Perfect Digital Invariants number

Last Updated : 25 May, 2021

A positive integer is called a Perfect Digital Invariant Number if the sum of some fixed power of their digits is equal to the number itself.
For any number, abcd… = pow(a, n) + pow(b, n) + pow(c, n) + pow(d, n) + …. ,
where n can be any integer greater than 0.

Check if N is Perfect Digital Invariant Number or not

Given a number N, the task is to check if the given number N is Perfect Digital Invariant Number or not. If N is a Perfect Digital Invariant Number then print “Yes” else print “No”.
Example:

Input: N = 153
Output: Yes
Explanation:
153 is a Perfect Digital Invariants number as for n = 3 we have
1³ + 5³ + 3³ = 153
Input: 4150
Output: Yes
Explanation:
4150 is a Perfect Digital Invariants number as for n = 5 we have
4⁵ + 1⁵ + 5⁵ + 0⁵ = 4150

Approach: For every digit in number N, calculate the sum of its digit power starting from a fixed number 1 until the sum of digit’s power of N exceeds N. If N is a Perfect Digital Invariant Number then print “Yes” else print “No”.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate x raised
// to the power y
int power(int x, unsigned int y)
{
    if (y == 0) {
        return 1;
    }
 
    if (y % 2 == 0) {
        return (power(x, y / 2)
                * power(x, y / 2));
    }
    return (x
            * power(x, y / 2)
            * power(x, y / 2));
}
 
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
bool isPerfectDigitalInvariant(int x)
{
    for (int fixed_power = 1;; fixed_power++) {
 
        int temp = x, sum = 0;
 
        // For each digit in temp
        while (temp) {
 
            int r = temp % 10;
            sum += power(r, fixed_power);
            temp = temp / 10;
        }
 
        // If satisfies Perfect Digital
        // Invariant condition
        if (sum == x) {
            return true;
        }
        // If sum exceeds n, then not possible
        if (sum > x) {
            return false;
        }
    }
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 4150;
 
    // Function Call
    if (isPerfectDigitalInvariant(N))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to calculate x raised
// to the power y
static int power(int x, int y)
{
    if (y == 0) 
    {
        return 1;
    }
  
    if (y % 2 == 0)
    {
        return (power(x, y / 2) * 
                power(x, y / 2));
    }
    return (x * power(x, y / 2) * 
                power(x, y / 2));
}
  
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
static boolean isPerfectDigitalInvariant(int x)
{
    for (int fixed_power = 1;; fixed_power++) 
    {
        int temp = x, sum = 0;
  
        // For each digit in temp
        while (temp > 0) 
        {
            int r = temp % 10;
            sum += power(r, fixed_power);
            temp = temp / 10;
        }
  
        // If satisfies Perfect Digital
        // Invariant condition
        if (sum == x) 
        {
            return true;
        }
        // If sum exceeds n, then not possible
        if (sum > x) 
        {
            return false;
        }
    }
}
  
// Driver Code
public static void main(String[] args)
{
    // Given Number N
    int N = 4150;
  
    // Function Call
    if (isPerfectDigitalInvariant(N))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 implementation of the above approach 
 
# Function to find the 
# sum of divisors 
def power(x, y):
    if (y == 0):
        return 1
    if (y % 2 == 0):
        return (power(x, y//2) * power(x, y//2)) 
    return (x * power(x, y//2) * power(x, y//2))
     
 
# Function to check whether the given 
# number is Perfect Digital Invariant 
# number or not 
def isPerfectDigitalInvariant(x):
    fixed_power = 0
    while True:
        fixed_power += 1
        temp = x 
        summ = 0
 
        # For each digit in temp 
        while (temp): 
            r = temp % 10
            summ = summ + power(r, fixed_power) 
            temp = temp//10
 
        # If satisfies Perfect Digital 
        # Invariant condition 
        if (summ == x): 
            return (True) 
           
        # If sum exceeds n, then not possible 
        if (summ > x):
            return (False)
 
# Driver code 
# Given Number N 
N = 4150
 
# Function Call 
if (isPerfectDigitalInvariant(N)):
    print("Yes")
else:
    print("No")
 
    # This code is contributed by vikas_g

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to calculate x raised
// to the power y
static int power(int x, int y)
{
    if (y == 0) 
    {
        return 1;
    }
 
    if (y % 2 == 0)
    {
        return (power(x, y / 2) * 
                power(x, y / 2));
    }
    return (x * power(x, y / 2) * 
                power(x, y / 2));
}
 
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
static bool isPerfectDigitalInvariant(int x)
{
    for(int fixed_power = 1;; fixed_power++) 
    {
        int temp = x, sum = 0;
 
        // For each digit in temp
        while (temp > 0) 
        {
            int r = temp % 10;
            sum += power(r, fixed_power);
            temp = temp / 10;
        }
 
        // If satisfies Perfect Digital
        // Invariant condition
        if (sum == x) 
        {
            return true;
        }
        // If sum exceeds n, then not possible
        if (sum > x) 
        {
            return false;
        }
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given number N
    int N = 4150;
 
    // Function call
    if (isPerfectDigitalInvariant(N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript implementation
 
// Function to calculate x raised
// to the power y
function power(x, y)
{
    if (y == 0) {
        return 1;
    }
  
    if (y % 2 == 0) {
        return (power(x, Math.floor(y / 2))
                * power(x, Math.floor(y / 2)));
    }
    return (x
            * power(x, Math.floor(y / 2))
            * power(x, Math.floor(y / 2)));
}
  
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
function isPerfectDigitalInvariant(x)
{
    for (var fixed_power = 1;; fixed_power++) {
  
        var temp = x, sum = 0;
  
        // For each digit in temp
        while (temp) {
  
            var r = temp % 10;
            sum += power(r, fixed_power);
            temp = Math.floor(temp / 10);
        }
  
        // If satisfies Perfect Digital
        // Invariant condition
        if (sum == x) {
            return true;
        }
        // If sum exceeds n, then not possible
        if (sum > x) {
            return false;
        }
    }
}
 
// Driver Code 
// Given Number N
var N = 4150;
 
// Function Call
if (isPerfectDigitalInvariant(N))
    document.write("Yes");
else
    document.write("No");
   
// This code is contributed by shubhamsingh10
</script>