POJ3122--Pie--二分

最新推荐文章于 2020-08-18 13:46:18 发布

原创最新推荐文章于 2020-08-18 13:46:18 发布 · 903 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#POJ3122 #pi #二分查找

二分查找专栏收录该内容

10 篇文章

订阅专栏

本文介绍了一个算法问题，即如何在给定多个不同大小的圆形蛋糕时，将蛋糕切成相同体积的块并分发给一定数量的人，同时确保每个人获得的蛋糕体积相等且最大化。通过二分查找的方法来寻找最优的蛋糕切分方案。

Description

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

注意一块饼能分给的人数是pi*r*r/s.二分查找适合的s即可

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
#define pi acos(-1.0)
double r[10008];
int n;
#define inf 1e-5
int F(double s)
{
	int sum=0;
	for(int i=1;i<=n;i++)
	{
		sum+=pi*r[i]*r[i]/s;
	}
	return sum;
}
inline double max(double a,double b)
{
	return a>b?a:b;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int f;
		scanf("%d%d",&n,&f);
		double maxsum=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%lf",&r[i]);
			maxsum=max(maxsum,r[i]);
		}
		double l=0,r=pi*maxsum*maxsum;
		while(l<r-inf)
		{
			double mid=(l+r)/2;
			if(F(mid)<=f)r=mid;
			else l=mid;
		}
		printf("%.4lf\n",l);
	}
	return 0;
}