UVa 107－The Cat in the Hat

Problem: 107 - The Cat in the Hat

Explain:
Given, Initial Height of a Cat, IH
The number of worker Cats of height 1, WC1

Objective:
Find, sum of all the cats' heights, SCH =?
Number of Not Working Cat, NWC =?



To understand this problem, Let IH = 216, WC1 = 125     

“The smallest Cats are of height one;
These are the cats that get the work done;” [Copied from Uva 107]

If, IH = 1 and WC1 = 1,
This Indicate, TC = 1, NWC = 0  
This also indicates the termination condition.
“The number of cats inside each (non-smallest) cat's hat is a constant, N. The height of these cats-in-a-hat is 1/(N+1) times the height of the cat whose hat they are in.” [Copied from Uva 107]
Let Constant N = 5,

Not Working Cat ,NWC	Height of Each Cat, Given, IH = 216	sum of all the cats' heights, SCH
		∑height = 216
1	The number of cats inside each (non-smallest) cat's (height = 216) hat is 5, Height = (216/(5+1) = 36	∑height = (216/6)*5 + 216=396
1+5	The number of cats inside each (non-smallest) cat's (height = 36) hat is 5, Height = 36/6 = 6	∑height = (216/62)*52 + 396 = 546
1+5+52 = 31	The number of cats inside each (non-smallest) cat's (height = 36) hat is 5, Height = 36/6 = 1	∑height = (216/63)*53 + 546 =671
	Break…..

Challenge:
Find N , i.e.
N^m = WC1 and
(N+1)^m = IH

 Code:
======

#include<iostream>
#include<cstdio>
#include<cmath>
#define ERROR 1e-8
using namespace std;

int main(){
  int H, num, m, N;
   #ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
   #endif


  while( scanf( "%d%d", &H, &num ) != EOF && !(H == 0 && num == 0)){
    if( H == 1 && num == 1 ){
      printf( "0 1\n");
      continue;
    }

    m = 1;
    while( H != (int)(pow(pow(num,1.0/m)+1.0, m)+ERROR) ) m++;
    N = (int)(pow(num,1.0/m)+ERROR);
    if( N != 1 ) printf( "%d %d\n", (1-num)/(1-N), (H-num)*N+H );
    else printf( "%d %d\n", m, (H-num)*N+H );
  }
  return 0;
}