Multiplying Two Base-36 Numbers, Java

Multiplying Two Base-36 Numbers

//'main' method must be in a class 'Rextester'.
//Compiler version 1.8.0_111

import java.util.*;
import java.lang.*;

/** 
 * Input:
 * BASE (основание системы исчисления (2 .. 36))
 * Строка A
 * Строка B
 * 
 * Output:
 * A*B
 */
class Rextester {
    private static final String alpha = "0123456789abcdefghijklmnopqrstuvwxyz";

public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int BASE = in.nextInt();
        String A = in.next();
        String B = in.next();
        int lenA = A.length();
        int lenB = B.length();
        int lenC = lenA + lenB + 1; // maximum length for result
        char[] C = new char[lenC];

// initializing the char array
        for (int i = 0; i < lenC; i++)
            C[i] = '0';

// going from right to left (from lesser digit to greatest)
        for (int i = lenA - 1; i >= 0; i--) {
            for (int j = lenB - 1; j >= 0; j--) {
                char charA = A.charAt(i); // character of A at i-th position
                char charB = B.charAt(j); // character of B at i-th position
                char charC = C[i + j + 2];
                char charHigher = C[i + j + 1];
                char charHighest = C[i + j];
                int digA = charToInt(charA); // A[i] in base-10
                int digB = charToInt(charB); // B[i] in base-10
                int digC = charToInt(charC);
                int digHigher = charToInt(charHigher);
                int digHighest = charToInt(charHighest);
                // sum of multiplication of both digits, existing value in current place
                // and overflow / super-overflow in the right places
                int mult = digA * digB + digC + digHigher * BASE + digHighest * BASE * BASE;
                int higher = mult / BASE; // setting overflow value
                int highest = higher / BASE; // setting super-overflow value
                mult = mult % BASE; // converting from base-10 to base-36
                // up to three digits can be produced by multiplication
                // so we write result to three places
                C[i + j + 2] = alpha.charAt(mult);
                C[i + j + 1] = alpha.charAt(higher % BASE);
                C[i + j] = alpha.charAt(highest);
            }
        }

// do not print first digits if they are 0
        int z = 0;
        while (C[z] == '0')
            z++;

for (int i = z; i < lenC; i++)
            System.out.print(C[i]);

// WHY????
        //        999
        //        999
        //        898001
        //AND NOT 998001
    }

private static int charToInt(char digit) {
        return alpha.indexOf(digit);
    }

}


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