uva1388 Graveyard

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 nd 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 le 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. Note:
Pictures show the rst three examples. Marked circles denote original
statues, empty circles denote new equidistant places, arrows denote
movement plans for existing statues.

可以规定有一个点不动【如果所有点都动答案一定不是最优】，把加入m个点后每两点之间的距离当成单位长度【为了之后便于计算】，然后对原来的n个点都移动到最近的位置上【也就是四舍五入后的位置】。因为原来任意两点距离大于1，所以一定不会有两个点移到一起。

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;int main(){    int n,m,i;    double ans,x;    while (scanf("%d%d",&n,&m)==2)    {        ans=0;        for (i=1;i<n;i++)        {            x=(double)(m+n)/n*i;            ans+=fabs(x-floor(x+0.5));        }        printf("%.4f\n",ans/(m+n)*10000);    }}