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.
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
//题目大意：就是把n个半径不同的pie分成面积相同的分给f+1个小朋友，求最大的面积
#include<stdio.h>
#include<math.h>
#define PI acos(-1.0)
int i;
int t,n,f;
double a[10100];
int judge(double x)
{
int cnt=0;
for(i=0;i<n;++i)
{
cnt+=(int)(a[i]/x);//这个地方wa了n次
}
if(cnt>=f) return 1;//找到的面积小了
return 0;
}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&f);
f++; //还有他自己
double max=0;
for(i=0;i<n;++i)
{
scanf("%lf",a+i);
a[i]=a[i]*a[i]*PI;
max+=a[i];
}
double left=0,right=max;
double mid;
while(right-left>=1e-7)
{
mid=(left+right)/2;
if(judge(mid)) left=mid;
else right=mid;
}
//right=(int)(right*10000)/10000.0;
printf("%.4f\n",mid);
}
return 0;
}
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