hdoj 1969 分蛋糕问题（二分法的应用）

这道题是典型的二分问题，由于没有看到其中的一个已知条件（分给每个人的蛋糕必须是整个大块的，不能东拼西凑出来的）

题目意思：我过生日请了f 个朋友来参加我的生日party，m个蛋糕，我要把它平均分给每个人（包括我），并且每个人只能从一块蛋糕得到自己的那一份，并且分得的蛋糕大小要一样，形状可以不一样，每块蛋糕都是圆柱，高度一样。

此题是一个二分题，下限是用最大的分，上限是sum/f+1。中间值是m，当cnt+=cnt+=(int)floor(p[i]/m);cnt<f+1,则r=m;否则l=m;

Pie

Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 190   Accepted Submission(s) : 72

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

 

  AC码如下

#include<stdio.h>
#define pi acos(-1.0)
#include<math.h>
int m,n,i,f,R;
double a[10010];
double mid;
int main()
{
	
	scanf("%d",&m);
	while(m--)
	{
		scanf("%d%d",&n,&f);
		f++;
		double max=0,s=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&R);
			a[i]=pi*R*R;
			s=s+a[i];      
			if(max<=a[i])
			max=a[i];     
		}
		double l=max/f ;   //最多每人分得的蛋糕量 
		double r=s/f ; 	  //最少每人分得的蛋糕量
		
		while(r-l>=1e-5)
		{   int cnt=0;
			mid=(l+r)/2;
 		    for(i=0;i<n;i++)
			cnt=cnt+(int)(a[i]/mid);  //这个就是判断到底能不能够分给n个人 
			if(cnt<f)                
			 r=mid;
			else
			 l=mid;
		}
	printf("%.4lf\n",mid);	//注意这个返回值，只能是l mid 
	}
	return 0;
}

解决本题的关键是： 1、找到最少和最多每个人分的蛋糕量。

                                2、 用每人分得的蛋糕量够不够分给n个人作为判断。

3、设置累加计算最多能分给几根人