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 <stdlib.h>
#include <string.h>
#include<math.h>
#include <stdio.h>
void shift(char* p, long m, int n)
{
	int r;
	while (m > 0)
	{
		r = m % n;			//求余
		if (r < 10)
		{
			*p = r + 48;    //小余10的数转换成字符后赋给P所指向的元素
		}
		else
		{
			*p = r + 55;	//10-15转化为“A~F”后赋给P所指向的元素
		}
		m = m / n;
		p++; 				//指针后移

	}
	*p = '/0';
}
int main(void)
{
	int x;
	long y;
	char a[32] = {'\0'};		//存放结果的数组

		scanf("%ld %d", &y,&x);

		printf("\n");
		
		shift(a, y, x);
		printf("%d", strlen(a)-1);
	


	
	return 0;
}