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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <iomanip>
#include <math.h>
#include <string.h>
using namespace std; using ll = long long; using pll = pair<ll, ll>;
using vl = vector<ll>; using vll = vector<vl>; using vpll = vector<pll>;
int ctoi(char c) {
	switch (c) {
	case '0': return 0; case '1': return 1; case '2': return 2;
	case '3': return 3; case '4': return 4; case '5': return 5;
	case '6': return 6; case '7': return 7; case '8': return 8;
	case '9': return 9;  default: return 0;
	}
}
bool pairCompare(const pll firstElof, pll secondElof)
{
	return firstElof.second < secondElof.second;
}
ll nod(ll F) { return log10(F) + 1; }
//**定義場所**//
//gcd, isPrime, eratos, modinv, bitSearch, dfs, bfs, dijkstra, UnionFind//
ll i, j, k, l, m, n; ll N, M, K, H, W, L; string S;
ll MOD = 1000000007; ll ans = 0; ll INF = 9999999999999;
vl flag, color; vll path;
//***********//
int main() {
	cin >> N >> K;
	while (N > 0) {
		N /= K;
		ans++;
	}
	cout << ans;
}