给定两个自然数N和M ，使用这两个自然数使用关系创建一个图，即一个数字连接到其除自身之外的最大因子。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除自身之外的最大因数，并通过连接这些因数创建图形，然后找到它们之间的最短路径。以下是步骤：
- 找到M的最大公因数并存储并设置为M 。
- 现在，直到M不等于 1 继续重复上述步骤并将生成的因子存储在数组mfactor[] 中。
- 重复步骤 1和步骤 2 ，将N作为数字并将生成的因子存储在数组nfactor[] 中。
- 现在，遍历数组mfactor[]和mfactor[]并打印最短路径。
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 chitranayalC#
// 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 Javascript
// Javascript program for the above approach
// Function to check the number is
// prime or not
function isprm(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 (let 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
function shortestpath(m, n)
{
// Use vector to store the factor
// of m and n
let mfactor = new Array()
let nfactor = new Array()
// Use map to check if largest common
// factor previously present or not
let fre = new Map();
// First store m
mfactor.push(m);
fre[m] = 1;
while (m != 1) {
// Check whether m is prime or not
if (isprm(m)) {
mfactor.push(1);
fre[1] = 1;
m = 1;
}
// Largest common factor of m
else {
for (let i = 2;
i <= Math.sqrt(m); i++) {
// If m is divisible by i
if (m % i == 0) {
// Store the largest
// common factor
mfactor.push(m / i);
fre[m / i] = 1;
m = (m / i);
break;
}
}
}
}
// For number n
nfactor.push(n);
while (fre[n] != 1) {
// Check whether n is prime
if (isprm(n)) {
nfactor.push(1);
n = 1;
}
// Largest common factor of n
else {
for (let i = 2;
i <= Math.sqrt(n); i++) {
if (n % i == 0) {
// Store the largest
// common factor
nfactor.push(n / i);
n = (n / i);
break;
}
}
}
}
// Print the path
// Print factors from m
for (let i = 0;
i < mfactor.length; i++) {
// To avoid duplicate printing
// of same element
if (mfactor[i] == n)
break;
document.write(mfactor[i] + " <--> ");
}
// Print the factors from n
for (let i = nfactor.length - 1;
i >= 0; i--) {
if (i == 0)
document.write(nfactor[i]);
else
document.write(nfactor[i] + " <--> ");
}
}
// Driver Code
// Given N and M
let m = 18, n = 19;
// Function Call
shortestpath(m, n);
// This code is contributed by _saurabh_jaiswal输出：
18 <--> 9 <--> 3 <--> 1 <--> 19时间复杂度： O(log (max(M, N))
辅助空间： O(N)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live