4
100
1101
1011
001
101
11001
10101
10011
01001
10001
10010

for all case answer 1001

I know that they have some extra line breaks
in the tests, so maybe test like

4
100
1101
1011


001
101
11001
10101

10011
01001


10001
10010


would be more helpful!

4
1100
0011
11010
01011


????
it is correct?

yeah .. thank you very much..

it's right .. the test will be like this:

4
1100


0011
...  there will be some spaces btween them..

everyone should be careful ... ^_^


Edited by author 30.08.2005 20:16

Edited by author 30.08.2005 20:16

tx very much :), that helps me much.

But I accepted.
I didn't check this.

But i don't think these tests are right?
Can you tell me whether it is right or not?
Or the answer.

It is known that the original words all HAVE (!!!) the following property: the sum of positions where symbols 1 are located is a multiple of (N+1) or equal to zero.

I think nothing need to be added

Thank you very much for that test!!!

(To JIeHuH*CCCP)
Thank you very much!
I got AC now...

Edited by author 17.10.2007 14:42

Edited by author 17.10.2007 14:42

Thanks a lot, KingPin. Your test really helped me to understnd that i've got an algorithm problem, not while reading)) Accepted finally!

you are so KIND that I don't know how to appreciate you!
Good Luck to everyone

Many thanks, very good tests

Thank you very much for this test. It helped me to find 3 mistakes) Although they all were really supid... So finally I got AC.

5
1100
1110
11110
11101
111100
111010
111001
011100
101100
110100
111000
0001
1001
1000
11001
10101
10011
010001
100001
100010
110001
101001
100101
100011
1010
0010
0110
0100
0101
11010
01110
01011
001010
010010
010100
101010
011010
010110
010101


this test is about all cases of n=5 ,the answer is
11100   *  11
10001   *  13
01010   *  15

Thank you!