Given a number N. The task is to find the largest good number in the divisors of given number N. A number X is defined as the good number if there is no such positive integer a > 1 such that a^2 is a divisor of X.
Input: N = 10 Output: 10 In 1, 2, 5, 10. 10 is the largest good number Input: N = 12 Output: 6 In 1, 2, 3, 4, 6, 12. 6 is the largest good number
Approach: Find all prime divisors of N. Assume they are p1, p2, …, pk (in O(sqrt(n)) time complexity). If the answer is a, then we know that for each 1 <= I <= k, obviously, a is not divisible by pi^2 (and all greater powers of pi). So a <= p1 × p2 ×… × pk. And we know that p1 × p2 × … × pk is itself a good number. So, the answer is p1 × p2 ×…× pk.
Below is the implementation of above approach:
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- Program to find count of numbers having odd number of divisors in given range
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