pi

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.<br><br>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. <br><br>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.<br>

 

Input

One line with a positive integer: the number of test cases. Then for each test case:<br>---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.<br>---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.<br>

 

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<br>3 3<br>4 3 3<br>1 24<br>5<br>10 5<br>1 4 2 3 4 5 6 5 4 2<br>

 

Sample Output

25.1327<br>3.1416<br>50.2655<br>

这道题真心没辙了。。  wa n次然后又时间超限n次

题意是分圆柱形派 高一样所以只要分面积，总共有f+1个人分但是每个人分一块，或者多个人分一块，一个人不能分多块，求出每个人做多分的大小

思路主要就是二分上界是最大半径的平方

一直wa我上网上找了下感觉思路并没有错误。。然后代码对比下真心没看出来那个地方错了。。

我的代码：

#include <iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double pi=acos(-1.0);
int T,n,f;
double b[10003];
int juge(double x)
{
    int total =0;
    for(int i=1; i<=n; i++)
    {
        total+=int(b[i]/x);
    }
    if(total>f)
        return 1;
    else return 0;
}
int main()
{
    double l,r,mid,rad;
    cin>>T;
    while(T--)
    {
        cin>>n>>f;
        double sum=0;
        for(int i =1; i<=n; i ++)
        {
            cin>>rad;
            b[i]=rad*rad*pi;
            sum+=b[i];
        }
        l=0;
        double r=sum/(f+1);
        while(r-l>0.0001)                 {
            mid=(l+r)/2;
            if(juge(mid))
                l=mid;
            else                r=mid;
        }
        cout<<fixed<<setprecision(4)<<mid<<endl;
    }
    return 0;
}

附上正确ac代码

#include <iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double pi=acos(-1.0);
int T,n,f;
double b[10003];
int juge(double x)
{
    int total =0;
    for(int i=1; i<=n; i++)
    {
        total+=int(b[i]/x);
    }
    if(total>f)
        return 1;
    else return 0;
}
int main()
{
    double l,r,mid,rad;
    cin>>T;
    while(T--)
    {
        cin>>n>>f;
        double sum=0;
        for(int i =1; i<=n; i ++)
        {
            cin>>rad;
            b[i]=rad*rad*pi;
            sum+=b[i];
        }
        l=0;
        double r=sum/(f+1);
        while(r-l>0.0001)                 {
            mid=(l+r)/2;
            if(juge(mid))
                l=mid;
            else                r=mid;
        }
        cout<<fixed<<setprecision(4)<<mid<<endl;
    }
    return 0;
}