2019_GDUT_新生专题I选集 A-尺取 POJ-3061

题目：

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

分析：尺取法，两个变量模拟尺头和尺尾，从第一个元素将尺尾不断后移开始找，当找到第一串符合条件的子序列后保存答案，再将尺头后移一位，重新查找并将查找到的答案与已有答案比较，直到尺头移到了最后一个元素或者尺尾已经到最后一个元素但是当前子序列和不足s为止。
代码：

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
	int t,n,s,a[100001];
	cin>>t;
	while (t--)
	{
		cin>>n>>s;
		for (int i=1;i<=n;i++)
		scanf("%d",a+i);
		int l=1,r=1;
		int ans=2147483647,sum=0;
		while (l<=n)
		{
			while (sum<s && r<=n)
			{
				sum+=a[r];
				r++;
			}
			if (sum<s)
			break;
			if (r-l<ans) ans=r-l;
			sum-=a[l];
			l++;
		}
		if (ans>n) cout<<0<<endl;
		else
		cout<<ans<<endl;
	}
}