Pie(派)

最新推荐文章于 2020-08-18 13:46:18 发布

Sharing_Li

最新推荐文章于 2020-08-18 13:46:18 发布

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分类专栏：算法练习文章标签：算法 C

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该博客探讨了一个涉及分配派给朋友的问题，其中要求确保每个人都能得到一定体积的派。问题涉及到输入参数N（派的数量）和F（朋友的数量），以及每个派的半径ri。解决方案采用了二分搜索算法来找到最大的可能体积，使得所有朋友都能获得满足条件的派块，答案需精确到小数点后三位。

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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。

用到二分法，代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
const double PI = 3.14159;
const int MAX = 10000;
int N,F;
double A[MAX];
bool fun(double area)
{
	int count = 0;
	for (int i = 0; i < N; i++)
	{
		count += int(A[i] / area);//取结果的整数(即每块饼可以分配的人数)相加
	}
	return count >= F + 1;
}
int main()
{
	int T;
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d",&N,&F);
		double temp = -1;
		for (int i = 0; i < N; i++)
		{
			int r;
			scanf("%d",&r);
			A[i] = PI * r * r;
			temp = __max(temp,A[i]);
		}
		double L = 0, R = temp;
		while (R - L > 1e-5)//二分法
		{
			double M = (L + R) / 2;
			if (fun(M))
			{
				L = M;
			}
			else
			{
				R = M;
			}
		}
		printf("%.3lf\n",L);
	}
	return 0;
}