POJ 3061 Subsequence

Subsequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8128 Accepted: 3141

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

AC碼：
#include
#include
#include
using namespace std;
int a[100000],n,S,sum[100001],t;
void solve()
{
  for(int i = 0;i>t;
          while(t--)
          {
            cin>>n>>S;
          for(int i=0;i注：上面的代碼中使用了lower_bound這個STL函數。這個函數從已排好序的序列a中利用二分搜索找出指向滿足ai>=k的最小的指針。類似的函數還有upper_bound，這個函數求出的是指向滿足ai>k的ai的最小的指針.