HDU 5073 Galaxy (2014 Anshan D简单数学)
HDU 5073 Galaxy (2014 Anshan D简单数学)
题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5073
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 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.
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
2
3 2
-1 0 1
4 2
-2 -1 1 2
Sample Output
0
0.5
题意:
给你n个在x轴点,其中k个可以任意变换坐标点,问此时每个点距离质心的距离平方和最小是多少?
题解:
首先将所有的点排序。然后每次o(n)扫描一遍。首先我们知道质心是每个坐标的和的平均值。那么从头到尾扫描每次删除起始点,添加最后点的下一个点。我们只需将这个公式拆开即可化简。注意n==k的时候输出0
代码:
#include <bits/stdc++.h>
const int N = 100005 ;
double pos[N] ;
int main()
{
int t ;
scanf("%d",&t) ;
while(t--){
int n , k ;
scanf("%d %d",&n,&k) ;
for(int i = 1 ; i <= n ; i ++){
scanf("%lf",pos+i) ;
}
if(n == k){
printf("0\n") ;
continue ;
}
std::sort(pos+1,pos+1+n) ;
double sum = 0 ;
double psum = 0 ;
for(int i = 1 ; i <= n-k ; i ++){
sum += pos[i] ;
psum += pos[i]*pos[i] ;
}
double avg = sum/(n-k) ;
double ans = psum + (n-k)*avg*avg - 2*avg*sum ;
for(int i = 1 ; i <= k ; i ++){
sum -= pos[i] ;
sum += pos[n-k+i] ;
psum -= pos[i]*pos[i] ;
psum += pos[n-k+i]*pos[n-k+i] ;
avg = sum/(n-k) ;
double temp = psum + (n-k)*avg*avg - 2*avg*sum ;
ans = std::min(ans,temp) ;
}
printf("%.10lf\n",ans) ;
}
return 0 ;
}
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题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5073
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 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.
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
2
3 2
-1 0 1
4 2
-2 -1 1 2
Sample Output
0
0.5
题意:
给你n个在x轴点,其中k个可以任意变换坐标点,问此时每个点距离质心的距离平方和最小是多少?
题解:
首先将所有的点排序。然后每次o(n)扫描一遍。首先我们知道质心是每个坐标的和的平均值。那么从头到尾扫描每次删除起始点,添加最后点的下一个点。我们只需将这个公式拆开即可化简。注意n==k的时候输出0
代码:
#include <bits/stdc++.h>
const int N = 100005 ;
double pos[N] ;
int main()
{
int t ;
scanf("%d",&t) ;
while(t--){
int n , k ;
scanf("%d %d",&n,&k) ;
for(int i = 1 ; i <= n ; i ++){
scanf("%lf",pos+i) ;
}
if(n == k){
printf("0\n") ;
continue ;
}
std::sort(pos+1,pos+1+n) ;
double sum = 0 ;
double psum = 0 ;
for(int i = 1 ; i <= n-k ; i ++){
sum += pos[i] ;
psum += pos[i]*pos[i] ;
}
double avg = sum/(n-k) ;
double ans = psum + (n-k)*avg*avg - 2*avg*sum ;
for(int i = 1 ; i <= k ; i ++){
sum -= pos[i] ;
sum += pos[n-k+i] ;
psum -= pos[i]*pos[i] ;
psum += pos[n-k+i]*pos[n-k+i] ;
avg = sum/(n-k) ;
double temp = psum + (n-k)*avg*avg - 2*avg*sum ;
ans = std::min(ans,temp) ;
}
printf("%.10lf\n",ans) ;
}
return 0 ;
}
Galaxy Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total S ...
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5073 解题报告:在一条直线上有n颗星星,一开始这n颗星星绕着重心转,现在我们可以把其中的任意k颗星星移 ...
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主题链接:http://acm.hdu.edu.cn/showproblem.php? pid=5073 Galaxy Time Limit: 2000/1000 MS (Java/Others) ...
Galaxy Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5073 Mean: 在一条数轴上,有n颗卫星,现在你可以改变k颗 ...
Galaxy Time Limit: 2000/1000 MS (J ...
Description Good news for us: to release the financial pressure, the government started selling gala ...
0.5 题意:有n(n<=5e4)个质点位于一维直线上,现在你可以任意移动其中k个质点,且移动到任意位置,设移动后的中心为e,求最小的I=(x[1]-e)^2+(x[2]-e)^2+(x[3]- ...
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PS:前面忘了给大家讲解后台需要做的 ,同步分页的时候,我们只需要定义一个方法,给前台传递一个page对象,前台接收到直接展示即可:异步分页要多一步,首先还是写一个方法,传递初始对象,后面的ajax返 ...
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