ONEZERO - Ones and zeros

Certain positive integers have their decimal representation consisting only of ones and zeros, and having at least one digit one, e.g. 101. If a positive integer does not have such a property, one can try to multiply it by some positive integer to find out whether the product has this property.

Input

Number K of test cases (K is approximately 1000);
In each of the next K lines there is one integer n (1 ≤ n ≤ 20000)

Output

For each test case, your program should compute the smallest multiple of the number n consisting only of digits 1 and 0 (beginning with 1).

Example

Input:
3
17
11011
17

Output:
11101
11011
11101

Added by:Paweł Dobrzycki
Date:2005-05-26
Time limit:8s
Source limit:4096B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS PERL6 VB.NET
Resource:II Polish Olympiad in Informatics, Ist Stage

hide comments
2016-08-12 19:26:40
Got AC in one, 0,98
Check modular arithmetic
https://brilliant.org/wiki/modular-arithmetic/
2016-07-14 13:26:43
@candide how would that make a difference if we do bfs ending with the left most digits or bfs ending with the right most digits ?
2016-07-05 19:41:32
Used the Queue n the Modulo(n states) n Backtracked for number
AWESOME

Last edit: 2016-07-05 19:44:07
2016-05-26 17:13:42
easy one !! 2nd on bfs ! :)
2016-05-16 01:58:46 candide
@kartikay singh: "Wonder,how ppl did it in 0.00s??"

Response: I did using a bfs ending with the left most digits of the number we are asked for. On the contrary, the most common and intuitive bfs method ends with the right most digit, and even optimised the later bfs needs 0.02s or more (checked 2017/03). Perphaps there exists a more mathematical (and ad hoc) method.

Last edit: 2017-03-25 00:58:41
2016-04-28 21:35:28
bfs and backtracking got me an AC :)
2016-01-12 12:01:40 minhthai
very nice problem. that example tests though :v
for java, you need to optimize a bit your bfs
2015-12-29 10:01:35 karan
weak test cases. try for 999 and 9999.
2015-12-26 21:21:39 kshitij tripathi
Nice Problem :)
2015-12-10 10:44:24 vipul grover
finally after so many attempts :)
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