FCTRHELL - Factor y Hell
Factorial(N) in base B : The number of trailing zeros.
Factorial(19) in base 9×10^0=9 can be written 725735500635080000, ending with 4 zeros. Factorial(43) in base 2×10^1=20 can be written 59HHHFECFCCEGH5G7I7A3A8G88F8CD8G000000000, ending with 9 zeros. What about working with serious constraints and tricky cases ? Factorial(N) will be a huge one, the base will be dummy too and have the special form : B×10^E.
Input
The input begins with the number T of test cases in a single line. In each of the next T lines there are three integers : N, B, E.
Output
For each test case, print the number of zeros at the end of Factorial(N) written in base B×10^E.
Example
Input: 3 19 9 0 43 2 1 10000 100 10 Output: 4 9 208
Constraints
1 <= T < 2000 1 <= N < 10^1000 1 <= B < 10^9 0 <= E < 10^9
Informations
Don't worry about the 'special' base 1 (B=1 and E=0), it is absent from input. About distribution : random input (N : log-uniform, B : uniform, E : uniform) in their range. Some tricky cases are added. It is recommended to solve FACTBASE first, and find a way to solve FCTRL much faster than common solutions. Time limit is ×13.6 my best Python3 time, or ×1.33 my "basic" one.
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(Tjandra Satria Gunawan)(曾毅昆):
2013-02-03 21:57:49
great problem! but I'm weak to deal with big integer.. (B and E is no problem but N) and I'm weak to work with python too (because it's slow).. Hard as hell! :-O
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Added by: | Francky |
Date: | 2013-02-03 |
Time limit: | 3s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | Own problem |