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『ACM C++』 Codeforces | 1003C - Intense Heat

2019-03-04 09:53:17来源：博客园阅读 ()

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The heat during the last few days has been really intense. Scientists from all over the Berland study how the temperatures and weather change, and they claim that this summer is abnormally hot. But any scientific claim sounds a lot more reasonable if there are some numbers involved, so they have decided to actually calculate some value which would represent how high the temperatures are.

Mathematicians of Berland State University came up with a special heat intensity value. This value is calculated as follows:

Suppose we want to analyze the segment of

We denote the average temperature of a segment of some consecutive days as the arithmetic mean of the temperature measures during this segment of days. So, if we want to analyze the average temperature from day

You have been hired by Berland State University to write a program that would compute the heat intensity value of a given period of days. Are you up to this task?

Input

The first line contains two integers

The second line contains

Output

Print one real number — the heat intensity value, i. e., the maximum of average temperatures over all segments of not less than

Your answer will be considered correct if the following condition holds:

input

4 3
3 4 1 2

output

2.666666666666667

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（一）题目分析：

　　　　正常div3难度题型，难度处理可能是在求和统计上，题目大概含义就是：

　　　　求n个数中至少连续k个数的平均值的最大值，如：n=4,k=3 给定的n个数是3 4 1 2，则求[3,4,1],[4,1,2],[3,4,1,2]三个区间的平均值的最大值。

　　　　我用的方法比较简单，就是简单的取值求和，今天下午打完题之后这道题还特地搜了一下，发现还有其他的解法：前缀和 / 尺缩

（二）AC代码：

　　　　因为代码比较简单，依旧不分块了~

#include<stdio.h>
#define max(a,b) (a>b) ? a : b
using namespace std;
int n,k,sum;
int temp[5001];
double ans,all;
int main()
{
    scanf("%d%d",&n,&k);
    ans = 0;
    for(int i = 1;i<=n;i++) scanf("%d",&temp[i]);

    for(int i = 1;i<=n;++i)
    {
        sum = 0;
        for(int j = i;j<=n;++j)
        {
            sum+= temp[j];
            if (j - i + 1 >= k) 
            {
                all = sum * 1.0 / (j - i + 1);
                if(all>ans) ans = all;
            }
        }
    }
    printf("%.15f\n",ans);
    return 0;
}