Galaxy (HDU 5073 mathematics)

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 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 input23 2-1 0 14 2-2-1 1 2

Sample output00.5

Source2014 Asia Anshan Regional Contest question: 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 value of the Chinese sub-questions. Train of Thought: refer to this great god for detailed derivation process ~ Code:

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <string> 7 #include <map> 8 #include <stack> 9 #include <vector>10 #include <set>11 #include <queue>12 #pragma comment (linker,"/STACK:102400000,102400000")13 #define maxn 5005014 #define MAXN 200515 #define mod 100000000916 #define INF 0x3f3f3f3f17 #define pi acos(-1.0)18 #define eps 1e-619 typedef long long ll;20 using namespace std;21 22 double a[maxn];23 24 int main()25 {26     int n,k,cas;27     double t,sum1,sum2;28     scanf("%d",&cas);29     while (cas--)30     {31         scanf("%d%d",&n,&k);32         sum1=0.0;33         sum2=0.0;34         for (int i=1;i<=n;i++)35             scanf("%lf",&a[i]);36         if (n==1||n==k||k==(n-1))37         {38             printf("0.00000000000\n");39             continue;40         }41         sort(a+1,a+n+1);42         for (int i=1;i<=n-k;i++)43         {44             sum1+=a[i];45             sum2+=a[i]*a[i];46         }47         double ans=(n-k)*sum2-sum1*sum1;48         for (int i=1;i<=k;i++)49         {50             sum1=sum1+a[n-k+i]-a[i];51             sum2=sum2+a[n-k+i]*a[n-k+i]-a[i]*a[i];52             double temp=(n-k)*sum2-sum1*sum1;53             if (temp<ans)54                 ans=temp;55         }56         printf("%.11f\n",(double)ans/(double)(n-k));57     }58     return 0;59 }

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Galaxy (HDU 5073 mathematics)