Subsequence（尺取法）

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本文介绍了一种利用尺取法解决特定问题的算法：在一系列正整数中找到和大于或等于给定值S的最短连续子序列。通过逐步增加或减少子序列的元素来调整和，直至找到满足条件的最短子序列。

题目

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.

输入

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.

输出

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

基本思路

本题意思为，求和大于或等于S的最小区间，根据题意可知这道题应该使用尺取法。

尺取法

尺取法顾名思义用到了“尺子”，我们可以先从第一个开始，逐个加起来与S进行比较，就像尺子一样慢慢增长，如果和大于S则我们考虑能否缩小尺子，减去最前面的数，再与S比较，每次纪录尺子的长度，所有长度中最小值及为所求。

源代码为

#include <iostream>
#include <string.h>
using namespace std;
int a[200000];
int sum;
int n,s;
void ruler()
{
    sum=0;
    int i=0;
    int r=0;
    int l=5555555;
    while(1)
    {
        while(sum<s&&i<n)
        {
            sum+=a[i++];
        }
        if(sum<s)
        {
            break;
        }
        l=min(l,i-r);
        sum-=a[r++];
    }
    if(l==5555555)
    {
        cout<<0<<endl;
        return ;
    }
    else
    {
        cout<<l<<endl;
        return ;
    }

}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(a,0,sizeof(a));
        cin>>n>>s;
        for(int i=0; i<n; i++)
        {
            cin>>a[i];
        }
        ruler();
    }
    return 0;
}