杜德尼数字

Question

给定整数n ，任务是检查n是否为 Dudeney 数字。 Dudeney 数字是一个正整数，是一个完美的立方体，因此其十进制数字的总和等于该数字的立方体根。

例子：

Input: N = 19683
Output: Yes
19683 = 27³ and 1 + 9 + 6 + 8 + 3 = 27

Input: N = 75742
Output: No

方法：

• 检查 n 是否是一个完美的立方体，如果不是，则它不能是 Dudeney 数。
• 如果 n 是一个完美的立方体，则计算其数字的总和。如果其数字的总和等于其立方根，则为 Dudeney 数字，否则为非。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function that returns true if
// n is a Dudeney number
bool isDudeney(int n)
{
    int cube_rt = int(round((pow(n, 1.0 / 3.0))));
  
    // If n is not a perfect cube
    if (cube_rt * cube_rt * cube_rt != n)
        return false;
  
    int dig_sum = 0;
    int temp = n;
    while (temp > 0) {
  
        // Last digit
        int rem = temp % 10;
  
        // Update the digit sum
        dig_sum += rem;
  
        // Remove the last digit
        temp /= 10;
    }
  
    // If cube root of n is not equal to
    // the sum of its digits
    if (cube_rt != dig_sum)
        return false;
  
    return true;
}
  
// Driver code
int main()
{
    int n = 17576;
    if (isDudeney(n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

---

Java

// Java implementation of the approach
import java.lang.Math;
  
class GFG
{
      
// Function that returns true if
// n is a Dudeney number
static boolean isDudeney(int n)
{
    int cube_rt = (int)(Math.round((Math.pow(n, 1.0 / 3.0))));
  
    // If n is not a perfect cube
    if (cube_rt * cube_rt * cube_rt != n)
        return false;
  
    int dig_sum = 0;
    int temp = n;
    while (temp > 0)
    {
  
        // Last digit
        int rem = temp % 10;
  
        // Update the digit sum
        dig_sum += rem;
  
        // Remove the last digit
        temp /= 10;
    }
  
    // If cube root of n is not equal to
    // the sum of its digits
    if (cube_rt != dig_sum)
        return false;
  
    return true;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 17576;
    if (isDudeney(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by Code_Mech.

---

Python3

# Python implementation of the approach
  
# Function that returns true if 
# n is a Dudeney number
def isDudeney(n):
    cube_rt = int(round((pow(n, 1.0 / 3.0))))
  
    # If n is not a perfect cube
    if cube_rt * cube_rt * cube_rt != n:
        return False
  
    dig_sum = 0
    temp = n
    while temp>0:
          
        # Last digit
        rem = temp % 10
          
        # Update the digit sum
        dig_sum += rem
          
        # Remove the last digit
        temp//= 10
  
    # If cube root of n is not equal to 
    # the sum of its digits
    if cube_rt != dig_sum:
        return False
  
    return True
  
# Driver code
if __name__ == '__main__':
      
    n = 17576
    if isDudeney(n):
        print("Yes")
    else:
        print("No")

---

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function that returns true if
// n is a Dudeney number
static bool isDudeney(int n)
{
    int cube_rt = (int)(Math.Round((Math.Pow(n, 1.0 / 3.0))));
  
    // If n is not a perfect cube
    if (cube_rt * cube_rt * cube_rt != n)
        return false;
  
    int dig_sum = 0;
    int temp = n;
    while (temp > 0)
    {
  
        // Last digit
        int rem = temp % 10;
  
        // Update the digit sum
        dig_sum += rem;
  
        // Remove the last digit
        temp /= 10;
    }
  
    // If cube root of n is not equal to
    // the sum of its digits
    if (cube_rt != dig_sum)
        return false;
  
    return true;
}
  
// Driver code
public static void Main()
{
    int n = 17576;
    if (isDudeney(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed 
// by Akanksha Rai