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 <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;

const char wall = '#';
vector<vector<char>> fields;
vector<vector<bool>> used;
vector<vector<ll>> dist;
const int dx[4] = {0, -1, 0, 1};
const int dy[4] = {1, 0, -1, 0};

ll N;
vector<ll> A;

ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return b/gcd(a,b)*b;}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  // Globalなvectorのリサイズ
  //A.resize(N);

  ll n, k;
  cin >> n >> k;
  rep(i, 10000) {
    if (pow(k, i) > n) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}