Can you please simplify the algorithm? Solution is not clear from the
posted submissions !!!

On 10 June 2012 20:32, KK <[email protected]> wrote:

> This problem is of ACM-ICPC kanpur online round 2012.
> you can find the solution 
> here<http://www.codechef.com/ACMKAN11/problems/ARTHMNCY>
> .
>
>
> On Sunday, 10 June 2012 16:37:33 UTC+5:30, payel roy wrote:
>>
>> Find the number of fractions a/b such that-
>>
>> 1. *gcd(a, b) = 1*
>> 2. *0 < a/b < 1*
>> 3. *a * b = (n!) ^ (n!)*
>>
>> Where *"n!"* denotes the factorial of n and "^" denotes power, *i.e.
>> (2!) ^ (2!) = 4*.
>
>
> On Sunday, 10 June 2012 16:37:33 UTC+5:30, payel roy wrote:
>>
>> Find the number of fractions a/b such that-
>>
>> 1. *gcd(a, b) = 1*
>> 2. *0 < a/b < 1*
>> 3. *a * b = (n!) ^ (n!)*
>>
>> Where *"n!"* denotes the factorial of n and "^" denotes power, *i.e.
>> (2!) ^ (2!) = 4*.
>
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