检查N是否可以表示为3个不同数字的乘积

给定数字N。打印三个乘积等于N的不同数字(> = 1) 。如果无法找到三个数字，则打印-1。
例子：

Input: 64 
Output: 2 4 8 
Explanation: 
(2*4*8 = 64)
Input: 24 
Output: 2 3 4 
Explanation: 
(2*3*4 = 24)
Input: 12 
Output: -1 
Explanation: 
No such triplet exists 
 

方法：

1. 使用本文讨论的方法制作一个存储给定数字的所有除数的数组
2. 设三个数字为a，b，c初始化为1
3. 遍历除数数组并检查以下条件：
   • a =值在除数数组的第一个索引处的值。
   • b的值=除数数组第二和第三索引的值的乘积。如果除数数组只有一个或两个元素，则不存在这样的三元组
   • 找到a和b后，c的值=除数数组中所有其余元素的乘积。
4. 检查最终条件，使a，b，c的值必须不同且不等于1。

下面是实现代码：

CPP

// C++ program to find the
// three numbers
#include "bits/stdc++.h"
using namespace std;
 
// function to find 3 distinct number
// with given product
void getnumbers(int n)
{
    // Declare a vector to store
    // divisors
    vector divisor;
 
    // store all divisors of number
    // in array
    for (int i = 2; i * i <= n; i++) {
 
        // store all the occurence of
        // divisors
        while (n % i == 0) {
            divisor.push_back(i);
            n /= i;
        }
    }
 
    // check if n is not equals to -1
    // then n is also a prime factor
    if (n != 1) {
        divisor.push_back(n);
    }
 
    // Initialize the variables with 1
    int a, b, c, size;
    a = b = c = 1;
    size = divisor.size();
 
    for (int i = 0; i < size; i++) {
 
        // check for first number a
        if (a == 1) {
            a = a * divisor[i];
        }
 
        // check for second number b
        else if (b == 1 || b == a) {
            b = b * divisor[i];
        }
 
        // check for third number c
        else {
            c = c * divisor[i];
        }
    }
 
    // check for all unwanted condition
    if (a == 1 || b == 1 || c == 1
        || a == b || b == c || a == c) {
        cout << "-1" << endl;
    }
    else {
        cout << a << ' ' << b
             << ' ' << c << endl;
    }
}
 
// Driver function
int main()
{
    int n = 64;
    getnumbers(n);
}

Java

// Java program to find the
// three numbers
import java.util.*;
 
class GFG{
  
// function to find 3 distinct number
// with given product
static void getnumbers(int n)
{
    // Declare a vector to store
    // divisors
    Vector divisor = new Vector();
  
    // store all divisors of number
    // in array
    for (int i = 2; i * i <= n; i++) {
  
        // store all the occurence of
        // divisors
        while (n % i == 0) {
            divisor.add(i);
            n /= i;
        }
    }
  
    // check if n is not equals to -1
    // then n is also a prime factor
    if (n != 1) {
        divisor.add(n);
    }
  
    // Initialize the variables with 1
    int a, b, c, size;
    a = b = c = 1;
    size = divisor.size();
  
    for (int i = 0; i < size; i++) {
  
        // check for first number a
        if (a == 1) {
            a = a * divisor.get(i);
        }
  
        // check for second number b
        else if (b == 1 || b == a) {
            b = b * divisor.get(i);
        }
  
        // check for third number c
        else {
            c = c * divisor.get(i);
        }
    }
  
    // check for all unwanted condition
    if (a == 1 || b == 1 || c == 1
        || a == b || b == c || a == c) {
        System.out.print("-1" +"\n");
    }
    else {
        System.out.print(a +" "+ b
                +" "+ c +"\n");
    }
}
  
// Driver function
public static void main(String[] args)
{
    int n = 64;
    getnumbers(n);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program to find the
# three numbers
 
# function to find 3 distinct number
# with given product
def getnumbers(n):
      
     # Declare a vector to store
    # divisors
    divisor = []
 
    # store all divisors of number
    # in array
    for i in range(2, n + 1):
 
        # store all the occurence of
        # divisors
        while (n % i == 0):
            divisor.append(i)
            n //= i
 
    # check if n is not equals to -1
    # then n is also a prime factor
    if (n != 1):
        divisor.append(n)
 
    # Initialize the variables with 1
    a, b, c, size = 0, 0, 0, 0
    a = b = c = 1
    size = len(divisor)
 
    for i in range(size):
 
        # check for first number a
        if (a == 1):
            a = a * divisor[i]
 
        # check for second number b
        elif (b == 1 or b == a):
            b = b * divisor[i]
 
        # check for third number c
        else:
            c = c * divisor[i]
 
    # check for all unwanted condition
    if (a == 1 or b == 1 or c == 1
        or a == b or b == c or a == c):
        print("-1")
    else:
        print(a, b, c)
 
# Driver function
 
n = 64
getnumbers(n)
 
# This code is contributed by mohit kumar 29

C#

// C# program to find the
// three numbers
using System;
using System.Collections.Generic;
 
class GFG{
   
// function to find 3 distinct number
// with given product
static void getnumbers(int n)
{
    // Declare a vector to store
    // divisors
    List divisor = new List();
   
    // store all divisors of number
    // in array
    for (int i = 2; i * i <= n; i++) {
   
        // store all the occurence of
        // divisors
        while (n % i == 0) {
            divisor.Add(i);
            n /= i;
        }
    }
   
    // check if n is not equals to -1
    // then n is also a prime factor
    if (n != 1) {
        divisor.Add(n);
    }
   
    // Initialize the variables with 1
    int a, b, c, size;
    a = b = c = 1;
    size = divisor.Count;
   
    for (int i = 0; i < size; i++) {
   
        // check for first number a
        if (a == 1) {
            a = a * divisor[i];
        }
   
        // check for second number b
        else if (b == 1 || b == a) {
            b = b * divisor[i];
        }
   
        // check for third number c
        else {
            c = c * divisor[i];
        }
    }
   
    // check for all unwanted condition
    if (a == 1 || b == 1 || c == 1
        || a == b || b == c || a == c) {
        Console.Write("-1" +"\n");
    }
    else {
        Console.Write(a +" "+ b
                +" "+ c +"\n");
    }
}
   
// Driver function
public static void Main(String[] args)
{
    int n = 64;
    getnumbers(n);
}
}
 
// This code is contributed by Rajput-Ji

输出：

2 4 8

时间复杂度： O( (log N)* sqrt(N) )