ACM学习历程—HDU 5073 Galaxy（数学）-白红宇

ACM学习历程—HDU 5073 Galaxy（数学）

阅读量：5251 次

发布时间：2019-06-14

本文共 3271 字，大约阅读时间需要 10 分钟。

Description

Good 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 Rho Galaxy. There are n stars in Rho Galaxy, 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 except 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 w _i is the weight of star i, d _i 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.

Input

The 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.

Output

For 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

3 2

-1 0 1

4 2

-2 -1 1 2

Sample Output

0.5

题目大意就是在n个数里面找n-k个数，然后让他们的方差*(n-k)最小。

首先D(x) = E(x^2) – E(x)^2

但是方差还有个定义：

由这个式子可以发现是一个关于an的二次函数，当前n-1个点的方差知道时，第n个点加入时，当第n个点越远离前n-1个点的重心，整体的方差越大。

于是对所有点排序，每次都连续取n-k个点，取里面最小的。

代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            #include 
             
              #include 
              
               #define LL long longusing namespace std;const int maxN = 50005;int n, k, a[maxN], d[maxN<<1], top;void quickSort(){ int len = 0; for (int i = 0; i <= top; ++i) { while (d[i]) { a[len++] = i-maxN; d[i]--; } }}void input(){ memset(d, 0, sizeof(d)); scanf("%d%d", &n, &k); int tmp; for (int i = 0; i < n; ++i) { scanf("%d", &tmp); tmp += maxN; d[tmp]++; if (i == 0 || top < tmp) top = tmp; } k = n-k;}void work(){ double ans; if (k == 0) ans = 0; else { quickSort(); double e2 = 0, e = 0; for (int i = 0; i < k; ++i) { e2 += (LL)a[i]*a[i]; e += a[i]; } ans = e2/k-e/k*e/k; for (int i = k; i < n; ++i) { e2 += (LL)a[i]*a[i]-(LL)a[i-k]*a[i-k]; e += a[i]-a[i-k]; ans = min(ans, e2/k-e/k*e/k); } } printf("%.10lf\n", ans*k);}int main(){ //freopen("test.in", "r", stdin); int T; scanf("%d", &T); for (int times = 0; times < T; ++times) { input(); work(); } return 0;}

转载于:https://www.cnblogs.com/andyqsmart/p/4907678.html

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