HDOJ 1969 Pie（二分）

Pie

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

题目大意：一个人有N个披萨和F个朋友，给出每个披萨的半径，要求分配这些披萨，每个人（包括主人自己）都有一块大小相同的披萨，且每个人的披萨不能由几块小披萨组合而成。

解题思路：令条件C(x) = 可以得到F + 1块面积为x的披萨，则问题变成了求满足C(x)条件的最大x。由于面积为Si的披萨最多可以分成floor(Si / x)块披萨，因此C(x) = (floor(Si / x)的总和是否大于或等于F + 1)。可以先对半径进行排序，然后取最大面积作为二分的上界即可。(PS：一共WA了5次，第一次没有算上主人自己，后面四次全是因为ub取小了，汗。。。)

代码如下：

#include <cstdio>
#include <cmath>
#include <algorithm>

#define PI acos(-1.0)
#define EPS 1e-6

using namespace std;
const int maxn = 10005;

int  radii[maxn];
int N,F;
bool judge(double x)
{
	int num = 0;
	for(int i = 0;i < N;i++){
		num += (int)(PI * radii[i] * radii[i] / x);
	}
	return num >= F + 1;
}

int main()
{
	double lb,ub;
	int i,j,k;
	scanf("%d",&k);
	while(k--){
		scanf("%d %d",&N,&F);
		for(i = 0;i < N;i++){
			scanf("%d",&radii[i]);
		}
		sort(radii,radii + N);
		lb = 0,ub = radii[N - 1] * radii[N - 1] * PI;
		while(ub - lb > EPS){
			double mid = (lb + ub) / 2;
			if(judge(mid))
				lb = mid;
			else 
				ub = mid;
		}
		printf("%.4lf\n",lb);
	}
	return 0;
}