Selenium号

Deserium Number：如果数字相对于从1到数字的幂的幂的数字总和等于该数字本身，则称该数字为Deserium Number。

例子 ：

Input : 135
Output : Yes
1^1 + 3^2 + 5^3 = 135

Input : 9
Output : Yes
9^1 = 9

Input  : 125
Output : No
1^1+2^2+5^3 != 125

这个想法很简单。
1)计算给定数字中的数字。
2)从最右边的数字遍历到最左边，并计算幂和。
3)如果幂的和等于给定的数字，则返回true。

C++

// C++ program to check whether a number
// is Deserium number or not
#include 
#include 
  
using namespace std;
  
// Returns count of digits in n.
int countDigits(int n)
{
    int c = 0;
      
    do {
        c++;
        n = n / 10;
    } while (n != 0);
      
    return c;
}
  
// Returns true if x is Diserium
bool isDeserium(int x)
{
    int temp = x;
    int p = countDigits(x);
  
    // Compute powers of digits
    // from right to left.
    int sum = 0;
    while (x != 0) {
        int digit = x % 10;
        sum += pow(digit, p);
        p--;
        x = x / 10;
    }
  
    // If sum of powers is same as
    // given number.
    return (sum == temp);
}
  
// Driver code
int main()
{
    int x = 135;
      
    if (isDeserium(x))
        cout << "Yes";
    else
        cout << "No";
          
    return 0;
}
  
// This code is contributed by vt_m.

Java

// Java program to check whether a number
// is Deserium number or not
import java.util.Scanner;
class Deserium {
  
    // Returns count of digits in n.
    static int countDigits(int n)
    {
        int c = 0;
        do {
            c++;
            n = n / 10;
        } while (n != 0);
        return c;
    }
  
    // Returns true if x is Diserium
    static boolean isDeserium(int x)
    {
        int temp = x;
        int p = countDigits(x);
  
        // Compute powers of digits 
        // from right to left.
        int sum = 0;
        while (x != 0) {
            int digit = x % 10;
            sum += Math.pow(digit, p);
            p--;
            x = x / 10;
        }
  
        // If sum of powers is same as
        // given number.
        return (sum == temp);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int x = 135;
        if (isDeserium(x))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# Python3 program to check whether 
# a number is Deserium number or not
  
# Returns count of digits in n.
def countDigits(n):
  
    c = 0
    while (n != 0):
        c += 1
        n = int( n / 10)
      
    return c
  
# Returns true if x is Diserium
def isDeserium(x):
  
    temp = x
    p = countDigits(x)
  
    # Compute powers of digits
    # from right to left.
    sum = 0
    while (x != 0): 
        digit = int(x % 10)
        sum += pow(digit, p)
        p -= 1
        x = int(x / 10)
      
  
    # If sum of powers is same as
    # given number.
    return (sum == temp)
  
# Driver code
x = 135
if (isDeserium(x)):
    print("Yes")
else:
    print("No")
          
# This code is contributed by Smitha Dinesh Semwal.

C#

// C# program to check whether a number
// is Deserium number or not
using System;
  
class Deserium {
  
    // Returns count of digits in n.
    static int countDigits(int n)
    {
        int c = 0;
        do {
            c++;
            n = n / 10;
        } while (n != 0);
        return c;
    }
  
    // Returns true if x is Diserium
    static bool isDeserium(int x)
    {
        int temp = x;
        int p = countDigits(x);
  
        // Compute powers of digits
        // from right to left.
        int sum = 0;
        while (x != 0) {
            int digit = x % 10;
            sum += (int)Math.Pow(digit, p);
            p--;
            x = x / 10;
        }
  
        // If sum of powers is same as
        // given number.
        return (sum == temp);
    }
  
    // Driver code
    public static void Main()
    {
        int x = 135;
          
        if (isDeserium(x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by vt_m.

输出 ：

Yes