Description
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
n1000, 1m1000)<tex2html_verbatim_mark> . The length of the alley along the park perimeter is exactly 10 000 feet.
Output
<tex2html_verbatim_mark>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
题目大意：在一个周长10000的圆上等间距分布n个雕塑，现又加入m个（位置可以随意）,希望所有n+m个塑像在圆上均匀分布。这就需要移动一些原有的塑像。要求n个塑像移动的总距离最小，输入n,m输出最小距离，小数点后4位。
解题思路：总是有一个雕塑没有移动，于是假设这个为原点，逆时针给n个点标号，表示到原点的距离【这里是比例距离】接下来我们把每个点移动到离它最近的距离，如果没有2个雕像移动到相同的位置，那么这样的移动一定是最优的。代码中，坐标为pos的雕塑移动的目的坐标位置是：floor(pos+0.5），就是四舍五入的结果。就是坐标缩小的好处。
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 using namespace std; 5 int main(){ 6 int n,m; 7 while(cin>>n>>m){ 8 double ans=0.0; 9 for(int i=1;i<n;i++){ 10 double pos=(double)i/n*(n+m);//计算每个需要移动的雕塑的坐标 11 ans+=fabs(pos-floor(pos+0.5))/(n+m);//累加移动距离 12 } 13 printf("%.4lf ",ans*10000);//等比例放大 14 }return 0; 15 }