Subsequence poj 3061 尺取法

Subsequence
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14262 Accepted: 5986
代码：http://blog.csdn.net/cherish_k/article/details/52953245
尺取法： http://blog.chinaunix.net/uid-24922718-id-4848418.html
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

同学说这个时间复杂度是f(n) ,

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;int a[5000000];int main(){    int t,n,s;    cin>>t;    while(t--)    {        memset(a,0,sizeof(a));        scanf("%d%d",&n,&s);        for(int i = 1; i <= n; i ++)        {            scanf("%d",&a[i]);        }        int r = 0,l = 0,sum = 0,ans =214748364;        while(r <= n)        {            r ++;            if(sum < s)            {                sum+=a[r];            }            while(sum >= s)            {                ans = min(r-l+1,ans);                sum -= a[l];//左端点左移                l++;            }        }        if(ans >= n+1)        cout <<"0"<<endl;        else        cout <<ans <<endl;    }}

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