hpu暑假训练B - Pie 【二分法】

最新推荐文章于 2021-05-20 14:23:15 发布

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本文介绍了一种算法，用于解决如何将多个不同大小的圆柱形蛋糕平均分配给特定数量的人的问题，确保每个人获得相同体积的蛋糕，同时考虑蛋糕不能被切割成多块的限制。

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My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input

One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.

Output

For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).

Sample Input

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

Sample Output

25.1327
3.1416
50.2655

解析：我过生日请了f 个朋友来参加我的生日party，m个蛋糕，我要把它平均分给每个人（包括我），并且每个人只能从一块蛋糕得到自己的那一份，并且分得的蛋糕大小要一样，形状可以不一样，每块蛋糕都是圆柱，高度一样。

此题是一个二分题，下限是用最大的分，上限是sum/f+1。中间值是m，当cnt+=cnt+=(int)floor(p[i]/m);cnt<f+1,则r=m;否则l=m;二分的结束条件是（l+0.00001）。

程序如下：

#include<cstdio>
#include<cmath>
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int n,f,r[10003],i;
		double s[10003],max,sum=0,L,R,M,cnt;
		scanf("%d%d",&n,&f);
		for(i=0;i<n;i++)
		    scanf("%d",&r[i]);
		f++;
		max=0;
		for(i=0;i<n;i++)
		{
			s[i]=acos(-1.0)*r[i]*r[i];
			if(s[i]>max)
			   max=s[i];
			sum+=s[i];
		}
		L=max/f;
	    R=sum/f;
	    while(L+0.00001<R)
        {
        M=(L+R)/2;
        cnt=0;
        for(i=0;i<n;i++)
          cnt+=(int)floor(s[i]/M);
        if(cnt<f) 
		   R=M;
        else
		   L=M;
        }
      printf("%.4lf\n",L);
	}
	return 0;
}