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FACVSPOW - Factorial vs Power |
Consider two integer sequences f(n) = n! and g(n) = an, 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 <= 106
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
Added by: | Spooky |
Date: | 2009-11-01 |
Time limit: | 1s |
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/ |
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2018-06-23 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: 2018-06-23 18:17:04 |
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2018-05-22 23:49:37
waste of time!! |
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2018-03-13 19:48:17
did it without Stirling's formula , using prefix sum and log function. |
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2017-09-19 13:29:47
easy one with stirling's formula |
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2017-02-18 23:23:01
u can skip the stirling's approximation,however use of log will surely help,,and yes use double instead of float.. |
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2016-02-20 00:31:48
ln(n!) = (n*ln(n))-n+(1/2*ln(2*pi*n)) !!!!!This is helpful here |
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2015-06-02 14:29:58 Ankush
DId it after 2 attempts :D Loved it |
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2015-01-14 20:56:23 /* Nitin Jaiman */
Stirling formula is working just think about high and low values of binary search. |
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2014-12-27 18:21:47 ashish kumar
a lot of WA using stirling approx.. but got AC without stirling. just simple coding with binary search......1.39sec |
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2014-07-15 17:24:27 vinod y
use printf, scanf instead of cin, cout Last edit: 2014-07-15 17:24:42 |