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IDC1948 - Identity crisis |
For every given number n we define x(n) as distance from n to the first number after n in form of 99...99. For example x(100)=899, x(45)=54, etc. Given several n numbers you have to find the Zp, where x(n) ≡ n (mod p).
Input
First line of input contains one number T (T < 20) - the number of test cases. In each of the next T lines contains one number each to represent n (0 < n < 30000000).
Output
In each line you have to write one number - the least p > 1 that x(n) ≡ n (mod p). If there is no such p the line should contain -1.
Example
Input: 2 234 5 Output: 3 -1
Explanation
x(234)=765. 765 mod 3=0, 234 mod 3=0 => 765 ≡ 234 (mod 3)
Added by: | Konrad Krystecki |
Date: | 2014-02-04 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
hide comments
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2019-12-29 06:42:52
Find smallest p>1 such that x(n) %p == n %p. Hopefully translating the statement to human is not a spoiler. |
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2016-07-19 20:34:37
300th on spoj..<3..CREDITS :@pulkit_gulati |
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2016-02-06 18:21:19 ROHIT GUPTA
bad problem. two same solution with different header , giving ac in c WA in c++ |
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2015-10-27 06:52:47
AC in 1 goo...easy one ..simple maths+sieve ;) |
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2015-10-21 11:53:08 kamran siddique
Easy One... |
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2015-09-28 05:32:51 ROHIT RAJ
AC in 1 go !! Learned a lot. nice question |
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2015-01-14 06:35:30 deadbrain
Very Poor Explanation.... Could have been much better.... Think beyond what is written and dont be distracted by the comments... |
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2015-01-12 14:47:58 D
yupiee! finally solved. |
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2014-07-12 23:47:33 zicowa
and also for 99 consider 99 |
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2014-07-12 23:46:13 zicowa
finally GOT AC go with the standard defination for modulo congurency and abs(a-b) |