CodeWars 일백 열 다섯 번째 문제

A Rule of Divisibility by 13

public static long thirt(long n) {
		
    long remainder = n;

    do {
        n = remainder;
        remainder = makeRemainder(makeSequence(n));

    } while(n != remainder);

    return remainder;
}

private static int makeRemainder(int[] sequence) {

    int remainder = 0;
    int[] cycle = {1, 10, 9, 12, 3, 4};

    for(int i = 0, j = 0; i < sequence.length; i++, j++) {

        if(j == cycle.length) {
            j = 0;
        }

        remainder += sequence[i] * cycle[j];
    }

    return remainder;
}

private static int[] makeSequence(long n) {

    int[] sequence = new int[(int)Math.log10(n) + 1];

    for(int i = 0; i < sequence.length; i++) {
        sequence[i] = (int) (n % 10);
        n = n / 10;
    }

    return sequence;
}

• 문제는 굉장히 쉬웠다.
• 하지만 위의 코드를 더 간단히… 재귀로 풀면 쉽게 풀 수 있다.

private static final int[] seq = new int[]{1,10,9,12,3,4};
    
public static long thirt(long n) {

    long v = 0, m = n;
    int p = 0;
    
    while (m>0) {
        v += (m%10)*seq[p++%6];
        m /= 10;
    }
    
    return v == n ? v : thirt(v);

}