UVA 1388 - Graveyard (推理，参考系)

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

题意:圆周上有n个墓碑，现在要加m个，要求仍平均摆，问最少移动距离。

思路:参考了大白上的思路。上面有两点证明，一个是有一个墓碑肯定不用移动，和所有墓碑肯定不会移动到同一个点，弄清楚就很好写了。

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>

int n, m;
double ans;

int main() {
while (~scanf("%d%d", &n, &m)) {
ans = 0;
for (int i = 1; i <= n; i ++) {
double pos = double(i) / n * (n + m);
ans += fabs(floor(pos + 0.5) - pos);
}
printf("%.4lf\n", ans / (n + m) * 10000);
}
return 0;
}