生成N整除的所有可能的排列

给定一个数字字符串S ，任务是打印该字符串的所有可被N整除的排列。

例子：

Input: N = 5, S = “125” 
Output: 125 215
Explanation: 
All possible permutations are S are {125, 152, 215, 251, 521, 512}. 
Out of these 6 permutations, only 2 {125, 215} are divisible by N (= 5).

Input: N = 7, S = “4321” 
Output: 4312 4123 3241

方法：想法是生成所有可能的排列，并针对每个排列检查是否可以被N整除。对于发现可以被N整除的每个排列，请打印它们。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to Swap two
// characters
void swap_(char& a, char& b)
{
    char temp;
    temp = a;
    a = b;
    b = temp;
}
 
// Function to generate all permutations
// and print the ones that are
// divisible by the N
void permute(char* str, int l, int r, int n)
{
    int i;
 
    if (l == r) {
 
        // Convert string to integer
        int j = atoi(str);
 
        // Check for divisibility
        // and print it
        if (j % n == 0)
            cout << str << endl;
 
        return;
    }
 
    // Print all the permutations
    for (i = l; i < r; i++) {
 
        // Swap characters
        swap_(str[l], str[i]);
 
        // Permute remaining
        // characters
        permute(str, l + 1, r, n);
 
        // Revoke the swaps
        swap_(str[l], str[i]);
    }
}
 
// Driver Code
int main()
{
    char str[100] = "125";
    int n = 5;
    int len = strlen(str);
 
    if (len > 0)
        permute(str, 0, len, n);
 
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to Swap two
// characters
static void swap_(char []a, int l, int i)
{
    char temp;
    temp = a[l];
    a[l] = a[i];
    a[i] = temp;
}
 
// Function to generate all permutations
// and print the ones that are
// divisible by the N
static void permute(char[] str, int l,
                         int r, int n)
{
    int i;
 
    if (l == r)
    {
         
        // Convert String to integer
        int j = Integer.valueOf(String.valueOf(str));
 
        // Check for divisibility
        // and print it
        if (j % n == 0)
            System.out.print(String.valueOf(str) + "\n");
 
        return;
    }
 
    // Print all the permutations
    for(i = l; i < r; i++)
    {
         
        // Swap characters
        swap_(str, l, i);
 
        // Permute remaining
        // characters
        permute(str, l + 1, r, n);
 
        // Revoke the swaps
        swap_(str, l, i);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "125";
    int n = 5;
    int len = str.length();
 
    if (len > 0)
        permute(str.toCharArray(), 0, len, n);
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 Program to implement
# the above approach
 
# Function to generate all
# permutations and print
# the ones that are
# divisible by the N
def permute(st, l, r, n):
 
    if (l == r):
 
        # Convert string
        # to integer
        p = ''.join(st)
        j = int(p)
 
        # Check for divisibility
        # and print it
        if (j % n == 0):
            print (p)
 
        return
 
    # Print all the
    # permutations
    for i in range(l, r):
 
        # Swap characters
        st[l], st[i] = st[i], st[l]
 
        # Permute remaining
        # characters
        permute(st, l + 1, r, n)
 
        # Revoke the swaps
        st[l], st[i] = st[i] ,st[l]
 
# Driver Code
if __name__ == "__main__":
 
    st = "125"
    n = 5
    length = len(st)
 
    if (length > 0):
      p = list(st)
      permute(p, 0, length, n);
 
# This code is contributed by rutvik_56

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
  
// Function to Swap two
// characters
static void swap_(char []a, int l,
                            int i)
{
    char temp;
    temp = a[l];
    a[l] = a[i];
    a[i] = temp;
}
  
// Function to generate all permutations
// and print the ones that are
// divisible by the N
static void permute(char[] str, int l,
                         int r, int n)
{
    int i;
  
    if (l == r)
    {
         
        // Convert String to integer
        int j = Int32.Parse(new string(str));
  
        // Check for divisibility
        // and print it
        if (j % n == 0)
            Console.Write(new string(str) + "\n");
  
        return;
    }
  
    // Print all the permutations
    for(i = l; i < r; i++)
    {
         
        // Swap characters
        swap_(str, l, i);
  
        // Permute remaining
        // characters
        permute(str, l + 1, r, n);
  
        // Revoke the swaps
        swap_(str, l, i);
    }
}
  
// Driver Code
public static void Main(string[] args)
{
    string str = "125";
    int n = 5;
    int len = str.Length;
  
    if (len > 0)
        permute(str.ToCharArray(), 0, len, n);
}
}
 
// This code is contributed by rutvik_56

输出：

125
215

时间复杂度： O(N！)
辅助空间： O(N)