POJ3061 Subsequence 尺取法（0（n））

Subsequence

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​​

题意：给定一个序列，找出最短的子序列长度，使得其和大于或等于S。
分析：

挑战上的解释

这个题目区间和明显是有趋势的：单调变化，所以根据题目要求很容易求解，但是在使用之间需要对区间前缀和进行预处理计算。

#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXN 100010
using namespace std;
typedef long long ll;
ll a[MAXN];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;ll s;
        scanf("%d%lld",&n,&s);
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        int l=1,r=1;
        int ans=1e9;ll sum=0;
        while(1)
        {
            while(r<=n&&sum<s) sum+=a[r++];
            if(sum<s) break;
            ans=min(ans,r-l);
            sum-=a[l++];
        }
        if (ans==1e9) ans=0;  
        printf("%d\n", ans); 
    }
}