Hdu5073 2014 Anshan field competition D Question

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 5073

Galaxy Time Limit: 2000/1000 MS (Java/others) memory limit: 262144/262144 K (Java/Others)
Total submission (s): 1421 accepted submission (s): 324
Special Judge

Problem descriptiongood 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 rock galaxy. there are n stars in neogalaxy, 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 enough t 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 WI is the weight of star I, Di 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.
Inputthe 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.
Outputfor 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

23 2-1 0 14 2-2 -1 1 2

Sample output

00.5

Source2014 Asia Anshan Regional Contest

Question: The number of N points on the number axis can be moved to any K points. To minimize the variance of the number after moving, ask the minimum value.

Analysis: first, we can think that the point to be moved must be some of the points at both ends, that is, the remaining points are continuous, and the moving position must be the average position of the remaining points. In this way, we can enumerate the starting point of the continuous n-k Number and find the variance of these numbers.

If it is n ^ 2, it will definitely time out. Note here (Xi-X ') ^ 2 = Xi ^ 2 + x' ^ 2-2 * x * x' (x' is the average), we maintain the sum of the squares of the n-k numbers and the sum, at the next point, the sum of squares and sum of values can be calculated by a sliding window plus or minus. Then the final variance is equal to sum (x ^ 2) + (sum (X)/m) ^ 2 * m-2 * sum (x) * sum (X)/m

Sum (x ^ 2)-sum (x) * sum (X)/m. Note that n-k is set to 0.

/** * @author neko01 *///#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <string.h>#include <iostream>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define min3(a,b,c) min(a,min(b,c))#define max3(a,b,c) max(a,max(b,c))#define pb push_back#define mp(a,b) make_pair(a,b)#define clr(a) memset(a,0,sizeof a)#define clr1(a) memset(a,-1,sizeof a)#define dbg(a) printf("%d\n",a)typedef pair<int,int> pp;const double eps=1e-9;const double pi=acos(-1.0);const int INF=0x3f3f3f3f;const LL inf=(((LL)1)<<61)+5;const int N=50005;double a[N];int main(){    int n,k,t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        for(int i=1;i<=n;i++)        {            scanf("%lf",&a[i]);        }        sort(a+1,a+1+n);        int m=n-k;        if(m==0)        {            puts("0");            continue;        }        double ans,ave,s1=0,s2=0;        for(int i=1;i<=m;i++)        {            s2+=a[i];            s1+=a[i]*a[i];        }        ans=s1-s2*s2/m;        for(int i=1;i<=k;i++)        {            s1=s1-a[i]*a[i]+a[i+m]*a[i+m];            s2=s2-a[i]+a[i+m];            ans=min(ans,s1-s2*s2/m);        }        printf("%.9f\n",ans);    }    return 0;}

Hdu5073 2014 Anshan field competition D Question