Codeforces 808B——Average Sleep Time——思维

B. Average Sleep Time

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts k days!

When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last n days. So now he has a sequence a1, a2, ..., an, where ai is the sleep time on the i-th day.

The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider k consecutive days as a week. So there will be n - k + 1 weeks to take into consideration. For example, if k = 2, n = 3 and a = [3, 4, 7], then the result is .

You should write a program which will calculate average sleep times of Polycarp over all weeks.

Input

The first line contains two integer numbers n and k (1 ≤ k ≤ n ≤ 2·105).

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Output average sleeping time over all weeks.

The answer is considered to be correct if its absolute or relative error does not exceed 10 - 6. In particular, it is enough to output real number with at least 6 digits after the decimal point.

Examples

input

3 23 4 7

output

9.0000000000

input

1 110

output

10.0000000000

input

8 21 2 4 100000 123 456 789 1

output

28964.2857142857

Note

In the third example there are n - k + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.

读题读了蛮久的，给你两个数字n，k（n >= k），即给你n个数字，选择其中连续的k个数字有（n - k + 1）种取法，然后求出每种选法的区间数字和，再把每次的区间数字和加起来除以（n - k + 1）。

直接模拟过程，先算出最前面的一段，两个指针，一个指向当前区间最左边的元素，即下一个区间将剔除的数据，另一个指向下一个区间即将包含的数据，同时移动维护求和就好了。刚开始没往这里想，想计算每个数字被用到的次数再求和瞎搞，结果失败了。。。不应该。

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<set>#include<queue>#include<map>#include<stack>#define max3(x, y, z) max((x), max((y), (z)))#define min3(x, y, z) min((x), min((y), (z)))#define pb push_back#define mp make_pair#define LL long longusing namespace std;const int N = 200020;int main(){    int n, m;    int ma[N];    int t;    LL sum;     ///爆int    double ans;    while(scanf("%d%d", &n, &m) == 2){        sum = 0;        t = n - m + 1;        for(int i = 0; i < n; i ++){            scanf("%d", &ma[i]);        }        for(int i = 0; i < m; i ++){            sum += ma[i];        }        LL q = sum; ///刚开始没用LL  WA了一发        for(int i = m, j = 0; i < n; i ++, j ++){       ///i，j双指针移动            q = q + ma[i] - ma[j];     ///新的区间的数字和            sum += q;        }        ans = sum * 1.0 / t;         printf("%.10lf\n", ans);    }    return 0;}