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EIGHTS - Triple Fat Ladies |
Pattern Matchers have been designed for various sorts of patterns. Mr. HKP likes to observe patterns in numbers. After completing his extensive research on the squares of numbers, he has moved on to cubes. Now he wants to know all numbers whose cube ends in 888.
Given a number k, help Mr. HKP find the kth number (indexed from 1) whose cube ends in 888.
Input
The first line of the input contains an integer t, the number of test cases. t test cases follow.
Each test case consists of a single line containing a single integer k (1 <= k <= 2000000000000).
Output
For each test case, output a single integer which denotes the kth number whose cube ends in 888. The result will be less than 263.
Example
Input: 1 1 Output: 192
Added by: | Matthew Reeder |
Date: | 2006-10-30 |
Time limit: | 1.197s |
Source limit: | 30000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | Al-Khawarizm 2006 |
hide comments
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2018-06-27 09:00:44
@sreejoy4242 why separate case for k==1 |
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2018-06-27 08:55:31
getting wrong answer using simple maths |
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2018-06-09 19:30:53
The test case may be long. Last edit: 2018-06-09 19:31:32 |
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2018-06-04 14:21:39
first run a bruteforce program to print you the first few elements of the sequence (using string manipulation). notice pattern, get ac in 7 lines (or less) |
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2018-05-31 11:21:42
test |
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2018-04-06 20:57:47
can anyone tell me how to figure out that the numbers will be in AP?. if you can please e-mail me at sanjib.sah@yahoo.com Thank You :) |
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2018-03-11 11:21:18
@sreejoy4242 good observation!! if k==1 192 else 192+(k-1)*250 Last edit: 2018-03-11 11:21:53 |
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2018-03-09 15:07:54
There should be another test case. Thanks for suggesting arithmetic progression. |
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2018-03-06 11:57:34
Total time waste |
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2018-03-03 08:26:06
AC in one go.. :) |