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Common Boardlines = [] for line in stdin: for i in range(0, len(line.split())): lines.append(line.split()[i]) lines = [int(i) for i in lines] lines_rev = list(reversed(lines)) numbers = [] for i in range(0, len(lines_rev)): numbers.append(lines_rev[i]**0.5) for i in range(len(numbers)): print(f"{numbers[i]:.4f}") I've got many RE16 with using python 3.4. Did anybody else get it? How did you solve it? I don't know why , but I just stopped using division in my solution and got AC! a = int(input()) f = [] for i in range(a): s = int(input()) f.append(s) if f.count(3) >=1: print('None') else: if f.count(5) == len(f): print('Named') elif sum(f) / len(f) >=4.5: print('High') else: print('Common') Can anyone give tests data or some discription? Edited by author 01.11.2023 14:24 Test 1 is the first sample test from the problem statement. 1000000000 1000000000 300 answer: 97395891 Hello, I am trying this question, however, I am unable to understand how to construct the solution. My Dp is weak so I thought to practice here, however I am having a tough time. Can someone please help in describing how to start thinking about such problems? we have 12 hours on clocks If you have WA10, make sure you don't have 'I' as a card value instead of 'T'. For some reason it was very easy for me to mix these letters in the statement... u can search for intersections instead of looking for cutouts, just invert the segments of the second function use min max instead of a lot of conditions and you get a very simple code :) Simpler is to brute-force My idea is to reverse the operations of adding characters and doubling a string. I have an iterative algorithm in O(n * log(n)). At each iteration, I count dp - the length of the maximum substring ending with index i, obtained by the doubling operation. Then I look for the maximum among the maximum substrings and truncate the current string to leave this substring. But this is a greedy solution. Is it possible to come up with a counterexample to it? admins, could you pls fix this? My first attempt use string hashing to check the palindrome with Segment Tree lazy prop. O(Q*K*logK*logN), esp. for the update query and it's TLE 11 However, using Manacher algorithm with Segment Tree you can achieve O(Q*(K+lgN)) Little help: - WA#3 : You're likely answering too much. (K involved) - WA#9 : N=10^5, overflow somewhere Help! What is wrong? [code deleted] Edited by moderator 05.11.2023 02:38 maybe this test can halp you? 5 00000 01111 01111 01110 00000 I also had WA 16, turns out this was the reason: 5 11100 10000 00011 11000 00000 In this test, i assumed that you can move away parts with 1's, thus separating 0's from everything. But it turns out to be wrong, you're only allowed to move zeros, and ones stay fixed in place. var N: integer; begin readln(n); if (abs(n)>10000) then writeln('Îøèáêà ââîäà') else if N>=0 then writeln(((1+n)*n)/2) else writeln(((1+n)*(abs(n)+2))/2); end. var a: integer; b: real; begin readln(a); if (abs(a) < 10001) then begin if (a < 0) then b := (a + 1) * (-a + 2) / 2 else b := 0; if (a > 0) then b := (a + 1) * a / 2 else if (a < 0) then b := (a + 1) * (-a + 2) / 2 else b := 1; writeln(b); end; end. Я так понимаю, ты используешь тип real для b, потому что / не работает. Следует помнить, что для целочисленного деления нужно использовать div, а не /. То есть, вариант 1 — заменить b: real на b: integer и все / на div. Вариант 2 — можно продолжать использовать /, но вместо writeln(b) следует написать writeln(b:0:0) (второе :0 — количество знаков после запятой для вывода). Также, проверки типа if (abs(a) < 10001) не нужны — если в задаче сказано о таком ограничении на входные данные, значит в тестах так честно и будет, и не нужно это проверять. А у меня тоже такая же проблема: var N: integer; begin readln(n); if (abs(n)>10000) then writeln('Îøèáêà ââîäà') else if N>=0 then writeln(((1+n)*n)/2) else writeln(((1+n)*(abs(n)+2))/2); end. IS that wrong answer? Read article: convex hull of n*(n-1)/2 intersection points of n lines Implementation is very easy. But the proof is cool to read. spell the months correctly |
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