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Обсуждение задачи 1331. Владислава

[censored] checkers! AC at last! -0 != 0
Послано Alias (Alexander Prudaev) 6 апр 2007 22:35
if program writes -0, your [censored] checker says WA#3
inspite of accuracy  in 2 digits. abs(-0-0) < 0.001 !

Edited by moderator 06.04.2007 22:54
Warning for you (+)
Послано Dmitry 'Diman_YES' Kovalioff 6 апр 2007 22:54
I am tired of you who offends Timus Online Judge again and again.

If you continue writing your posts to this board such a way, I will just remove them completely.

Edited by author 06.04.2007 22:55
Re: Warning for you (+)
Послано Alias (Alexander Prudaev) 6 апр 2007 23:28
After your edit, observers could have an impression that I have used an abusive word.
It's not true, it was word "stupid"
Checker is correct. In the 3rd test your output was -1.#J and 1.#R. (-)
Послано Sandro (USU) 7 апр 2007 01:18
Re: Checker is correct. In the 3rd test your output was -1.#J and 1.#R. (-)
Послано Alias (Alexander Prudaev) 7 апр 2007 22:05
you are wrong

there is fragment of my program :

double Norm(double x)
//    if (x>0.001)
        return x;
//    else
//        return 0;

with these commets program gets WA#3
without - AC
if you want, i can send you my program entirely

Edited by author 07.04.2007 22:15
Checker is really correct. (+)
Послано Sandro (USU) 8 апр 2007 00:56
Your WA and AC solution have are wrong both but tests are weak. You have WA after rejudge.