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

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;
}


只捉到一只哎 伤心~

• 千千 Chrome | 57.0.2950.5 Windows 10/11

C++提交答案错误，但是G++通过了