POJ 3122 Pie (二分)

Pie
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16185 Accepted: 5505 Special Judge

 

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

每个pie都是高为1的圆柱体,输入这N个pie的半径,要公平地把pie分给每一个人,所有人得到pie的尺寸一样,但是形状可以不同,并且每一个人得到的pie必须来源于一个pie

就是说如果有3个pie, 尺寸分别为1,2,3。

如果要给每人尺寸为2的pie,那么最多分给2个人,而不是3个人

因为第一个pie尺寸为1,小于2,扔掉

第二个pie尺寸为2,等于2,刚好分给一个人

第三个pie尺寸为3,切出尺寸为2的一份,分给一个人,剩下的尺寸为1的就扔掉

 

思路

我们知道圆的面积与其半径的平方成正比例,因此给每一个人分配的大小便与所有pie半径平方直接相关咯!

又因为每一个人不可以同时从两个pie得到属于自己的那部分,所以我们可以固定当前所要分给每个人的半径,然后判断所有的pie,是否可以刚好分给所有人,若满足,可以尝试增大半径,若不满足,尝试减小半径,所以在这里采用二分的思想。

 

AC 代码

#include <stdio.h>
#include <math.h>
#define PI 3.1415926535897932

double a[10050];
int n,f;
void solve(double sum)
{
    double low=0,high=sum/f;    //high是当前pie平均分配时的值,也就是最大值
    while(high-low>1e-8)    //二分
    {
        double mid = (low+high)/2.0;
        int count=0;
        for(int i=0; i<n; i++)  //判断每一个pie看可以分成几个块
            count+=(int)(a[i]/mid);
        if(count>=f)low=mid;
        else high=mid;
    }
    printf("%.4lf\n",low*PI);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        double sum=0.0;
        scanf("%d%d",&n,&f);
        ++f;    //加上他自己
        for(int i=0; i<n; i++)
        {
            scanf("%lf",a+i);
            a[i]*=a[i];     //计算平方,因为圆的面积与半径平方成正比例
            sum+=a[i];      //总的半径平方和
        }
        solve(sum);
    }
    return 0;
}

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