HDU 5073 Galaxy (2014 Anshan onsite competition)

Galaxy Time Limit: 2000/1000 MS (Java/others) memory limit: 262144/262144 K (Java/Others)
Total submission (s): 577 accepted submission (s): 132
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

Expand the squared difference formula.

(Xi-x average) ^ 2 = Xi ^ 2 + x average ^ 2-2 * Xi * x average

Therefore, variance = Σ Xi ^ 2 + N * x mean-2 * Σ Xi * x mean

Let's set V = n-k. First, calculate the Σ Xi ^ 2 of the first V, calculate the average value, and then calculate the variance. After each forward processing, subtract the previous one and add

The next one calculates the average value and returns the variance, so that the result can be obtained at O (n.

Code:

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>long long a[50005];using namespace std;int main(){    int t,n,k;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        for(int i = 1;i<=n;i++)        cin>>a[i];        if(n==k)        {           printf("0\n");           continue;        }        sort(a+1,a+n+1);        long long sum = 0;        long long sums = 0;         for(int i = 1;i<=n-k;i++)         {                sum += a[i];                sums += a[i] * a[i];         }        double ave  = sum*1.0/(n-k);        double mini = sums + (n-k)*ave*ave - 2*sum*ave;        for(int i = 1 ;i <= k; i++)        {                sum = sum - a[i]+a[n-k+i];                sums = sums - a[i]*a[i]+ a[n-k+i]*a[n-k+i];                ave = sum*1.0/(n-k);                double temp  = sums + (n-k)*ave*ave - 2*sum*ave;                if(temp < mini)                {                    mini = temp;                }        }        printf("%.10f\n",mini);}}

PS: CDD has used a wrong greedy strategy, so it can also be a. It has not been specially determined to be n = K. This is really a face-oriented world o

HDU 5073 Galaxy (2014 Anshan onsite competition)