hdoj 1969 （poj 3122&&acm）Pie

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8519    Accepted Submission(s): 3104

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
   
   
   
   
   
   
   
   

    
    
    
    

       就是有n个蛋糕分给f+1个人，每人只能分一块（可以是一整个蛋糕），每个人分的体积相等,下面输入N个蛋糕的半径，高均为1. 
    
    
    
    
    

      二分法，[0,maxs]然后一分为二，并比较s=mid时可以分配给的人数与ｆ＋１的大小，确定下一个二分区间。 
    
    
    
    
    

      #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #define PI acos(-1)／／反函数就是圆周率的值 using namespace std; double r[10010],s[10010]; int main() {     int t,n,f;     double left,right,mid;     scanf("%d",&t);     while(t--)     {         double maxm;         scanf("%d%d",&n,&f);         f+=1;         for(int i=0;i<n;i++)         {             scanf("%lf",&r[i]);             s[i]=1.0*r[i]*r[i];         }         sort(s,s+n);         maxm=s[n-1];／／最大底面积         int ans; 
    
    
    
    
    
 
      ／／         left=0.0;         right=maxm;         while(right-left>(1e-6))         {             mid=(left+right)/2;             ans=0;             for(int i=0;i<n;i++)             {                 ans+=(int)s[i]/mid;             }             if(ans<f)                 right=mid;             else                 left=mid;         }         printf("%.4lf\n",mid*PI);     }     return 0; }