Queries to print all divisors of number

Given a positive integer ‘n’ and query ‘q’. Print all the divisors of number ‘n’.

Input: 6
Output: 1 2 3 6
Explanation
Divisors of 6 are: 1, 2, 3, 6

Input: 10
Output: 1 2 5 10

Naive approach is to iterate through 1 to sqrt(n) for every query ‘q’ and print the divisors accordingly. See this to understand more. Time complexity of this approach is q*sqrt(n) which is not sufficient for large number of queries.

Efficient approach is to use factorization by using sieve base approach.

• Create a list of consecutive integers from 1 to ‘n’.
• For any number ‘d’, iterate through all the multiples of ‘d’ i.e., d, 2d, 3d, … etc. Meanwhile push the divisor ‘d’ for every multiples.
  
  

  
  
  

  // C++ program to print divisors of // number for multiple query #include <iostream> #include <vector> using namespace std;    const int MAX = 1e5;    // Initialize global divisor vector // array of sequence container vector<int> divisor[MAX + 1];    // Sieve based approach to pre- // calculate all divisors of number void sieve() {     for (int i = 1; i <= MAX; ++i) {         for (int j = i; j <= MAX; j += i)             divisor[j].push_back(i);     } }    // Utility function to print divisors // of given number inline void printDivisor(int& n) {     for (auto& div : divisor[n])         cout << div << " "; }    // Driver code int main() {     sieve();        int n = 10;     cout << "Divisors of " << n << " = ";     printDivisor(n);        n = 30;     cout << "\nDivisors of " << n << " = ";     printDivisor(n);     return 0; }

  Output 
  Divisors of 10 = 1 2 5 10 
  Divisors of 30 = 1 2 3 5 6 10 15 303
  

  Time complexity: O(len) for each query, where len is equal to total divisors of number ‘n’.
  Auxiliary space: O(MAX)

  

  
  
  
  
  
  My Personal Notes arrow_drop_up

  Save