二分查找——D:Pie 解题报告

                                                                                            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：把N个派分给F+1个人（自己也要），输入N，F，和每个派的半径（派的高度为1）。分派要满足一下要求：1.每个人分的派体积一样，2.任何人不能同时吃到两块派，但是可以得到一整个派，3.派可以剩余

解题思路：设每人最多能得到的派为x，首先可以确定x不会超过最大的一块派，所以x的上限就为max（派i）,进入二分查找最大的x，二分思路如下：如果可以按照x分，尝试下更大的（low=mid）,如果不能，就尝试小一些的（high=mid）.分派思路：从第一块派开始，如果这块派剩余大小比x大，则这块派减去x，人数-1，否则用下一块派继续这一过程，直到派用完（x太大）或者人数为0（可以按照x分）。

#include 
#include 
#include 
using namespace std;

double m=0,y;
double pi[10005],temp[10005];
const double P=acos(-1.0);//求π
const double EPS=1e-7;
int N,F;

double MAX(double a,double b)
{
    return a>b?a:b;
}


int f(double x)//判断是否可以按x分
{
    int i,num=F+1;
    for(i=0;i0)
    {

        if(temp[i]>=x)
        {//这一块还可以分
            temp[i]-=x;
            num--;
        }
        else
        {
            i++;//这一块分不了了，下一块
        }
        if(i>=N)//x过大
            return 0;
    }
    return 1;
}

double bs()
{
    double lo=0,hi=m;
    double mi;
    int k=100;
    while(hi-lo>EPS)
    {
        mi=(lo+hi)/2.0;//浮点数问题就别移位运算了
        if(f(mi)==0)//如果x过大，尝试下小的
            hi=mi;
        else        //如果可按照x分，尝试下更大的
            lo=mi;
        //cout<>repeat;
    while(repeat--)
    {
        cin>>N>>F;
        for(int i=0;i>pi[i];
            pi[i]=pi[i]*pi[i]*P;
            m=MAX(pi[i],m);
            //cout<