Smith Number - GeeksforGeeks


Smith Number - GeeksforGeeks
Given a number n, the task is to find out whether this number is smith or not. A Smith Number is a composite number whose sum of digits is equal to the sum of digits in its prime factorization.

const int MAX  = 10000;
 
// array to store all prime less than and equal to 10^6
vector <int> primes;
 
// utility function for sieve of sundaram
void sieveSundaram()
{
    // In general Sieve of Sundaram, produces primes smaller
    // than (2*x + 2) for a number given number x. Since
    // we want primes smaller than MAX, we reduce MAX to half
    // This array is used to separate numbers of the form
    // i+j+2ij from others where 1 <= i <= j
    bool marked[MAX/2 + 100] = {0};
 
    // Main logic of Sundaram. Mark all numbers which
    // do not generate prime number by doing 2*i+1
    for (int i=1; i<=(sqrt(MAX)-1)/2; i++)
        for (int j=(i*(i+1))<<1; j<=MAX/2; j=j+2*i+1)
            marked[j] = true;
 
    // Since 2 is a prime number
    primes.push_back(2);
 
    // Print other primes. Remaining primes are of the
    // form 2*i + 1 such that marked[i] is false.
    for (int i=1; i<=MAX/2; i++)
        if (marked[i] == false)
            primes.push_back(2*i + 1);
}
 
// Returns true if n is a Smith number, else false.
bool isSmith(int n)
{
    int original_no = n;
 
    // Find sum the digits of prime factors of n
    int pDigitSum = 0;
    for (int i = 0; primes[i] <= n/2; i++)
    {
        while (n % primes[i] == 0)
        {
            // If primes[i] is a prime factor,
            // add its digits to pDigitSum.
            int p = primes[i];
            n = n/p;
            while (p > 0)
            {
                pDigitSum += (p % 10);
                p = p/10;
            }
        }
    }
 
    // If n!=1 then one prime factor still to be
    // summed up;
    if (n != 1 && n != original_no)
    {
        while (n > 0)
        {
            pDigitSum = pDigitSum + n%10;
            n = n/10;
        }
    }
 
    // All prime factors digits summed up
    // Now sum the original number digits
    int sumDigits = 0;
    while (original_no > 0)
    {
        sumDigits = sumDigits + original_no % 10;
        original_no = original_no/10;
    }
 
    // If sum of digits in prime factors and sum
    // of digits in original number are same, then
    // return true. Else return false.
    return (pDigitSum == sumDigits);
}
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