poj 3122 Pie 二分（最大化平均值）

Pie

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

题意好麻烦：有n个pi，f个朋友，还有一个自己，给出每个pi的大小，每个pi可以切成几个蛋糕，而且所有pi切出的蛋糕都必须一样大，每个人至少分一个蛋糕，求最大的蛋糕的体积。高均为1.

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define maxn 25100000
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
int a[maxn];
int n,f;
bool ok(double x){
   int ans=0;
   for(int i=0;i<n;i++){
      ans+=a[i]*a[i]*pi/x;
   }
   if(ans<f+1)return true;
   else return false;
}
int main()
{
    int t;
    freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&f);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        double ua=0.0;
        double ub=1000000000;
        double mid;
       /*
        while(ub-ua>eps){        //两种方法均可，此题对精度要求比较低
            mid=(ua+ub)/2;
            if(ok(mid))ub=mid;
            else ua=mid;
        }*/
        for(int i=0;i<100;i++){
             mid=(ua+ub)/2;
            if(ok(mid))ub=mid;
            else ua=mid;
        }
        printf("%.4f\n",mid);
    }
}