POJ3061---Subsequence（尺取法）

[align=center]Subsequence[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9998		Accepted: 4052

尺取法，像尺子一样；一段一段的找；
题意：找最小区间和大于等于给定的数；

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

#include <iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#include<queue>

#include<stack>

#define LL long long

using namespace std;

const int maxn=100010;

#define INF 0x3f3f3f3f

LL a[maxn],sum[maxn];

int main()

{

     int t,n,m, ans,r1;

     LL cnt1;

     while(~scanf("%d",&t))

     {

         while(t--)

         {

             scanf("%d%d",&n,&m);

             a[0]=sum[0]=0;

             for(int i=1;i<=n;i++)

             {

                 scanf("%lld",&a[i]);

                 sum[i]=sum[i-1]+a[i];

             }

             a[n+1]=0;

             r1=1;cnt1=0;

             ans=INF;

             if(sum
<m)

             {

                 printf("0\n");

                 continue;

             }

             for(int i=1;i<=n;i++)

             {

                 while(r1<i||(r1<=n+1&&cnt1<m))

                 {

                     cnt1+=a[r1];

                     r1++;

                 }

                if(r1>n+1)

                    break;

                    if(cnt1>=m)

                    ans=min(ans,r1-i);

                    cnt1-=a[i];

             }

             printf("%d\n",ans);

         }

     }

    return 0;

}