检查一个数字是否可以表示为两个正完全立方体的乘积

给定一个正整数N ，任务是检查给定的数字N 是否可以表示为两个正完全立方体的乘积。如果可能，则打印“是” 。否则，打印“否” 。

例子：

Input: N = 216
Output: Yes
Explanation:
The given number N(= 216) can be represented as 8 * 27 =  2³ * 3³.
Therefore, print Yes.

Input: N = 10
Output: No

方法：解决给定问题的最简单方法是将从1到N 的立方根的所有数字的完美立方体存储在 Map 中，并检查N 是否可以表示为 Map 中存在的两个数字的乘积。

请按照以下步骤解决问题：

• 初始化一个有序映射，比如cubes ，它以排序的顺序存储完美的立方体。
• 遍历地图并检查是否存在任何乘积为N 的对，然后打印“是” 。否则，打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if N can
// be represented as the product
// of two perfect cubes or not
void productOfTwoPerfectCubes(int N)
{
    // Stores the perfect cubes
    map cubes;
 
    for (int i = 1;
         i * i * i <= N; i++)
        cubes[i * i * i] = i;
 
    // Traverse the Map
    for (auto itr = cubes.begin();
         itr != cubes.end();
         itr++) {
 
        // Stores the first number
        int firstNumber = itr->first;
 
        if (N % itr->first == 0) {
 
            // Stores the second number
            int secondNumber = N / itr->first;
 
            // Search the pair for the
            // first number to obtain
            // product N from the Map
            if (cubes.find(secondNumber)
                != cubes.end()) {
                cout << "Yes";
                return;
            }
        }
    }
 
    // If N cannot be represented
    // as the product of the two
    // positive perfect cubes
    cout << "No";
}
 
// Driver Code
int main()
{
    int N = 216;
    productOfTwoPerfectCubes(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
import java.util.*;
 
class GFG{
     
// Function to check if N can
// be represented as the product
// of two perfect cubes or not
static void productOfTwoPerfectCubes(int N)
{
     
    // Stores the perfect cubes
    Map cubes = new HashMap<>();
 
    for(int i = 1; i * i * i <= N; i++)
        cubes.put(i * i * i,i);
 
    // Traverse the Map
    for(Map.Entry itr: cubes.entrySet())
    {
         
        // Stores the first number
        int firstNumber = itr.getKey();
 
        if (N % itr.getKey() == 0)
        {
             
            // Stores the second number
            int secondNumber = N / itr.getKey();
 
            // Search the pair for the
            // first number to obtain
            // product N from the Map
            if (cubes.containsKey(secondNumber))
            {
                System.out.println("Yes");
                return;
            }
        }
    }
 
    // If N cannot be represented
    // as the product of the two
    // positive perfect cubes
    System.out.println("No");
}
 
// Driver code
public static void main(String[] args)
{
    int N = 216;
     
    productOfTwoPerfectCubes(N);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to check if N can
# be represented as the product
# of two perfect cubes or not
def productOfTwoPerfectCubes(N):
 
    # Stores the perfect cubes
    cubes = {}
 
    i = 1
     
    while i * i * i <= N:
        cubes[i * i * i] = i
        i += 1
 
    # Traverse the Map
    for itr in cubes:
 
        # Stores the first number
        firstNumber = itr
 
        if (N % itr == 0):
 
            # Stores the second number
            secondNumber = N // itr
 
            # Search the pair for the
            # first number to obtain
            # product N from the Map
            if (secondNumber in cubes):
                print("Yes")
                return
 
    # If N cannot be represented
    # as the product of the two
    # positive perfect cubes
    print("No")
 
# Driver Code
if __name__ == "__main__":
     
    N = 216
     
    productOfTwoPerfectCubes(N)
 
# This code is contributed by mohit ukasp

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
 // Function to check if N can
// be represented as the product
// of two perfect cubes or not
static void productOfTwoPerfectCubes(int N)
{
    // Stores the perfect cubes
    Dictionary cubes = new Dictionary();
 
    for (int i = 1; i * i * i <= N; i++){
       cubes.Add(i * i * i, i);
    }
 
    // Traverse the Map
    foreach(KeyValuePair kvp in cubes)
   {
 
        // Stores the first number
        int firstNumber = kvp.Key;
 
        if (N % kvp.Key == 0) {
 
            // Stores the second number
            int secondNumber = N / kvp.Key;
 
            // Search the pair for the
            // first number to obtain
            // product N from the Map
            if (cubes.ContainsKey(secondNumber)) {
                Console.Write("Yes");
                return;
            }
        }
    }
 
    // If N cannot be represented
    // as the product of the two
    // positive perfect cubes
    Console.Write("No");
}
 
// Driver Code
public static void Main()
{
    int N = 216;
    productOfTwoPerfectCubes(N);
 
}
}
 
// This code is contributed by ipg2016107..

Javascript

 
// Javascript program for the above approach
 
// Function to check if N can
// be represented as the product
// of two perfect cubes or not
function productOfTwoPerfectCubes(N)
{
     
    // Stores the perfect cubes
    let cubes = new Map();
  
    for(let i = 1; i * i * i <= N; i++)
        cubes.set(i * i * i, i);
  
    // Traverse the Map
    for(let [key, value] of cubes.entries())
    {
         
        // Stores the first number
        let firstNumber = key;
  
        if (N % key == 0)
        {
              
            // Stores the second number
            let secondNumber = N / key;
  
            // Search the pair for the
            // first number to obtain
            // product N from the Map
            if (cubes.has(secondNumber))
            {
                document.write("Yes<br>");
                return;
            }
        }
    }
  
    // If N cannot be represented
    // as the product of the two
    // positive perfect cubes
    document.write("No<br>");
}
 
// Driver code
let N = 216;
      
productOfTwoPerfectCubes(N);
 
// This code is contributed by avanitrachhadiya2155
 

输出：

Yes

时间复杂度： O(N ^1/3 * log(N))
辅助空间： O(N ^1/3 )

高效方法：上述方法也可以根据观察进行优化，即只有完美立方体可以表示为 2 个完美立方体的乘积。

Let the two numbers be x and y such that x³ * y³= N — (1)
Equation (1) can be written as:
=> (x*y)³ = N
Taking cube root both sides, 
=> x*y = (N)^1/3 — (2)
For equation (2) to be true, N should be a perfect cube.

因此，问题简化为检查N是否为完美立方体。如果发现是真的，打印“是” ，否则打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if the number N
// can be represented as the product
// of two perfect cubes or not
void productOfTwoPerfectCubes(int N)
{
    int cube_root;
    cube_root = round(cbrt(N));
 
    // If cube of cube_root is N
    if (cube_root * cube_root
            * cube_root
        == N) {
 
        cout << "Yes";
        return;
    }
 
    // Otherwise, print No
    else {
        cout << "No";
        return;
    }
}
 
// Driver Code
int main()
{
    int N = 216;
    productOfTwoPerfectCubes(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
 
class GFG{
     
// Function to check if the number N
// can be represented as the product
// of two perfect cubes or not
public static void productOfTwoPerfectCubes(double N)
{
    double cube_root;
    cube_root = Math.round(Math.cbrt(N));
 
    // If cube of cube_root is N
    if (cube_root * cube_root * cube_root == N)
    {
        System.out.println("Yes");
        return;
    }
 
    // Otherwise, print No
    else
    {
        System.out.println("No");
        return;
    }
}
 
// Driver Code
public static void main(String args[])
{
    double N = 216;
     
    productOfTwoPerfectCubes(N);
}
}
 
// This code is contributed by SoumikMondal

蟒蛇3

# Python3 program for the above approach
 
# Function to check if the number N
# can be represented as the product
# of two perfect cubes or not
def productOfTwoPerfectCubes(N):
     
    cube_root = round((N) ** (1 / 3))
    print(cube_root)
 
    # If cube of cube_root is N
    if (cube_root * cube_root * cube_root == N):
        print("Yes")
        return
     
    # Otherwise, prNo
    else:
        print("No")
        return
 
# Driver Code
if __name__ == '__main__':
     
    N = 216
     
    productOfTwoPerfectCubes(N)
 
# This code is contributed by mohit kumar 29

C＃

// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG{
     
// Function to check if the number N
// can be represented as the product
// of two perfect cubes or not
public static void productOfTwoPerfectCubes(double N)
{
    double cube_root;
    cube_root = Math.Round(Math.Cbrt(N));
 
    // If cube of cube_root is N
    if (cube_root * cube_root * cube_root == N)
    {
        Console.Write("Yes");
        return;
    }
 
    // Otherwise, print No
    else
    {
        Console.Write("No");
        return;
    }
}
 
// Driver Code
static public void Main()
{
    double N = 216;
     
    productOfTwoPerfectCubes(N);
}
}
 
// This code is contributed by mohit kumar 29.

Javascript

 
// Javascript program for the above approach
 
// Function to check if the number N
// can be represented as the product
// of two perfect cubes or not
function productOfTwoPerfectCubes(N)
{
    var cube_root;
    cube_root = Math.round(Math.cbrt(N));
 
    // If cube of cube_root is N
    if (cube_root * cube_root * cube_root == N)
    {
        document.write("Yes");
        return;
    }
 
    // Otherwise, print No
    else
    {
        document.write("No");
        return;
    }
}
 
// Driver code
var N = 216;
productOfTwoPerfectCubes(N);
 
// This code is contributed by Ankita saini
 

输出：

Yes

时间复杂度： O(N ^1/3 )
辅助空间： O(1)