Subsequence (尺取法)

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本文介绍了一种算法，用于寻找给定整数序列中，连续元素子序列的最小长度，这些元素之和大于或等于指定数值S。输入包括多个测试案例，每个案例由序列长度N、目标和S及序列本身组成。输出则是每个案例对应的最小长度，若不存在符合条件的子序列则输出0。

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

先取前x个数(r++)，直到大于S，减去该区间最前面的一个数（收缩 l++），再次判断是否大于S，重复操作，直至t==n或取得的区间无法大于S 停止。

代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
int num[100002];
int main(){
    int t;
    int n,s;
    int INF=999999;
    int l,r,sum,mi;
    cin>>t;
    while(t--){
        
        cin>>n>>s;
        mi=INF;
        l=0;r=0;sum=0;
        for(int j=0;j<n;j++)
            cin>>num[j];
        while(true){
            while(sum<s&&r<n)
                sum+=num[r++];
            if(sum<s)
                break;
            sum-=num[l];
            mi=min(mi,r-l);
            l++;
        }
        if(mi==INF)
            cout<<0<<endl;
        else
            cout<<mi<<endl;
    }
    
    
    return 0;
}