POJ-3154 Graveyard（四舍五入法）

最新推荐文章于 2020-11-23 22:29:13 发布

碳酸钙的01妖精

最新推荐文章于 2020-11-23 22:29:13 发布

阅读量238

点赞数

分类专栏： ACM 思考题

本文链接：https://blog.youkuaiyun.com/xigongdali/article/details/83048121

版权

ACM 思考题专栏收录该内容

7 篇文章

订阅专栏

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

Input file 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

Write a single real number to the output file — the minimal sum of travel distances of all statues (in feet). The answer must be precise to at least 4 digits after decimal point.

Sample Input

sample input #1
2 1

sample input #2
2 3

sample input #3
3 1

sample input #4
10 10

Sample Output

sample output #1
1666.6667

sample output #2
1000.0

sample output #3
1666.6667

sample output #4
0.0

Hint

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

思路：定下末状态： m+n个点假设两点之间的距离为1 （m+n==10000,最后转换回来）

之前状态 n个点。（m+n）/n*i是当前点的坐标，找最近的整点即可。

#include<iostream>
#include<cstdio>
#include<cmath>

using namespace std;

int n,m;
double pos,ans,tmp;

int main()
{
	while(cin>>n>>m)
	{
		ans=0;
		for(int i=1;i<n;i++)
		{
			pos=(n+m)*1.0/n*i;//离散化，原来的n个点在后来的n+m个点数轴的坐标
			tmp=abs(floor(pos+0.5)-pos);//四舍五入法的妙用
			ans+=tmp;
		}
		ans=ans*10000/(n+m);//将坐标还原
		printf("%.4f\n",ans);
	}
	return 0;
}
--------------------- 
作者：@Herobrine 
来源：优快云 
原文：https://blog.youkuaiyun.com/qq_41636123/article/details/83033219?utm_source=copy