http://acm.hdu.edu.cn/showproblem.php?pid=5073

推公式即可，质心公式segma(xi*wi)/segma(wi)

最终剩下的一定是连续n-k个星 

然后枚举左边需要移除几个星即可

计算I的时候展开来算

比较坑的地方在于，星星的位置如果是int型，一定记得Double计算的时候 *1.0或者直接将位置数组声明为double  否则WA到死。。。

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double pi = acos(-1.0);
const int INF = 100000000;
const int MAXN = 50000+50;
const double EPS = 1e-14;/////

double pos[MAXN];
int n,k;
double  sum[MAXN],sumpos[MAXN];

double solve()
{
    double tmp=0.0;
    double ans=1000000000000000000.0;
    if(n==k)return 0.0;
    for(int left=0;left<=k;left++)
    {
        tmp=sum[n-k+left]-sum[left];
        double spp=sumpos[n-k+left]-sumpos[left];
        double d;
        d=spp/(1.0*n-k);
        double ret=tmp-2*spp*d+(n-k)*d*d;
        if(left)
        {
            ans=min(ans,ret);
        }
        else
            ans=ret;
    }
    return ans;
}

int main()
{
    //IN("hdu5073.txt");
    int ncase;
    scanf("%d",&ncase);
    while(ncase--)
    {
        scanf("%d%d",&n,&k);
        sum[0]=0.0;
        sumpos[0]=0.0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf",&pos[i]);
        }
        sort(pos+1,pos+n+1);
        for(int i=1;i<=n;i++)
        {
            sum[i]=sum[i-1]+pos[i]*pos[i];
            sumpos[i]=sumpos[i-1]+pos[i];
        }
        printf("%.14lf\n",solve());
    }
    return 0;
}

hdu 5073 2014鞍山现场赛题 物理题