POJ 3122 Pie （贪心+二分）

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.

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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

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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.

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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

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题目倒是不难，比较容易想，但是这个精确度确实是有点恶心...WA了6，7次。首先需要注意什么地方要用double，什么地方要转int，其次也学会了在小数范围内用二分，之前一直WA竟然是二分判断的时候小数点后0不够多,,,加了一个0后立马AC...醉惹，那个排序貌似并没有什么用...原来32ms，排序后16ms...

 #include <iostream>
 #include <string.h>
 #include <stdio.h>
 #include <algorithm>
 #include <cstdio>
 #include <queue>
 #pragma warning ( disable : 4996 )

 using namespace std;

 int Max( int x, int y ) { return x>y?x:y; }
 int Min( int x, int y ) { return x>y?y:x; }

 const int inf = 0x3f3f3f3f;
 const int vspot = 1e4 + ;
 const int espot = 1e5 + ;
 const double PI = 3.1415926535897932;

 int N, F;
 double rad[vspot];

 bool cmp( int a, int b )
 {
     return a>b;
 }

 int main()
 {
     int all;
     scanf( "%d",&all );
     while (all--)
     {
         scanf( "%d %d", &N, &F ); F++;
         int x, fin;
         double sum = ;
         for( int i = ; i < N; i++ )
             { scanf( "%d", &x ); rad[i] = x*x; sum += rad[i]; }

         double rhs = sum/F, lhs = 0.0;
         double mid;
         //sort( rad, rad+N, cmp );

         while ( rhs - lhs > 0.0000001 )
         {
             fin = ;
             mid = (rhs+lhs)/;
             for ( int i = ; i < N; i++ )
             {
                 //if ( rad[i] < mid )
                 //    break;
                 fin += (int)(rad[i]/mid);
             }
             if ( fin >= F )
                 lhs = mid;
             else
                 rhs = mid;
         }

         printf( "%.4lf\n", PI*mid ); 

     }
     return ;
 }