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: Accepted)

#include<stdio.h>
#include <stdlib.h>
#include<string.h>
#include<math.h>
#define loop(i, l, r) for(i=l; i<r; i++)/*Instead of function, (for).*/
#define rep(i,r) for(i=0; i<r; i++)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long int lli;
/*Return Combination ,(nCr)*/
int combi(int n, int r)
{
	if(r==0 || r==n) return 1;
	else return combi(n-1,r) + combi(n-1,r-1);
}
/*Return gcd(a, b)*/
int gcd(int a, int b){
	if(b == 0) return a;
	else return gcd(b, a % b);
}
int main(void)
{
	int n, k, s, cut=0;
	scanf("%d%d", &n, &k);
	while(n != 0){
		n/=k;
		cut++;
	}
	printf("%d", cut);
	return 0;
}