【POJ - 3061】Subsequence

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本文介绍了一种寻找序列中连续子序列最小长度的算法，该子序列的和大于或等于给定值S。通过逐步累加序列元素并比较目标值来实现，适用于编程竞赛和实际问题解决。

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15792		Accepted: 6663

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

从第一个数开始加，不需要排序，加到大于s时，再减去第一个数，减到小于s后再继续向后加。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[100010];
int main()
{
	int m,n,i,s;
	scanf("%d",&m);
	while(m--)
	{
		int flag;
		int left,right;
		scanf("%d%d",&n,&s);
		a[0]=0;                   //设a[0],从a[0]开始加
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			a[i]=a[i]+a[i-1];     //前i个数的和
		}
		if(a[n]<s)
		printf("0\n");
		else 
		{
		    left=1;right=n;
			while(left<right-1)
			{
				flag=0;
			    int mid=(left+right)/2;
				for(i=mid;i<=n;i++)
				{
					if((a[i]-a[i-mid])>=s)
					{
						flag=1;
						break;
					}
				}
				if(flag==1)
					right=mid;
				else  left=mid;
			}
			printf("%d\n",right);
		}
	} 
return 0;
}