武大校赛资格赛 求方差

Problem 1570 - G - April disease

Time Limit: 1000MS   Memory Limit: 65536KB   
Total Submit: 210  Accepted: 105  Special Judge: No

Description

Holding a contest is an interesting mission, and apparently arranging the problems is not exceptional, either.

You may consider that there are N problems prepared, and it's required to choose M problems to form the problem set of the upcoming contest.

What's more, the evaluating function of all the problems choosed is as follows:

Where Xi represents the hardness of the i_th problem and X represents the average of the M problems.

However, the contest principal Alice has lost himself in April disease, and just wants to hold the worst contest ever.(Sounds horrible...)

Now he wonders, what's the minimum value of s^2 can be.

Input

The first line of each test case contains two numbers N and M. ( 1 <= M <= N <= 30 )

The following line will contain N numbers Xi. （ 1 <= Xi <= 10^9 ）

Output

For each test case, output your answer in one line.
(Please keep three decimal places.)

Sample Input

4 3
6 10 1 2
5 2
1 8 3 4 9

Sample Output

4.667
0.250

 

#include<algorithm>#include<iostream>#include<limits.h>#include<stdlib.h>#include<string.h>#include<cstring>#include<iomanip>#include<stdio.h>#include<bitset>#include<cctype>#include<math.h>#include<string>#include<time.h>#include<vector>#include<cmath>#include<queue>#include<stack>#include<list>#include<map>#include<set>#define LL long longusing namespace std;const LL mod = 1e9 + 7;const double PI = acos(-1.0);const int M = 105;double a[M];double calc(int l, int r){    double sum = 0;    for(int i = l; i <= r; ++i)        sum += a[i];    int len = r - l + 1;    double x = (double)(sum / len);    double ans = 0;    for(int i = l; i <= r; ++i){        ans += (a[i] - x) * (a[i] - x);    }    ans /= len;    return ans;}int main(){    int m, n;    while( cin >> n >> m ){        for(int i = 0; i < n; ++i)            cin >> a[i];        sort(a,a + n);        double ans = 999999999;        for(int i = 0; i <= n - m; ++i){            ans = min(ans,calc(i,i + m - 1));        }        printf("%.3f\n",ans);    }    return 0;}