zzuoj 10412: G.Interference Signal 【暴力枚举】

10412: G.Interference Signal

Time Limit: 2 Sec  Memory Limit: 128 MB
Submit: 2  Solved: 2
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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

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

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

比赛时这道题坑死，热身赛时听学长们说的TLE不提醒，一直以为是超时了。最后才发现不需要四舍五入。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 4000000+10
using namespace std;
int a[2010];
int aver[MAX];
int main()
{
    int t;
    int n, m;
    int i, j, k, l;
    int sum = 0;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(i = 1; i <= n; i++)
        scanf("%d", &a[i]);
        k= 0;
        for(i = 1; i <= n; i++)
        {
            for(l = m; l <= n; l++)
            {
                if(i + l > n+1)
                break;
                sum = 0;
                for(j = i; j < i+l; j++)
                {
                    sum += a[j];
                }
                aver[k++] = sum *1000 / l;
            }
        }
        sort(aver, aver+k);
        printf("%d\n", aver[k-1]);
    }
    return 0;
}