Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of digits that N has in base K.

Notes

For information on base-K representation, see Positional notation - Wikipedia.

Constraints

All values in input are integers.

1 \leq N \leq 10^9

2 \leq K \leq 10

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of digits that N has in base K.

Sample Input 1

11 2

Sample Output 1

4

In binary, 11 is represented as 1011.

Sample Input 2

1010101 10

Sample Output 2

7

Sample Input 3

314159265 3

Sample Output 3

18

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using static System.Console;
using static System.Math;
 
public class Hello{
    public static void Main(){
        int[] read = inta();
        long a = read[0];
        long b = read[1];
        long c= 1;
        int k = 0;
        while(c< a){
            c*= b;
            k++;
        }
        if(k==0){
            k++;
        }
        WriteLine(k);
    
    
    
    
    
    
    }
    //各種読取
    static string rstr(){
        return ReadLine();
    }
    static int rint(){
        return int.Parse(ReadLine());
    }
    static long rlong(){
        return long.Parse(ReadLine());
    }
    static string[] stra(){
        return ReadLine().Split(' ');
    }
    static int[] inta(){
        string[] read_str_array = ReadLine().Split(' ');
        int[] return_int_array = new int[read_str_array.Length];
        for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_int_array[countup_i] = int.Parse(read_str_array[countup_i]);
        }
        return return_int_array;
    }
    static int[,] inta2(int num_array,int in_array){
        int[,] int_array2 = new int[num_array,in_array];
        for(int i=0;i<num_array;i++){
            int[] temp_array = inta();
            for(int j=0;j<in_array;j++){
                int_array2[i,j] = temp_array[j];
            }
        }
        return int_array2;
    }
    // A/B小数点以下切上
    static int divideup(int divided,int divid){
        int temp_divide = divided/divid;
        if(divided % divid > 0){
            temp_divide++;
        }
        return temp_divide;
    }
}