(sum>m)
            {
                ans = min(ans, i-l+1);
                sum -= g[l++];
            }
        }
        if(ans == n+1)ans = 0;
        printf("%d\n",ans);
    }
    return 0;
}