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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.NoSuchElementException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BDigits solver = new BDigits();
        solver.solve(1, in, out);
        out.close();
    }

    static class BDigits {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            long n = in.nextLong();
            long k = in.nextLong();
            out.println(String.valueOf(convert(n, k)).length());
        }

        long convert(long x, long n) {
            long y = 0;
            long i = 0;
            long z;
            while (x > 0) {
                z = x % n;
                y += z * Math.pow(10, i);
                x = x / n;
                i++;
            }
            return y;

        }

    }

    static class Scanner {
        private final InputStream in;
        private final byte[] buffer;
        private int ptr;
        private int buflen;

        public Scanner(InputStream in) {
            this.in = in;
            this.buffer = new byte[1024];
            this.ptr = 0;
            this.buflen = 0;
        }

        private boolean hasNextByte() {
            if (ptr < buflen) return true;
            else {
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (buflen <= 0) return false;
            }
            return true;
        }

        private byte readByte() {
            if (hasNextByte()) return buffer[ptr++];
            return -1;
        }

        private boolean isPrintableChar(byte c) {
            return '!' <= c && c <= '~';
        }

        private void skipUnprintable() {
            while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
                ptr++;
            }
        }

        public boolean hasNext() {
            skipUnprintable();
            return hasNextByte();
        }

        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            byte b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) throw new NumberFormatException();
            while (true) {
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                } else if (b == -1 || !isPrintableChar(b)) {
                    return minus ? -n : n;
                } else {
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }

    }
}