第八届河南省ACM省赛 - G - Interference Signal

河南省第八届ACM大学生程序设计大赛

G.Interference Signal

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 maximum average

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

Source

第八届河南省赛

比赛的时候一直WA，WA到比赛结束都不知道错在了哪，出来后小伙伴们说当时也WA，不知道错哪了，最后抱着试一试的心态，用%d输出，居然A了，太坑了。。 不能用%.f输出，只能%d，这次算是记住了。其实思路很简单，就暴力。。 枚举出所有的情况找最大的就好了，坑点就是输出！！！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main() {
    int k;
    scanf("%d", &k);
    while(k--){
        int n, m;
        int a[2010];
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);

        double max = 0, sum = 0;
        double avg, c = 0; 
        for(int i = 0; i+m <= n; i++){
            sum = 0;
            c = 0;
            for(int j = i; j < n; j++){
                if(c<m){
                    sum+=a[j];
                    c+=1;
                }
                else{
                    avg = sum/c;
                    if(avg > max)
                        max = avg; 
                    sum+=a[j];
                    c+=1;
                    if(j==n-1){
                        avg = sum/c;
                        if(avg > max)
                            max = avg;
                    }
                }
            }
            if(c == m){
                avg = sum/c;
                if(avg > max)   max = avg;
            } 
        } 
        int im = (int)(max*1000);
        printf("%d\n", im);
    } 
    return 0;
}