下面有没有一种更简洁的方法来完成同样的任务？向下舍入到最接近的1或5mod6

import math

def round_down_to_1_or_5_mod_6(number):
    rounded_number = math.floor(number/6) * 6 + 1
    return rounded_number

# 示例使用
number = 10
rounded_number = round_down_to_1_or_5_mod_6(number)
print(rounded_number)

import java.lang.Math;

public class Main {
    public static int roundDownTo1or5Mod6(int number) {
        int roundedNumber = (int) (Math.floor(number/6) * 6 + 1);
        return roundedNumber;
    }

    // 示例使用
    public static void main(String[] args) {
        int number = 10;
        int roundedNumber = roundDownTo1or5Mod6(number);
        System.out.println(roundedNumber);
    }
}

#include <iostream>
#include <cmath>

int roundDownTo1or5Mod6(int number) {
    int roundedNumber = std::floor(number/6) * 6 + 1;
    return roundedNumber;
}

// 示例使用
int main() {
    int number = 10;
    int roundedNumber = roundDownTo1or5Mod6(number);
    std::cout << roundedNumber << std::endl;
    return 0;
}