POJ 3061 Subsequence

最新推荐文章于 2020-02-07 22:32:37 发布

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本文介绍了一个关于寻找连续子序列的最小长度的问题，使得该子序列的元素之和大于或等于给定的目标值S。文章提供了完整的代码实现，并通过样例输入输出展示了如何解决这一问题。

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8695		Accepted: 3441

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

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int n,ss;
int a[100005],s[100005];
int main(){
	int t;
	cin>>t;
	while(t--){
		memset(a,0,sizeof(a));
		memset(s,0,sizeof(s));
		int kk=100005;
		cin>>n>>ss;
		for(int i=1;i<=n;i++){
			cin>>a[i];
			s[i]=s[i-1]+a[i];
		}
		if(s[n]<ss)
			cout<<"0"<<endl;
		else{
			for(int i=1;s[i]+ss<=s[n];i++){
				int k=lower_bound(s+i,s+n+1,s[i]+ss)-s;
				kk=min(kk,k-i);				
			}
			cout<<kk<<endl;
		}
	}
	return 0;
}