子序列（LA 2678） (二分查找)_子序列 la2678题解-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_41061455/article/details/79561097

本文介绍了一种算法，用于解决给定正整数序列中找到最短连续子序列的问题，该子序列的和大于等于指定值S。通过使用前缀和与二分查找的方法，实现了高效求解。

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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

2 
3

题意：从给定的一段序列中找最短的连续子列, 使得他们的和大于等于S。

在这里，其实有个前缀和技巧，又由于都是正整数，所以前缀和是非递减的，直接二分查找。

#include<bits/stdc++.h>
using namespace std;
int b[100005];
int a[100005];
int main()
{
    int n,s;
    while(scanf("%d%d",&n,&s)==2)
    {
        b[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=b[i-1]+a[i];
        }
        int ans=n+1;
        for(int j=1;j<=n;j++)
        {
            int i=lower_bound(b,b+j,b[j]-s)-b;
            if(i>0) ans=min(ans,j-i+1);
        }
        printf("%d\n",ans==n+1?0:ans);
    }
    return 0;
}