1068 birthday cake

Description

July 17 is the birthday of mr. w, ACM-THU for this to create a volume of N π M-layer birthday cake, each layer is a cylinder. Set the number of I (1 <= I <= m) layers from bottom to top to a column with Ri radius and HI height. When Siri + 1 and HI> Hi + 1. To minimize the cost of using cream on a cake, we hope that the surface of the cake (except the bottom of the bottom layer) has the minimum area Q. To make Q = s π, program the N and M given to find out the cake production scheme (appropriate RI and HI values) and minimize S. (All data except Q is a positive integer)

Input

The test data contains multiple groups of data each with two rows. The first behavior n (n <= 10000) indicates that the volume of the cake to be prepared is N π; the second behavior M (M <= 20) indicates that the number of layers of the cake is M.

Output

Only one row is a positive integer S (if there is no solution, S = 0 ).

Sample Input

1002

Sample output

68

Prompt

Cylindrical formula: Volume V = π r2h, side area a' = 2 π RH, bottom area A = π r2

Source

Noi1999 by aluvin

This question is searched by DFS to ensure that the correct value can be found within the specified time.

 # include 
   
     # define in (a, B) (a 
    
      N | S + mins [level-1]> bests | 2 * (N-V) /R + S> = bests) // If you replace level-1 in the brackets with level, it is 32 msreturn; int I, j, HH; for (I = R-1; i> = level; I --) {If (Level = m) S = I * I; HH = in (n-v-MINV [level-1]) /(I * I), h-1); For (j = HH; j> = level; j --) DFS (V + I * J, S + 2 * I * j, level-1, I, j) ;}} int main () {int I; MINV [0] = 0; mins [0] = 0; for (I = 1; I <= 20; I ++) {MINV [I] = MINV [I-1] + I * I; mins [I] = mins [I-1] + 2 * I;} while (scanf ("% d", & N, & M) = 2) {bests = 0x7fffffff; DFS (0, 0, m, n + 1, n + 1); If (bests = 0x7fffffff) printf ("0 \ n "); elseprintf ("% d \ n", bests) ;}return 0 ;}