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);
}
}