UVA 294 - Divisors

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In this problem, it's easy to get TLE while using a general approach. here you have to follow one rule which is

       1. The total number of divisors of a number is just doubled of divisors till the square root of that number though there's some special case which is if the square root of given number is an integer then the total number of divisors is 2*number of divisors till sqrt(number) -1 otherwise 2*number of divisors till sqrt(number).

for example let a given number is 6 then sqrt(6) = 2.4494 . here 1 and 2 (2 divisors till sqrt(6)) is the divisors till 2.4494(sqrt(6)) and the square root is not an integer(2.4494) so the total number of divisors is 2*2=4.

again, let a given number is 25 then sqrt(25) = 5 . here 1 and 5 (2 divisors till sqrt(25)) is the divisors till 5(sqrt(25)) and the square root is an integer(5) so total number of divisors is 2*2-1=3.
similarly, you will have to find each number's total number of divisors. then output the number which has the largest divisors.

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A solution in c++