NWERC 2006 / UVa 12097 Pie (二分法求精度)

12097 - Pie

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=456&page=show_problem&problem=3249

 

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 ≤ 10000: the number of pies and the number of friends.
• One line with N integers r_i with 1 ≤ r_i ≤ 10000: 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

思路：注意这题是让我们求小数形式的最优解而不是即约分数(最简分数)形式

那有更好的方法求吗？——回想高中解超越方程近似根所用的方法（二分法），我们就有了如下的代码：

/*0.029s*/

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

const double PI = acos(-1.0);
const int maxn = 10005;

int n, f;
double A[maxn];

inline bool ok(double area)
{
	int sum = 0;
	for (int i = 0; i < n; i++)
		sum += floor(A[i] / area);
	return sum >= f + 1;
}

int main(void)
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &n, &f);
		double maxa = -1.0;
		for (int i = 0; i < n; i++)
		{
			int r;
			scanf("%d", &r);
			A[i] = PI * r * r;
			maxa = max(maxa, A[i]);
		}
		double L = 0.0, R = maxa;
		while (R - L > 1e-5)//二分查找计算，达到要求精度时停止查找
		{
			double M = (L + R) / 2;
			if (ok(M))
				L = M;
			else
				R = M;
		}
		printf("%f\n", L);
	}
	return 0;
}