A number is said to be a Peterson number if the sum of factorials of each digit of the number is equal to the number itself.
Example:
Input : n = 145 Output = Yes Explanation: 145 = 5! + 4! + 1! = 120 + 24 +1 = 145 Input : n = 55 Output : No Explanation: 5! + 5! = 120 + 120 = 240 Since 55 is not equal to 240 It is not a Peterson number.
We will pick each digit (Starting from the last digit) of the given number and find its factorial. And add all factorials. Finally, we check if the sum of factorials is equal to number or not.
C++
// C++ program to determine whether the number is // Peterson number or not #include <iostream> using namespace std;
// To quickly find factorial of digits int fact[10]
= { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };
// Function to check if a number is Peterson // or not bool peterson( int n)
{ int num = n, sum = 0;
// stores the sum of factorials of
// each digit of the number.
while (n > 0) {
int digit = n % 10;
sum += fact[digit];
n = n / 10;
}
// Condition check for a number to
// be a Peterson Number
return (sum == num);
} // Driver Program int main()
{ int n = 145;
if (peterson(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
Java
//checks whether a number entered by user is peterson number or not import java.util.*;
class GFG
{ public static void main(String args[])
{
Scanner sc= new Scanner(System.in);
//taking input from the user
System.out.println( "Enter the number" );
int num=sc.nextInt();
int temp=num; //storing the number in a temporary variable
int f= 1 ,sum= 0 ;
while (num!= 0 ) //running while loop until number becomes zero
{
f= 1 ;
//extracting last digit of the number
//and storing in r
int r=num% 10 ;
//for loop to find the factorial of a digit
for ( int i= 1 ;i<=r;i++)
{
f=f*i;
}
sum=sum+f; //adding the factotial of the digits
num=num/ 10 ;
}
//checking if the sum of the factorial of digits
//is equal to the number or not
if (sum==temp)
System.out.println( "PETERSON NUMBER" );
else
System.out.println( "NOT PETERSON NUMBER" );
}
}
|
Python3
# Python3 code to determine whether the # number is Peterson number or not # To quickly find factorial of digits fact = [ 1 , 1 , 2 , 6 , 24 , 120 , 720 ,
5040 , 40320 , 362880 ]
# Function to check if a number # is Peterson or not def peterson(n):
num = n
sum = 0
# stores the sum of factorials of
# each digit of the number.
while n > 0 :
digit = int (n % 10 )
sum + = fact[digit]
n = int (n / 10 )
# Condition check for a number
# to be a Peterson Number
return ( sum = = num)
# Driver Code n = 145
print ( "Yes" if peterson(n) else "No" )
# This code is contributed by "Sharad_Bhardwaj".. |
C#
// C# program to determine whether the // number is Peterson number or not using System;
public class GFG {
// To quickly find factorial of digits
static int [] fact
= new int [10] { 1, 1, 2, 6, 24,
120, 720, 5040, 40320, 362880 };
// Function to check if a number is
// Peterson or not
static bool peterson( int n)
{
int num = n;
int sum = 0;
// stores the sum of factorials of
// each digit of the number.
while (n > 0) {
int digit = n % 10;
sum += fact[digit];
n = n / 10;
}
// Condition check for a number to
// be a Peterson Number
return (sum == num);
}
// Driver Program
static public void Main()
{
int n = 145;
if (peterson(n))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
} // This code is contributed by vt_m. |
PHP
<?php // PHP program to determine // whether the number is // Peterson number or not // To quickly find // factorial of digits $fact = array (1, 1, 2, 6, 24, 120, 720,
5040, 40320, 362880);
// Function to check if // a number is Peterson // or not function peterson( $n )
{ $num = $n ; $sum = 0;
// stores the sum of factorials of
// each digit of the number.
while ( $n > 0)
{
$digit = $n % 10;
$n = $n / 10;
}
// Condition check for
// a number to be a
// Peterson Number
return ( $sum == $num );
} // Driver Code
$n = 145;
if (peterson( $n ))
echo "Yes" ;
else
echo "No" ;
// This code is contributed by ajit ?> |
Javascript
<script> // Javascript program to determine whether // the number is Peterson number or not // To quickly find factorial of digits let fact = [ 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 ];
// Function to check if a number is // Peterson or not function peterson(n)
{ let num = n, sum = 0;
// stores the sum of factorials of
// each digit of the number.
while (n > 0)
{
let digit = n % 10;
sum += fact[digit];
n = parseInt(n / 10);
}
// Condition check for a number to
// be a Peterson Number
return (sum == num);
} // Driver code let n = 145; if (peterson(n))
document.write( "Yes" );
else document.write( "No" );
// This code is contributed by souravmahato348 </script> |
Output
Yes
Time Complexity: log10(n)
Auxiliary Space: O(1)