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.
题目大意
有n个数(10<n<1e6),求k个连续的数,使得这k个数的值大于或等于S,求k的最小值
解题思路
首先如果枚举左右端点来计算k个数的值,那么时间复杂度O(n^2),超时。

假设有一段数列
a[i],a[i+1],a[i+2],…,a[j-1]的和小于S,
a[i],a[i+1],a[i+2],…,a[j-1]+a[j]的和大于等于S,
那么以a[i]为开头的k此时是最小的。
所以此时应该以a[i+1]为开头,
如果此时a[i+1],a[i+2],…,a[j-1]+a[j]的和小于S,那么数列应该不断向右拓展直至和大于等于S,比较此时的k值。
然后再把开变成a[i+2]不断重复这个过程。
关于数列的和,用一个变量保存就可以了,一个数入队,那么该变量加上这个数,一个数出队,该变量减去这个数。时间复杂度O(n)
具体代码实现

#include <cstdio>
using namespace std;
int a[101010];
int main()
{
	int T;
	scanf("%d",&T);
	while (T--)
	{
		int n,s;
		scanf("%d%d",&n,&s);
		for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	    int num=0,sum=0;
	    for (int i=1;i<=n;i++)
	      {
	        sum+=a[i];
	        num++;
	        if (sum>=s) break;
	      }
	    int L=1; 
		int ans=num;
		if (sum<s) 
		{
		   printf("0\n"); 
		   continue;
	    }
		for (int i=num;i<=n;i++)
		  {
		  	if (i!=num) sum+=a[i];
		  	while (sum-a[L]>=s) 
			  {
			  	 sum-=a[L];
			     L++;
			     if (i-L+1<ans) ans=i-L+1;  
			  }
		  }
		printf("%d\n",ans);     
	} 
}