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 long

using 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;
}