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Показать все ветки Спрятать все ветки Показать все сообщения Спрятать все сообщения | Страница 1 | WA#6. What's wrong? | Afli | 1756. Полтора землекопа | 6 окт 2012 01:26 | 1 | i found the mistake. Edited by author 07.10.2012 17:02 Edited by author 07.10.2012 17:34 | -> Tests | Flint Storman | 1756. Полтора землекопа | 24 июл 2014 10:08 | 2 | -> Tests Flint Storman 16 сен 2012 18:56 Hope it will help: 5 5 6 5 5 5 5 5 0 2 5 4 3 3 3 1 41 17 16 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 37 41 17 15 47 47 47 47 47 47 47 47 47 47 47 47 47 47 39 31 13 20 21 21 21 21 21 21 21 21 21 21 21 21 21 21 21 21 21 21 21 4 3 3 6 2 2 2 2 1 0 5 6 4 8 8 8 6 5 2 3 4 4 2 3 2 9 1 1 1 1 1 1 0 0 0 3 4 10 2 2 2 2 2 2 0 0 0 0 2 3 9 1 1 1 1 1 1 0 0 0 6 2 13 1 1 1 1 1 1 1 1 1 1 1 1 0 4 5 10 2 2 2 2 2 2 2 2 2 2 This answers for tests right too. 5 5 6 5 4 4 4 4 4 2 5 4 3 3 2 2 41 17 16 44 44 44 44 44 44 44 44 44 43 43 43 43 43 43 43 41 17 15 47 47 47 47 47 47 47 46 46 46 46 46 46 46 46 31 13 20 21 21 21 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 3 3 6 2 2 2 1 1 1 5 6 4 8 8 7 7 5 2 3 4 3 3 3 2 9 1 1 1 1 1 1 0 0 0 3 4 10 2 2 1 1 1 1 1 1 1 1 2 3 9 1 1 1 1 1 1 0 0 0 | Complete and Correct solution of this Problem | Abbasi | 1756. Полтора землекопа | 18 мар 2012 15:19 | 1 | kindly provide me the correct solution of this in c#. | Hint | Anton | 1756. Полтора землекопа | 27 янв 2012 09:25 | 1 | Hint Anton 27 янв 2012 09:25 m * d1 is count of men needed to complete work in 1 day. You should divide this count into d2 days. On obvious way to do it with min men per day is to add men equally to d2 days one-by-one, while total men count is not 0. | No subject | Mehemmed Veliyev[Azerbaijan] | 1756. Полтора землекопа | 30 дек 2011 01:11 | 2 | No subject Mehemmed Veliyev[Azerbaijan] 30 дек 2011 01:05 I don't understand do i output 0 ,if on same days diggers don't work ? Edited by author 30.12.2011 01:10 | "It is possible that on some days nobody will work" | giorgi | 1756. Полтора землекопа | 24 окт 2012 19:06 | 3 | so in every case diggers can work only 1th day and after that just place 0? In most of the cases, no because the statement says you should finish the work with a minimal number of diggers I first do this just output m*d1, 0 but its wrong | WA #3 | Marat | 1756. Полтора землекопа | 19 апр 2012 00:21 | 6 | WA #3 Marat 21 июл 2011 17:35 Give me some tests, please. I think the reason for WA3 is in the following test 5 6 4 The evaluator accepts 8 8 7 7 but not 8 8 8 6. But why is 8 8 8 6 wrong. The max is still only 8. My program outputs 8 8 8 6 and got AC I had the same error. The mistake was in logic. In test 3 3 6 I had 2 2 2 2 1 1 but correct answer is 2 2 2 2 1 0 ,of course. I think you are right. but my program which WA#3 output 1 2 1 2 1 2 in test 3 3 6,I think it's right,too.... | If you have WA #15 | smolcoder | 1756. Полтора землекопа | 22 авг 2010 06:07 | 1 | 2 3 9 ans.: 1 1 1 1 1 1 0 0 0 NOT simple 1 1 1 1 1 1 | Clarification conditions | awpris | 1756. Полтора землекопа | 15 мар 2010 02:51 | 1 | Question removed Edited by author 15.03.2010 03:09 | WA #9 | Stankov | 1756. Полтора землекопа | 5 июл 2021 02:03 | 4 | WA #9 Stankov 13 мар 2010 15:21 I have Wa #9 please give some tests Re: WA #9 Oleg Strekalovsky [Vologda SPU] 26 апр 2010 19:57 Try to find minimum amount of peoples with complexity o(1) like that: if ((m * d1) % d2 == 0) { b = (m * d1) / d2; } else { b = (m * d1) / d2 + 1; } Edited by author 26.04.2010 19:58 i've changed for(int j=1; j<=i; ++j) cout<<floor(fr)<<' '; for(int j=i+1; j<=d2; ++j) cout<<ceil(fr)<<' '; to for(int j=1; j<=i; ++j) cout<<fixed<<setprecision(0)<<floor(fr)<<' '; for(int j=i+1; j<=d2; ++j) cout<<fixed<<setprecision(0)<<ceil(fr)<<' '; | Test | insi | 1756. Полтора землекопа | 7 май 2020 22:35 | 11 | Test insi 13 мар 2010 14:44 I have WA #2 .. Can somebody give me some tests, please ?:) Edited by author 13.03.2010 14:57 5 2 3 Ans: 3 3 4 5 6 4 Ans: 7 7 8 8 My program: 5 2 3 -> 1 1 8 5 6 4 -> 1 1 1 27 Are these answers correct ? find less diggers solution insi your answers are not correct. You need to keep an even distribution. Edited by author 27.09.2014 11:20 Edited by author 27.09.2014 11:21 Re: Test Yashar Abbasov 13 апр 2010 02:03 for test 5 6 4 is answer 7 7 7 9 correct? Edited by author 13.04.2010 02:05 Edited by author 13.04.2010 02:05 Re: Test dAFTc0d3r [Yaroslavl SU] 13 апр 2010 22:23 No. 8 8 7 7 id correct. Because max(7,7,7,9) = 9 > 8 = max(8,8,7,7) and we need to minimize that max. Also acceptable answer: 8 8 8 6 |
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