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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

int main(void){
    int N;
    int K;
    int sum=0;
    int a;
    int keisan=0;

    scanf("%d",&N);
    scanf("%d",&K);
    a = N;
    sum = 1;
    if(a > K){ 
    while(a > K){//割れる回数A＝A進数
        keisan = a%K;
        a = a/K+keisan;
        sum += 1;
    }   
    }else if(a == K){ 
        sum = 2;
    }
    printf("%d\n",sum);
    return 0;                                                                                    
}