【POJ】2061 - Subsequence（STL & 二分）

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该博客主要解析了POJ 2061题目的解法，内容涉及STL和二分查找技术。博主通过解释题意，给出了输入输出格式，并提供了样例及相应的代码实现，该代码利用动态规划和二分查找找到满足条件的最短子序列长度。

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Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11797		Accepted: 4949

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

Southeastern Europe 2006

题意：给出n个数和一个S，求这 n 个数的子序列大于等于 S 的最短序列长度是多少。

题解：求出当 i 为之前面所有数的和。然后从最后一个数开始，找出第一个满足条件的数的位置，求之间的距离更新 ans 值，最后得出最短的 ans 。

代码如下：

#include <cstdio>
#include <algorithm>
#define MAX 100000
using namespace std;
int main()
{
	int u;
	int sum[MAX+22];
	int ans;
	int n,S;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d %d",&n,&S);
		scanf ("%d",&sum[0]);
		for (int i = 1 ; i < n ; i++)
		{
			scanf ("%d",&sum[i]);
			sum[i] += sum[i-1];
		}
		if (sum[n-1] < S)
		{
			printf ("0\n");
			continue;
		}
		ans = n;
		int pos;		//存放序列开始的位置 
		for (int i = n - 1 ; i >= 0 ; i--)		//从最后面开始，把其当作序列尾 
		{
			if (sum[i] < S)
				break;
			pos = upper_bound(sum , sum + n , sum[i] - S) - sum;
			ans = min(ans , i - pos + 1);
		}
		printf ("%d\n",ans);
	}
	return 0;
}