poj 3122 Pie（二分法）

Pie

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

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

题意：

F+1个人分N个形状不一的蛋糕，使每个人分得一块体积相等的蛋糕，并且，这块蛋糕只能来源某一个蛋糕。

思路：

将最大的蛋糕二分，判断mid大小的分法能否分出F+1份即可。先预处理出蛋糕体积和最大的体积。

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iomanip>

using namespace std;
const double PI = acos(-1.0);
double cake[10010];
int T, N, F;

bool check(double num){
	int sum = 0;
	for(int i=0; i<N; i++){
		sum += int(cake[i]/num);
	}
	if(sum>=F)
		return true;
	else
		return false;
}


int main(){
	cin>>T;
	while(T--){
		cin>>N>>F;
		F++;
		double maxx = -1.0;
		memset(cake, 0, sizeof(cake));
		for(int i=0; i<N; i++){
			cin>>cake[i];
			cake[i] = cake[i]*cake[i]*PI;
			maxx = max(cake[i], maxx);
		}

		double low = 0.0;
		double high = maxx;

		while(low+(1e-5)<high){
			double mid = (low+high)/2.0;
			if(check(mid))
				low = mid;
			else
				high = mid;
		}
		cout<<setiosflags(ios::fixed)<<setprecision(4)<<low<<endl;
	}
	return 0;
}