Given a number n, find n’th smart number (1<=n<=1000). Smart number is a number which has at least three distinct prime factors. We are given an upper limit on value of result as MAX

For example 30 is 1st smart number because it has 2, 3, 5 as it’s distinct prime factors. 42 is 2nd smart number because it has 2, 3, 7 as it’s distinct prime factors.

Examples:

Input : n = 1 Output: 30 // three distinct prime factors 2, 3, 5 Input : n = 50 Output: 273 // three distinct prime factors 3, 7, 13 Input : n = 1000 Output: 2664 // three distinct prime factors 2, 3, 37

The idea is based on Sieve of Eratosthenes. We use an array to use an array prime[] to keep track of prime numbers. We also use the same array to keep track of the count of prime factors seen so far. Whenever the count reaches 3, we add the number to result.

- Take an array primes[] and initialize it with 0.
- Now we know that first prime number is i = 2 so mark primes[2] = 1 i.e; primes[i] = 1 indicates that ‘i’ is prime number.
- Now traverse the primes[] array and mark all multiples of ‘i’ by condition primes[j] -= 1 where ‘j’ is multiple of ‘i’, and check the condition primes[j]+3 = 0 because whenever primes[j] become -3 it indicates that previously it had been multiple of three distinct prime factors. If condition
**primes[j]+3=0**becomes true that means ‘j’ is a Smart Number so store it in a array result[]. - Now sort array result[] and return result[n-1].

Below is C++ implementation of above idea.

// C++ implementation to find n'th smart number #include<bits/stdc++.h> using namespace std; // Limit on result const int MAX = 3000; // Function to calculate n'th smart number int smartNumber(int n) { // Initialize all numbers as not prime int primes[MAX] = {0}; // iterate to mark all primes and smart number vector<int> result; // Traverse all numbers till maximum limit for (int i=2; i<MAX; i++) { // 'i' is maked as prime number because // it is not multiple of any other prime if (primes[i] == 0) { primes[i] = 1; // mark all multiples of 'i' as non prime for (int j=i*2; j<MAX; j=j+i) { primes[j] -= 1; // If i is the third prime factor of j // then add it to result as it has at // least three prime factors. if ( (primes[j] + 3) == 0) result.push_back(j); } } } // Sort all smart numbers sort(result.begin(), result.end()); // return n'th smart number return result[n-1]; } // Driver program to run the case int main() { int n = 50; cout << smartNumber(n); return 0; }

Output:

273

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