FACVSPOW  Factorial vs Power
Consider two integer sequences f(n) = n! and g(n) = a^{n}, where n is a positive integer. For any integer a > 1 the second sequence is greater than the first for a finite number of values. But starting from some integer k, f(n) is greater than g(n) for all n >= k. You are to find the least positive value of n for which f(n) > g(n), for a given positive integer a > 1.
Input
The first line of the input contains number t – the amount of tests. Then t test descriptions follow. Each test consist of a single number a.
Constraints
1 <= t <= 100000
2 <= a <= 10^{6}
Output
For each test print the least positive value of n for which f(n) > g(n).
Example
Input: 3 2 3 4 Output: 4 7 9
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a_ranjan:
20180623 18:16:04
No need for any advanced formula , just precompute the array of log (factorial) upto 10000000 and find the answer using binary search tehnique ! Last edit: 20180623 18:17:04 

excel_blaze:
20180522 23:49:37
waste of time!! 

waqar_ahmad224:
20180313 19:48:17
did it without Stirling's formula , using prefix sum and log function. 

vengatesh15:
20170919 13:29:47
easy one with stirling's formula 

holmesherlock:
20170218 23:23:01
u can skip the stirling's approximation,however use of log will surely help,,and yes use double instead of float.. 

sy_117:
20160220 00:31:48
ln(n!) = (n*ln(n))n+(1/2*ln(2*pi*n)) !!!!!This is helpful here 

Ankush :
20150602 14:29:58
DId it after 2 attempts :D Loved it 

/* Nitin Jaiman */:
20150114 20:56:23
Stirling formula is working just think about high and low values of binary search. 

ashish kumar:
20141227 18:21:47
a lot of WA using stirling approx..


vinod y:
20140715 17:24:27
use printf, scanf instead of cin, cout Last edit: 20140715 17:24:42 
Added by:  Spooky 
Date:  20091101 
Time limit:  0.985s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 NODEJS OBJC PERL6 SQLITE VB.NET 
Resource:  Advancement Autumn 2009, http://sevolymp.uuuq.com/ 