POJ - 3061 Subsequence （尺取法

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本文介绍了一种解决特定区间和问题的方法，通过编程实现寻找序列中连续元素子序列的最短长度，使得这些元素的总和大于或等于给定值S。文章提供了完整的AC代码示例，并介绍了其核心思路。

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题目链接

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 

给出一个序列,求区间和大于或者等于S的最短区间长度.

以前学过的一个方法，代码解释在自己电脑里，先把ac代码放上回头再更新笔记版代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
int a[100005];
int n,S;
void slove()
{
	int res=n+1;
	int s=0,sum=0,z=0;
	for(;;)
	{
		while(z<n&&sum<S)
		{
			sum+=a[z++];
		}
		if(sum<S)break;
		res=min(res,z-s);
		sum-=a[s++];
	}
	if(res>n)
		res=0;
	cout<<res<<endl;
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		cin>>n>>S;
		for(int i=0;i<n;i++)
		{
			cin>>a[i];
		}
		slove();
	}
}