HDU 5073 Galaxy (Mathematics + derivation)

Galaxy Time Limit: 2000/1000 MS (Java/others) memory limit: 262144/262144 K (Java/Others)
Total submission (s): 553 accepted submission (s): 126
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

 

There are N points on the number axis. Each point has a weight of 1. You can move K points to any position to minimize the Chinese sub-values.

Di indicates the distance between vertex I and the center of gravity (that is, the average distance) of the current N points.

Idea: select the continuous number with the length of n-k, and then the number of other K is moved to the center of the number of N-K.

The value of the continuous number with the Interval Length t = n-k: P = (a [I]-x) * (a [I]-x) + (A [I + 1]-x) * (a [I + 1]-x)

+... + (A [T-1 + I]-x) * (a [T-1 + I]-x); (where X is the average distance)

P = (a [I] * A [I] + A [I + 1] * A [I + 1] + ...... + A [T-1 + I] * A [T-1 + I])-2 * x * (a [I] + A [I + 1] + ...... + A [T-1 + I]) + T * x * X;

So that sum1 [I] indicates the sum of the squares of the I items before array A, and sum2 [I] indicates the sum of the I items before array;

That is, x = (sum2 [T + I-1]-sum2 [I-1])/T;

Sum1 [T + I-1]-sum1 [I-1] = A [I] * A [I] + A [I + 1] * A [I + 1] + .. .... + A [T-1 + I] * A [T-1 + I];

Sum2 [T + I-1]-sum2 [I-1] = A [I] + A [I + 1] + ...... + A [T-1 + I];

So P = sum1 [T + I-1]-sum1 [I-1]-(sum2 [T + I-1]-sum2 [I-1]) * (sum2 [T + I-1]-sum2 [I-1])/T;

Then I is enumerated, and the minimum result is the answer. (Precision issues)

 

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;const int maxn=50050;double sum1[maxn],sum2[maxn],a[maxn];int n,k;void input(){    scanf("%d %d",&n,&k);    for(int i=1; i<=n; i++) scanf("%lf",&a[i]);}void solve(){    double ans;    if(k==n)  ans=0.0;    else    {        sort(a+1,a+n+1);        sum1[0]=sum2[0]=0.0;        for(int i=1; i<=n; i++)        {            sum1[i]=sum1[i-1]+a[i]*a[i];            sum2[i]=sum2[i-1]+a[i];        }        int t=n-k;        for(int i=1; i+t<=n+1; i++)        {            double p=sum1[i+t-1]-sum1[i-1]-(sum2[i+t-1]-sum2[i-1])*(sum2[i+t-1]-sum2[i-1])/t;            if(i==1) ans=p;            else ans=min(ans,p);        }    }    printf("%.9lf\n",ans);}int main(){    int T;    scanf("%d",&T);    while(T--)    {        input();        solve();    }    return 0;}

HDU 5073 Galaxy (Mathematics + derivation)