Intense Heat（CodeForces - 1003C）（前缀和/尺取）

最新推荐文章于 2019-09-04 21:35:18 发布

原创最新推荐文章于 2019-09-04 21:35:18 发布 · 406 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#尺取 #前缀和

spiny 专栏收录该内容

46 篇文章

订阅专栏

本文解析了CodeForces-1003C题目——IntenseHeat，介绍了两种解决方法：前缀和算法及尺取算法，并提供了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

欢迎访问https://blog.youkuaiyun.com/lxt_Lucia～～

宇宙第一小仙女\(^o^)/～～萌量爆表求带飞=≡Σ((( つ^o^)つ~ dalao们点个关注呗～～

Intense Heat（CodeForces - 1003C）（前缀和/尺取）

Decription

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 nn consecutive days. We have measured the temperatures during these nn days; the temperature during ii-th day equals aiai.

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 temperaturefrom day xx to day yy, we calculate it as ∑i=xyaiy−x+1∑i=xyaiy−x+1 (note that division is performed without any rounding). The heat intensity value is the maximum of average temperatures over all segments of not less than kkconsecutive days. For example, if analyzing the measures [3,4,1,2][3,4,1,2] and k=3k=3, we are interested in segments [3,4,1][3,4,1], [4,1,2][4,1,2] and [3,4,1,2][3,4,1,2] (we want to find the maximum value of average temperatureover these segments).

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 nn and kk (1≤k≤n≤50001≤k≤n≤5000) — the number of days in the given period, and the minimum number of days in a segment we consider when calculating heat intensity value, respectively.

The second line contains nn integers a1a1, a2a2, ..., anan (1≤ai≤50001≤ai≤5000) — the temperature measures during given nn days.

Output

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

Your answer will be considered correct if the following condition holds: |res−res0|<10−6|res−res0|<10−6, where resresis your answer, and res0res0 is the answer given by the jury's solution.

Example

Input

4 3
3 4 1 2

Output

2.666666666666667

题意：

第一行输入n，k，n表示有连续的n天，k表示最少的连续天数。

第二行输入n天的温度分别为a[ i ]。

现在定义平均热度为，x，y表示任意从x天开始，到y天结束。即y-x+1>=k。求连续天数(>=k天)的最大的平均热度。

此题有两种思路：前缀和和尺取。

思路一：

前缀和。设dp数组来记录，dp[ i ] 表示前 i 天的热度和，然后双重循环求平均找最大。

代码一：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-10

using namespace std;

int main()
{
    int i,j,n,k,a[5001];
    double dp[5010],max1=-INF;
    memset(dp,0,sizeof(dp));
    scanf("%d%d",&n,&k);
    scanf("%d",&a[0]);
    dp[0]=a[0];
    for(i=1; i<=n-1; i++)
    {
        scanf("%d",&a[i]);
        dp[i]=dp[i-1]+a[i];   //dp[i]表示前i+1天的热度和
    }
    for(i=0; i<=n-1; i++)
    {
        for(j=i; j<=n-1; j++)
        {
            if(j-i+1>=k)
                max1=max(max1,(dp[j]-dp[i-1])/(j-i+1));
        }
    }
    printf("%.15lf",max1);
    return 0;
}

思路二：

尺取。先取k个数，分别求平均找最大，然后不断增加个数直到n个全取到。

代码二：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-10

using namespace std;

int main()
{
    int i,j,p,q,n,k,a[5001];
    double max1=-INF,sum;
    scanf("%d%d",&n,&k);
    for(i=0; i<=n-1; i++)
        scanf("%d",&a[i]);
    for(i=k; i<=n; i++)
    {
        sum=0;
        for(j=0; j<=i-1; j++)
        {
            sum+=a[j];
            max1=max(max1,sum/i);
        }
        q=0;
        p=i;
        while(p!=n)
        {
            sum-=a[q++];
            sum+=a[p++];
            max1=max(max1,sum/i);
        }
    }
    printf("%.15lf",max1);
    return 0;
}

宇宙第一小仙女\(^o^)/～～萌量爆表求带飞=≡Σ((( つ^o^)つ~ dalao们点个关注呗～～