Problem G: G. Interference Signal

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本文介绍了一种计算连续干扰信号块最大平均强度的方法，信号由一系列正整数表示，需包含至少M个整数。通过递归算法找到满足条件的最大平均强度，并提供了完整的C++实现代码。

Problem G: G. Interference Signal

Time Limit: 2 Sec Memory Limit: 128 MB
Submit: 17 Solved: 16
[ Submit][ Status][ Web Board]

Description

Dr.Kong’s laboratory monitor some interference signals. The interference signals can be digitized into a series of positive integer. May be, there are N integers a1,a2,…,an.

Dr.Kong wants to know the average strength of a contiguous interference signal block. the block must contain at least M integers.

Please help Dr.Kong to calculate the maximum average strength, given the constraint.

Input

The input contains K test cases. Each test case specifies:

* Line 1: Two space-separated integers, N and M.

* Lines2~line N+1: ai (i=1,2,…,N)

1 ≤ K≤ 8, 5 ≤ N≤ 2000, 1 ≤ M ≤ N, 0 ≤ ai ≤9999

Output

For each test case generate a single line containing a single integer that is 1000 times the maximal average value. Do not perform rounding.

Sample Input

2
10 6
6
4
2
10
3
8
5
9
4
1
5 2
10
3
8
5
9

Sample Output

6500
7333

代码如下：

#include<iostream>
#include<stdio.h>
using namespace std;
double A[2000], mk;
int N, M;
void FindmkAver(double sum, int x,int num)
{
    if (x == N)return;
 
    if (num >= M )
        if (((A[x] + sum)/ num) > mk)
            mk = (A[x] + sum) / num;
 
    FindmkAver(sum + A[x], x + 1, num + 1);
    FindmkAver(0, x + 1, 1);
}
int main()
{
    int K;
    cin >> K;
    while (K--)
    {
        mk = 0;
        cin >> N >> M;
        for (int i = 0; i < N; i++)
            cin >> A[i];
 
        for (int j = 0; j < N; j++)
            FindmkAver(0, j, 1);
 
        printf("%.f\n", mk*1000);
    }
    return 0;
}