POJ2061 Subsequence 2017-05-25 19:49 83人阅读 评论(0) 收藏

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14709		Accepted: 6210

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

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题目的意思是给出n个整数，求区间和大于等于m的最小区间长度

思路：由于和具有单调性，直接尺取法

#include <iostream>
#include <functional>
#include <stdio.h>
#include <queue>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
const int inf=0x3f3f3f3f;
int n,m;
int a[1000005];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
      scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
    scanf("%d",&a[i]);
    }
    int l=0,r=0;
    int sum=0;
    int ans=inf;
    while(1)
    {
        while(r<n&&sum<m)
        {
            sum+=a[r++];
        }
        if(sum<m) break;
        ans=min(ans,r-l);
        sum-=a[l++];
    }
    if(ans==inf)
        ans=0;
    printf("%d\n",ans);
  }
    return 0;
}