查找可以被3整除但不能被6整除的n的排列

给定一个整数$n$ 。任务是找到另一个n的整数，该整数可以被3整除，但不能被6整除。假设n被6整除，如果不可能，则打印-1。

例子：

Input: n = 336
Output: 363

Input: n = 48
Output: -1

对于要被6整除的数，它必须被3以及2整除，这意味着每个被3整除的偶数整数都可以被6整除。因此，可以被3整除但不能被6整除的整数是可以被3整除的奇数整数。
因此，如果整数n包含任何奇数整数，则存在可以被3整除但不能被6整除的置换，否则不存在这种置换。

算法：

1. 令LEN为整数长度(即ceil(log10(n)))。
2. 遍历LEN并检查n是偶数还是奇数。
   • 如果n为奇数，则返回n
   • 否则，将n旋转一次。并继续。
3. 如果LEN超过返回-1

下面是上述方法的实现：

C++

// C++ program to find permutation of n
// which is divisible by 3 but not
// divisible by 6
 
#include 
using namespace std;
 
// Function to find the permutation
int findPermutation(int n)
{
    // length of integer
    int len = ceil(log10(n));
 
    for (int i = 0; i < len; i++) {
        // if integer is even
        if (n % 2 != 0) {
            // return odd integer
            return n;
        }
        else {
            // rotate integer
            n = (n / 10) + (n % 10) * pow(10, len - i - 1);
            continue;
        }
    }
 
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
int main()
{
    int n = 132;
 
    cout << findPermutation(n);
 
    return 0;
}

Java

// Java program to find permutation
// of n which is divisible by 3
// but not divisible by 6
import java.lang.*;
import java.util.*;
 
class GFG
{
// Function to find the permutation
static int findPermutation(int n)
{
    // length of integer
    int len = (int)Math.ceil(Math.log10(n));
 
    for (int i = 0; i < len; i++)
    {
        // if integer is even
        if (n % 2 != 0)
        {
            // return odd integer
            return n;
        }
        else
        {
            // rotate integer
            n = (n / 10) + (n % 10) *
                (int)Math.pow(10, len - i - 1);
            continue;
        }
    }
 
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 132;
 
    System.out.println(findPermutation(n));
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Python3

# Python3 program to find permutation
# of n which is divisible by 3 but
# not divisible by 6
from math import log10, ceil, pow
 
# Function to find the permutation
def findPermutation(n):
     
    # length of integer
    len = ceil(log10(n))
 
    for i in range(0, len, 1):
         
        # if integer is even
        if n % 2 != 0:
             
            # return odd integer
            return n
        else:
             
            # rotate integer
            n = ((n / 10) + (n % 10) *
                  pow(10, len - i - 1))
            continue
         
    # return -1 in case no required
    # permutation exists
    return -1
 
# Driver Code
if __name__ == '__main__':
    n = 132
 
    print(int(findPermutation(n)))
 
# This code is contributed
# by Surendra_Gangwar

C#

// C# program to find permutation
// of n which is divisible by 3
// but not divisible by 6
using System;
 
class GFG
{
// Function to find the permutation
static int findPermutation(int n)
{
    // length of integer
    int len = (int)Math.Ceiling(Math.Log10(n));
 
    for (int i = 0; i < len; i++)
    {
        // if integer is even
        if (n % 2 != 0)
        {
            // return odd integer
            return n;
        }
        else
        {
            // rotate integer
            n = (n / 10) + (n % 10) *
                (int)Math.Pow(10, len - i - 1);
            continue;
        }
    }
 
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
public static void Main()
{
    int n = 132;
 
    Console.WriteLine(findPermutation(n));
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

输出：

213