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Program to check Strong Number
• Difficulty Level : Easy
• Last Updated : 21 Apr, 2021

Strong Numbers are the numbers whose sum of factorial of digits is equal to the original number. Given a number, check if it is a Strong Number or not.
Examples:

Input  : n = 145
Output : Yes
Sum of digit factorials = 1! + 4! + 5!
= 1 + 24 + 120
= 145

Input :  n = 534
Output : No

1) Initialize sum of factorials as 0.
2) For every digit d, do following
a) Add d! to sum of factorials.
3) If sum factorials is same as given
number, return true.
4) Else return false.

An optimization is to precompute factorials of all numbers from 0 to 10.

## C++

 // C++ program to check if a number is// strong or not.#include using namespace std; int f[10]; // Fills factorials of digits from 0 to 9.void preCompute(){    f[0] = f[1] = 1;    for (int i = 2; i<10; ++i)        f[i] = f[i-1] * i;} // Returns true if x is Strongbool isStrong(int x){    int factSum = 0;     // Traverse through all digits of x.    int temp = x;    while (temp)    {        factSum += f[temp%10];        temp /= 10;    }     return (factSum == x);} // Driver codeint main(){    preCompute();     int x = 145;    isStrong(x) ? cout << "Yes\n" : cout << "No\n";    x = 534;    isStrong(x) ? cout << "Yes\n" : cout << "No\n";    return 0;}

## Java

 // Java program to check if// a number is Strong or not class CheckStrong{    static int f[] = new int[10];      // Fills factorials of digits from 0 to 9.    static void preCompute()    {        f[0] = f[1] = 1;        for (int i = 2; i<10; ++i)            f[i] = f[i-1] * i;    }         // Returns true if x is Strong    static boolean isStrong(int x)    {        int factSum = 0;              // Traverse through all digits of x.        int temp = x;        while (temp>0)        {            factSum += f[temp%10];            temp /= 10;        }              return (factSum == x);    }         // main function    public static void main (String[] args)    {          // calling preCompute        preCompute();             // first pass        int x = 145;        if(isStrong(x))        {            System.out.println("Yes");        }        else            System.out.println("No");                     // second pass        x = 534;        if(isStrong(x))        {            System.out.println("Yes");        }        else            System.out.println("No");    }}

## Python

 # Python program to check if a number is# strong or not. f = [None] * 10 # Fills factorials of digits from 0 to 9.def preCompute() :    f[0] = f[1] = 1;    for i in range(2,10) :        f[i] = f[i-1] * i  # Returns true if x is Strongdef isStrong(x) :         factSum = 0    # Traverse through all digits of x.    temp = x    while (temp) :        factSum = factSum + f[temp % 10]        temp = temp / 10     return (factSum == x)     # Driver codepreCompute()x = 145if(isStrong(x) ) :    print "Yes"else :    print "No"x = 534if(isStrong(x)) :    print "Yes"else:    print "No" # This code is contributed by Nikita Tiwari.

## C#

 // C# program to check if// a number is Strong or notusing System; class CheckStrong{    static int []f = new int[10];     // Fills factorials of digits from 0 to 9.    static void preCompute()    {        f[0] = f[1] = 1;        for (int i = 2; i < 10; ++i)            f[i] = f[i - 1] * i;    }         // Returns true if x is Strong    static bool isStrong(int x)    {        int factSum = 0;             // Traverse through all digits of x.        int temp = x;        while (temp > 0)        {            factSum += f[temp % 10];            temp /= 10;        }             return (factSum == x);    }         // Driver Code    public static void Main ()    {        // calling preCompute        preCompute();             // first pass        int x = 145;        if(isStrong(x))        {            Console.WriteLine("Yes");        }        else            Console.WriteLine("No");                     // second pass        x = 534;        if(isStrong(x))        {            Console.WriteLine("Yes");        }        else            Console.WriteLine("No");    }} // This code is contributed by Nitin Mittal.



## Javascript



Output:

Yes
No

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