三、搜索和二分 [Cloned] F - 二分

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本文介绍了一道关于如何公平地将不同大小的圆形馅饼平均分配给多人的问题，并提供了使用二分查找法解决该问题的具体代码实现。

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原题：

y 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.

题意:

将n块馅饼分给（m+1）个人，每个人的大小必须相同，而且每个人的馅饼只能来自同一块馅饼，就是不能剁成很小的小块然后并均分，问怎么才能让大家分到的馅饼最多。

题解：

二分法的应用，因为他要分饼的精度为小数点后三位，所以可以选择小数点后五位（随便选的还可以更小），然后用选定的标准来分每块饼，如果够分就再加大，如果不够分就减小，直到符合截止标准。

代码：AC

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
double pi=acos(-1.0);
double p[10000+10];
double sum,maxn;
int i,n,f,t,k,cnt;
double l,r,m;
double max(double a,double b)
{
	if(a>b)
		return a;
	else
		return b;
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
      scanf("%d%d",&n,&f);
      f++;
      l=sum=0;
      for(i=0;i<n;i++)
      {
         scanf("%d",&k);
         p[i]=pi*k*k;
         l=max(p[i],l);
         sum+=p[i];
      }
      l=l/f;
      r=sum/f;
      while(l+0.00001<r)
      {
        m=(l+r)/2;
        cnt=0;
        for(i=0;i<n;i++)
          cnt+=(int)floor(p[i]/m);
        if(cnt<f) r=m;
        else l=m;
      }
      printf("%.4lf\n",l);
    }
    return 0;
}