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 yint 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 notbool 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 Codeint main(){ // Given Number N int N = 4150; // Function Call if (isPerfectDigitalInvariant(N)) cout << "Yes"; else cout << "No";} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to calculate x raised// to the power ystatic 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 notstatic 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 Codepublic 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 approachusing System;class GFG{// Function to calculate x raised// to the power ystatic 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 notstatic 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 Codepublic 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 yfunction 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 notfunction 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 Nvar N = 4150;// Function Callif (isPerfectDigitalInvariant(N)) document.write("Yes");else document.write("No"); // This code is contributed by shubhamsingh10</script> |
Output:
Yes
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