通过将除数从N到M连接来创建图，并找到最短路径

给定两个自然数N和M ，请使用这两个自然数创建关系图，该关系是一个数字与其自身以外的最大因子有关。任务是在创建图形后找到这两个数字之间的最短路径。

例子：

Input: N = 6, M = 18
Output: 6 <–> 3 <–> 9 <–> 18
Explanation:
For N = 6, the connection of graph is:
6 — 3 — 1
For N = 18, the connection of graph is:
18 — 9 — 3 — 1
Combining the above two graphs, the shortest path is given by:
6 — 3 — 9 — 18

Input: N = 4, M = 8
Output: 4 <–> 8

方法：想法是找到每个数字以外的最大因素，并通过连接这些因素来创建图，然后找到它们之间的最短路径。步骤如下：

1. 找出M的最大公因数并将其存储并将其设置为M。
2. 现在，直到M不等于1为止，请继续重复上述步骤，并将生成的因子存储在数组mfactor []中。
3. 通过将N作为数字来重复步骤1和步骤2 ，并将生成的因子存储在数组nfactor []中。
4. 现在，遍历数组mfactor []和mfactor []并显示最短路径。

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
  
// Function to check the number is
// prime or not
int isprm(int n)
{
    // Base Cases
    if (n <= 1)
        return 0;
    if (n <= 3)
        return 1;
    if (n % 2 == 0 || n % 3 == 0)
        return 0;
  
    // Iterate till [5, sqrt(N)] to
    // detect primarility of numbers
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return 0;
    return 1;
}
  
// Function to print the shortest path
void shortestpath(int m, int n)
{
    // Use vector to store the factor
    // of m and n
    vector mfactor, nfactor;
  
    // Use map to check if largest common
    // factor previously present or not
    map fre;
  
    // First store m
    mfactor.push_back(m);
    fre[m] = 1;
  
    while (m != 1) {
  
        // Check whether m is prime or not
        if (isprm(m)) {
            mfactor.push_back(1);
            fre[1] = 1;
            m = 1;
        }
  
        // Largest common factor of m
        else {
            for (int i = 2;
                 i <= sqrt(m); i++) {
  
                // If m is divisible by i
                if (m % i == 0) {
  
                    // Store the largest
                    // common factor
                    mfactor.push_back(m / i);
                    fre[m / i] = 1;
                    m = (m / i);
                    break;
                }
            }
        }
    }
  
    // For number n
    nfactor.push_back(n);
  
    while (fre[n] != 1) {
  
        // Check whether n is prime
        if (isprm(n)) {
            nfactor.push_back(1);
            n = 1;
        }
  
        // Largest common factor of n
        else {
            for (int i = 2;
                 i <= sqrt(n); i++) {
                if (n % i == 0) {
  
                    // Store the largest
                    // common factor
                    nfactor.push_back(n / i);
                    n = (n / i);
                    break;
                }
            }
        }
    }
  
    // Print the path
    // Print factors from m
    for (int i = 0;
         i < mfactor.size(); i++) {
  
        // To avoid duplicate printing
        // of same element
        if (mfactor[i] == n)
            break;
  
        cout << mfactor[i]
             << " <--> ";
    }
  
    // Print the factors from n
    for (int i = nfactor.size() - 1;
         i >= 0; i--) {
        if (i == 0)
            cout << nfactor[i];
        else
            cout << nfactor[i]
                 << " <--> ";
    }
}
  
// Driver Code
int main()
{
    // Given N and M
    int m = 18, n = 19;
  
    // Function Call
    shortestpath(m, n);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to check the number is
// prime or not
static int isprm(int n)
{
      
    // Base Cases
    if (n <= 1)
        return 0;
    if (n <= 3)
        return 1;
    if (n % 2 == 0 || n % 3 == 0)
        return 0;
  
    // Iterate till [5, Math.sqrt(N)] to
    // detect primarility of numbers
    for(int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return 0;
          
    return 1;
}
  
// Function to print the shortest path
static void shortestpath(int m, int n)
{
      
    // Use vector to store the factor
    // of m and n
    Vector mfactor = new Vector<>();
    Vector nfactor = new Vector<>();
  
    // Use map to check if largest common
    // factor previously present or not
    HashMap fre = new HashMap<>();
  
    // First store m
    mfactor.add(m);
    fre.put(m, 1);
  
    while (m != 1) 
    {
  
        // Check whether m is prime or not
        if (isprm(m) != 0)
        {
            mfactor.add(1);
            fre.put(1, 1);
            m = 1;
        }
  
        // Largest common factor of m
        else
        {
            for(int i = 2; 
                    i <= Math.sqrt(m); i++)
            {
  
                // If m is divisible by i
                if (m % i == 0) 
                {
                      
                    // Store the largest
                    // common factor
                    mfactor.add(m / i);
                    fre.put(m / i, 1);
                    m = (m / i);
                    break;
                }
            }
        }
    }
  
    // For number n
    nfactor.add(n);
  
    while (fre.containsKey(n) && fre.get(n) != 1)
    {
          
        // Check whether n is prime
        if (isprm(n) != 0)
        {
            nfactor.add(1);
            n = 1;
        }
  
        // Largest common factor of n
        else
        {
            for(int i = 2; 
                    i <= Math.sqrt(n); i++) 
            {
                if (n % i == 0) 
                {
  
                    // Store the largest
                    // common factor
                    nfactor.add(n / i);
                    n = (n / i);
                    break;
                }
            }
        }
    }
  
    // Print the path
    // Print factors from m
    for(int i = 0; i < mfactor.size(); i++)
    {
          
        // To astatic void duplicate printing
        // of same element
        if (mfactor.get(i) == n)
            break;
  
        System.out.print(mfactor.get(i) +
                         " <--> ");
    }
  
    // Print the factors from n
    for(int i = nfactor.size() - 1;
            i >= 0; i--)
    {
        if (i == 0)
            System.out.print(nfactor.get(i));
        else
            System.out.print(nfactor.get(i) +
                             " <--> ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given N and M
    int m = 18, n = 19;
  
    // Function call
    shortestpath(m, n);
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
import math
  
# Function to check the number is
# prime or not
def isprm(n):
  
    # Base Cases
    if (n <= 1):
        return 0
    if (n <= 3):
        return 1
    if (n % 2 == 0 or n % 3 == 0):
        return 0
  
    # Iterate till [5, sqrt(N)] to
    # detect primarility of numbers
    i = 5
    while i * i <= n:
        if (n % i == 0 or n % (i + 2) == 0):
            return 0
              
        i += 6
          
    return 1
  
# Function to print the shortest path
def shortestpath(m, n):
  
    # Use vector to store the factor
    # of m and n
    mfactor = []
    nfactor = []
  
    # Use map to check if largest common
    # factor previously present or not
    fre = dict.fromkeys(range(n + 1), 0)
  
    # First store m
    mfactor.append(m)
    fre[m] = 1
  
    while (m != 1):
  
        # Check whether m is prime or not
        if (isprm(m)):
            mfactor.append(1)
            fre[1] = 1
            m = 1
  
        # Largest common factor of m
        else:
            sqt = (int)(math.sqrt(m))
            for i in range(2, sqt + 1):
  
                # If m is divisible by i
                if (m % i == 0):
  
                    # Store the largest
                    # common factor
                    mfactor.append(m // i)
                    fre[m // i] = 1
                    m = (m // i)
                    break
  
    # For number n
    nfactor.append(n)
  
    while (fre[n] != 1):
  
        # Check whether n is prime
        if (isprm(n)):
            nfactor.append(1)
            n = 1
          
        # Largest common factor of n
        else:
            sqt = (int)(math.sqrt(n))
            for i in range(2, sqt + 1):
                if (n % i == 0):
  
                    # Store the largest
                    # common factor
                    nfactor.append(n // i)
                    n = (n // i)
                    break
  
    # Print the path
    # Print factors from m
    for i in range(len(mfactor)):
  
        # To avoid duplicate printing
        # of same element
        if (mfactor[i] == n):
            break
  
        print(mfactor[i], end = " <--> ")
  
    # Print the factors from n
    for i in range(len(nfactor) - 1, -1, -1):
        if (i == 0):
            print (nfactor[i], end = "")
        else:
            print(nfactor[i], end = " <--> ")
                  
# Driver Code
if __name__ == "__main__":
      
    # Given N and M
    m = 18
    n = 19
  
    # Function call
    shortestpath(m, n)
  
# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to check the number is
// prime or not
static int isprm(int n)
{
      
    // Base Cases
    if (n <= 1)
        return 0;
    if (n <= 3)
        return 1;
    if (n % 2 == 0 || n % 3 == 0)
        return 0;
  
    // Iterate till [5, Math.Sqrt(N)] to
    // detect primarility of numbers
    for(int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return 0;
          
    return 1;
}
  
// Function to print the shortest path
static void shortestpath(int m, int n)
{
      
    // Use vector to store the factor
    // of m and n 
    List mfactor = new List();
    List nfactor = new List();
  
    // Use map to check if largest common
    // factor previously present or not
    Dictionary fre = new Dictionary();
  
    // First store m
    mfactor.Add(m);
    fre.Add(m, 1);
  
    while (m != 1) 
    {
  
        // Check whether m is prime or not
        if (isprm(m) != 0)
        {
            mfactor.Add(1);
            if(!fre.ContainsKey(1))
                fre.Add(1, 1);
                  
            m = 1;
        }
  
        // Largest common factor of m
        else
        {
            for(int i = 2; 
                    i <= Math.Sqrt(m); i++)
            {
  
                // If m is divisible by i
                if (m % i == 0) 
                {
                      
                    // Store the largest
                    // common factor
                    mfactor.Add(m / i);
                    if(!fre.ContainsKey(m/i))
                        fre.Add(m / i, 1);
                          
                    m = (m / i);
                    break;
                }
            }
        }
    }
  
    // For number n
    nfactor.Add(n);
  
    while (fre.ContainsKey(n) && fre[n] != 1)
    {
          
        // Check whether n is prime
        if (isprm(n) != 0)
        {
            nfactor.Add(1);
            n = 1;
        }
  
        // Largest common factor of n
        else
        {
            for(int i = 2; 
                    i <= Math.Sqrt(n); i++) 
            {
                if (n % i == 0) 
                {
  
                    // Store the largest
                    // common factor
                    nfactor.Add(n / i);
                    n = (n / i);
                    break;
                }
            }
        }
    }
  
    // Print the path
    // Print factors from m
    for(int i = 0; i < mfactor.Count; i++)
    {
          
        // To astatic void duplicate printing
        // of same element
        if (mfactor[i] == n)
            break;
  
        Console.Write(mfactor[i] +
                        " <--> ");
    }
  
    // Print the factors from n
    for(int i = nfactor.Count - 1;
            i >= 0; i--)
    {
        if (i == 0)
            Console.Write(nfactor[i]);
        else
            Console.Write(nfactor[i] +
                            " <--> ");
    }
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given N and M
    int m = 18, n = 19;
  
    // Function call
    shortestpath(m, n);
}
}
  
// This code is contributed by 29AjayKumar

输出：

18 <--> 9 <--> 3 <--> 1 <--> 19

时间复杂度： O(log(max(M，N))
辅助空间： O(N)