题解 | #小乐乐与进制转换#

#include <stdio.h>
int main() {
    int n;
    double sum = 0, t = 1;
    scanf("%d", &n);
    for (int i = 0; n > 0; i++) {
        sum += (n % 6) * t;
        t *= 10;
        n /= 6;
    }
    printf("%.0lf", sum);
    return 0;
}