Pie

原题链接：http://poj.org/problem?id=3122

Problem 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

解题思路：

二分法

AC代码：

#include
#include
using namespace std;
const double pi = acos(-1.0);
double mid;
int T,N,F;					//T为测试数据组数
double maxS(double ri[],int n)
{
	double max = 0;
	for (int i = 0; i < n; i++)
	{
		max += (ri[i] * ri[i] * pi);
	}
	return max;
}
int main()
{
	cin >> T;
	for (int i = 1; i <= T; i++)
	{
		double ri[10010];
		cin >> N >> F;					//N为饼的个数，F为朋友数
		int M = F + 1;					//M为分饼的人

		for (int i = 0; i < N; i++)
		{
			cin >> ri[i];				//输入饼的半径
		}

		double L = 0;						//左边界
		double R = maxS(ri, N) / M;		//右边界(每个人理论能分到的最大面积)

		while ( R - L > 1e-6)
		{
			mid = (L + R) / 2;			//中值
			int sum = 0;
			for (int i = 0; i < N; i++)
			{
				sum += int(ri[i] * ri[i] * pi / mid);		//总共能分几人 
			}
			if (sum >= M)				//人数大于M，mid嫌小
			{
				L = mid;
			}
			else
			{
				R = mid;				//人数小于M，mid嫌大
			}
		}
		printf("%.4lf\n",mid);
	}
	return 0;
}