poj3061，求连续数字中达到给定数字最短数字 贪心

Subsequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8030 Accepted: 3092

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

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

23

Source

Southeastern Europe 2006

Subsequence
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8030 Accepted: 3092
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

思路：

先求出第一个连续的数字之和大于s，用一个数记录数字的长度，在从第一个开始减，同时记录连续数字的长度，若是减的数字小于s了，则在原来到达的数字后继续加。。

#include <iostream>

using namespace std;
long long  a[100000+10],b[100000+10];

int main()
{
   int m,i,j,n,s,min,sum,c,l;
   cin>>m;
   while(m--)
   {
      cin>>n>>s;
      for(i=0;i<n;i++)
        {
           cin>>a[i];
           b[i]=a[i];
        }

      for(i=0,j=0,min=1000000,sum=0,l=0;i<n;i++)
      {
         sum+=a[i];
         l++;
         while(sum>=s)
         {
            sum-=b[j++];
            l--;
            if(sum<s)
            {
               if(l<min)min=l+1;
               continue;
            }

         }



      }
      if(min==1000000)cout<<0<<endl;
      else cout<<min<<endl;
   }





    return 0;
}

2014-04-18