Given a number n, count total perfect divisors of n. Perfect divisors are those divisors which are square of some integer. For example a perfect divisor of 8 is 4.

Input : n = 16 Output : 3 Explanation : There are only 5 divisor of 16: 1, 2, 4, 8, 16. Only three of them are perfect squares: 1, 4, 16. Therefore the answer is 3 Input : n = 7 Output : 1

**Naive approach**

A brute force is find all the divisors of a number. Count all divisors that are perfect squares.

// Below is C++ code to count total perfect Divisors #include<bits/stdc++.h> using namespace std; // Utility function to check perfect square number bool isPerfectSquare(int n) { int sq = (int) sqrt(n); return (n == sq * sq); } // Returns count all perfect divisors of n int countPerfectDivisors(int n) { // Initialize result int count = 0; // Consider every number that can be a divisor // of n for (int i=1; i*i <= n; ++i) { // If i is a divisor if (n%i == 0) { if (isPerfectSquare(i)) ++count; if (n/i != i && isPerfectSquare(n/i)) ++count; } } return count; } // Driver code int main() { int n = 16; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n) << "\n"; n = 12; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n); return 0; }

Output:Total Perfect divisors of 16 = 3 Total Perfect divisors of 12 = 2

**Time complexity: **O(sqrt(n))

**Auxiliary space: **O(1)

**Efficient approach (For large number of queries)**

The idea is based on Sieve of Eratosthenes. This approach is beneficial if there are large number of queries. Following is the algorithm to find all perfect divisors up to n numbers.

- Create a list of n consecutive integers from 1 to n:(1, 2, 3, …, n)
- Initially, let d be 2, the first divisor
- Starting from 4(square of 2) increment all the multiples of 4 by 1 in perfectDiv[] array, as all these multiples contain 4 as perfect divisor. These numbers will be 4d, 8d, 12d, … etc
- Repeat the 3
^{rd}step for all other numbers. The final array of perfectDiv[] will contain all the count of perfect divisors from 1 to n

Below is C++ implementation of above steps.

// Below is C++ code to count total perfect // divisors #include<bits/stdc++.h> using namespace std; #define MAX 100001 int perfectDiv[MAX]; // Pre-compute counts of all perfect divisors // of all numbers upto MAX. void precomputeCounts() { for (int i=1; i*i < MAX; ++i) { // Iterate through all the multiples of i*i for (int j=i*i; j < MAX; j += i*i) // Increment all such multiples by 1 ++perfectDiv[j]; } } // Returns count of perfect divisors of n. int countPerfectDivisors(int n) { return perfectDiv[n]; } // Driver code int main() { precomputeCounts(); int n = 16; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n) << "\n"; n = 12; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n); return 0; }

Output:Total Perfect divisors of 16 = 3 Total Perfect divisors of 12 = 2

**Time complexity: **O(MAX * log(log (MAX)))

**Auxiliary space: **O(MAX)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.