Galaxy (hdu 5073 数学)

Galaxy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 827    Accepted Submission(s): 201

Special Judge

Problem Description

Good news for us: to release the financial pressure, the government started selling galaxies and we can buy them from now on! The first one who bought a galaxy was Tianming Yun and he gave it to Xin Cheng as a present.






To be fashionable, DRD also bought himself a galaxy. He named it Rho Galaxy. There are n stars in Rho Galaxy, and they have the same weight, namely one unit weight, and a negligible volume. They initially lie in a line rotating around their center of mass.



Everything runs well except one thing. DRD thinks that the galaxy rotates too slow. As we know, to increase the angular speed with the same angular momentum, we have to decrease the moment of inertia.



The moment of inertia I of a set of n stars can be calculated with the formula






where w_i is the weight of star i, d_i is the distance form star i to the mass of center.



As DRD’s friend, ATM, who bought M78 Galaxy, wants to help him. ATM creates some black holes and white holes so that he can transport stars in a negligible time. After transportation, the n stars will also rotate around their new center of mass. Due to financial
pressure, ATM can only transport at most k stars. Since volumes of the stars are negligible, two or more stars can be transported to the same position.



Now, you are supposed to calculate the minimum moment of inertia after transportation.

 

Input

The first line contains an integer T (T ≤ 10), denoting the number of the test cases.



For each test case, the first line contains two integers, n(1 ≤ n ≤ 50000) and k(0 ≤ k ≤ n), as mentioned above. The next line contains n integers representing the positions of the stars. The absolute values of positions will be no more than 50000.

 

Output

For each test case, output one real number in one line representing the minimum moment of inertia. Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.

 

Sample Input

2
3 2
-1 0 1
4 2
-2 -1 1 2

 

Sample Output

0
0.5

 

Source

2014 Asia AnShan Regional Contest

题意：数轴上有n个点。每一个点重量1 ，能够移动当中k个点到不论什么位置。 使得题中式子的值最小  。

思路：详细推导过程看这位大神~

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 50050
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

double a[maxn];

int main()
{
    int n,k,cas;
    double t,sum1,sum2;
    scanf("%d",&cas);
    while (cas--)
    {
        scanf("%d%d",&n,&k);
        sum1=0.0;
        sum2=0.0;
        for (int i=1;i<=n;i++)
            scanf("%lf",&a[i]);
        if (n==1||n==k||k==(n-1))
        {
            printf("0.00000000000\n");
            continue;
        }
        sort(a+1,a+n+1);
        for (int i=1;i<=n-k;i++)
        {
            sum1+=a[i];
            sum2+=a[i]*a[i];
        }
        double ans=(n-k)*sum2-sum1*sum1;
        for (int i=1;i<=k;i++)
        {
            sum1=sum1+a[n-k+i]-a[i];
            sum2=sum2+a[n-k+i]*a[n-k+i]-a[i]*a[i];
            double temp=(n-k)*sum2-sum1*sum1;
            if (temp<ans)
                ans=temp;
        }
        printf("%.11f\n",(double)ans/(double)(n-k));
    }
    return 0;
}

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