POJ 3061 Subsequence

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本文介绍了一种使用尺取法解决寻找最小长度连续子序列的问题，使得子序列的总和不小于给定值S。通过逐步调整子序列的起始和结束位置来找到满足条件的最短序列。

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Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15819		Accepted: 6675

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

题目链接：http://poj.org/problem?id=3061

题意：给定长度为n的整数数列以及整数S，求总和不小于S的连续子序列长度的最小值。如果杰不存在，则输出0。

解题思路：采用尺取法，设以a[s]开始的最初大于S的连续子序列为a[s]+a[s+1]+...+a[t-1]，这时a[s+1]+a[s+2]+...+a[t-2]<S，所以从a[s+1]开始总和最初超过S的连续子序列如果是a[s+1]+a[s+2]+...+a[t‘-1]的话，则必然有t<=t'。利用这一性质便可以设计出这样的算法：1、以s=t=sum=0初始化。2、只要依然有sum<S，就不断将sum增加a[t]，并将t增1。3、如果2中无法满足sum>=S的条件的话就跳出循环。否则更新ans=min(ans,t-s)。4、将sum减去a[s]，s增加1后回到2。（可以参考挑战程序设计竞赛）

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN = 1e5+7;
int n,S;
int a[MAXN];
void solve()
{
	int ans=n+1;
	int sum=0,s=0,t=0;
	for(;;)
	{
		while(t<n&&sum<S)
		{
			sum+=a[t++];
		}
		if(sum<S) break;
		ans=min(ans,t-s);
		sum-=a[s++];
	}
	if(ans>n) ans=0;
	printf("%d\n",ans);
}
int main(void)
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&S);
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
		}
		solve();
	}
	return 0;
}