第一章例题4墓地雕塑UVa1388(参照系的选取)

1388 - Graveyard

Time limit: 3.000 seconds

Programming contests became so popular in the year 2397 that the governor of New Earck -- the largest human-inhabited planet of the galaxy -- opened a special Alley of Contestant Memories (ACM) at the local graveyard. The ACM encircles a green park, and
holds the holographic statues of famous contestants placed equidistantly along the park perimeter. The alley has to be renewed from time to time when a new group of memorials arrives.

When new memorials are added, the exact place for each can be selected arbitrarily along the ACM, but the equidistant disposition must be maintained by moving some of the old statues along the alley.

Surprisingly, humans are still quite superstitious in 24th century: the graveyard keepers believe the holograms are holding dead people souls, and thus always try to renew the ACM with minimal possible movements of existing statues (besides, the holographic
equipment is very heavy). Statues are moved along the park perimeter. Your work is to find a renewal plan which minimizes the sum of travel distances of all statues. Installation of a new hologram adds no distance penalty, so choose the places for newcomers
wisely!

Input

The input file contains several test cases, each of them consists of a a line that contains two integer numbers:
n -- the number of holographic statues initially located at the ACM, and
m -- the number of statues to be added
(2

n

1000,
1

m

1000).
The length of the alley along the park perimeter is exactly 10 000 feet.

Output

For each test case, write to the output a line with a single real number -- the minimal sum of travel distances of all statues (in feet). The answer must be precise to at least 4 digits after decimal point.



Pictures show the first three examples. Marked circles denote original statues, empty circles denote new equidistant places, arrows denote movement plans for existing statues.

Sample Input

2 1
2 3
3 1
10 10

Sample Output

1666.6667
1000.0
1666.6667
0.0

/*我个人认为这道题是涉及到参照系的选取和double数据的取整

这道题我们要巧妙地选取n+m个雕塑时的情况为基本的坐标系，

并选取原点0坐标，其余的为1,2，。。。n+m-1；

原来n个点的时候，这n个点在n+m坐标系里的坐标

应该是0，k,2k,3k,....(n-1)k,其中k=(n+m)/n，是距离的比率

我们可以通过floor(pos+0.5)类似于四舍五入的方法选取在m+n坐标系里的最近点

再用自己的坐标和最近点的坐标作差并将所有的差值累加即可

*/

/*那么我们自然要问，会不会出现原来n坐标系中的两个点同时移动到m+n坐标系中的同一个点上呢，

答案是否定的，原因是，由于我们用的是floor(pos+0.5)的四舍五入方法选定临近点的，

如果n坐标系里的两个点在m+n坐标系里选取了同一个临近点，证明在m+n坐标系内两者的坐标差值要小于1，

但是我们知道，由于n坐标系更加稀疏，因此两个点之间的距离必然大于1，

因此不存在两个点选了同一个临近点的情况*/

#include<iostream>

#include<cstdio>

#include<cmath>

#include<iomanip>

using namespace std;

int n,m;

int main()

{

while(cin>>n>>m)

{

int i;

double ans=0.0;

double pos;//用来记录n坐标系内的点在n+m坐标系里的坐标

double k=(double)(n+m)/n;

for(i=1;i<n;i++)

{

pos=k*i;

ans+=fabs(pos-floor(pos+0.5));/*四舍五入法选取最近点*/

}

ans*=(double)10000/(n+m);

cout<<setiosflags(ios::fixed)<<setprecision(4)<<ans<<endl;

}

return 0;

}