poj 3061 Subsequence

最新推荐文章于 2020-10-08 19:47:33 发布

尘飞acmer

最新推荐文章于 2020-10-08 19:47:33 发布

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分类专栏： acm初级训练文章标签： poj 二分尺取法

本文链接：https://blog.youkuaiyun.com/motefly_acmer/article/details/37904105

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Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8356		Accepted: 3241

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

书上说这叫尺取法，听起来让我有点忧伤。。。貌似是说：保存数组两端下标，依据实际情况交替推进端点得到结果。

恩恩就是这样。

/*
Problem: 3061		User: motefly
Memory: 1500K		Time: 79MS
Language: G++		Result: Accepted
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=100005;
int n,s,a[maxn];

int sum[maxn+1];

void solve()
{
    for(int i=0;i<n;i++)
    {
        sum[i+1]=sum[i]+a[i];
    }

    if(sum[n]<s)
    {
        printf("0\n");
        return;
    }
    int res=n;
    for(int i=0;sum[i]+s<=sum[n];i++)
    {
        int t=lower_bound(sum+i,sum+n,sum[i]+s)-sum;
        res=min(res,t-i);
    }
    printf("%d\n",res);
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%d%d",&n,&s);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        solve();
    }
    return 0;
}