It seems that unary '+' must be removed before the 1st output.
Int the 3rd and 4th example
2 * -3 + -6 - +4 = r
->
2 * -3 + -6 - 4 = r
If you parse the string and store the ops this won't be a problem. But I use a quite dumb algorithm in my code(find a operator and then scan forward and backward ...
Search found 4 matches
- Thu Nov 07, 2013 11:48 am
- Forum: Volume 3 (300-399)
- Topic: 397 - Equation Elation
- Replies: 8
- Views: 5841
- Wed Oct 09, 2013 9:21 pm
- Forum: Volume 118 (11800-11899)
- Topic: 11889 - Benefit
- Replies: 27
- Views: 14854
Re: 11889 - Benefit
There is no need to factor a or c.
c % a != 0 -> NO SOLUTION
factoring gcd(c/a, a) is enough
c % a != 0 -> NO SOLUTION
factoring gcd(c/a, a) is enough
- Sun Oct 06, 2013 7:49 pm
- Forum: Volume 101 (10100-10199)
- Topic: 10142 - Australian Voting
- Replies: 82
- Views: 45787
Re: 10142 - Australian Voting
It seems that the verdict result is inaccurate -- TLE could be judged as RE.
I got lots of RE first and I moved to a smarter way, then got AC.
But I don't think my first solution will cause a runtime error.
I got lots of RE first and I moved to a smarter way, then got AC.
But I don't think my first solution will cause a runtime error.
- Thu Oct 03, 2013 10:29 pm
- Forum: Volume 7 (700-799)
- Topic: 785 - Grid Colouring
- Replies: 27
- Views: 18564
Re: 785 - Grid Colouring
Hello everyone:
I got RE with this problem the first time and then I enlarge the array of std::string and got AC.
So I think the really tricky part of this problem is that the blank line above and under the grid. They are not counted within the 30 lines. So an array of 30 rows could result in RE.
By ...
I got RE with this problem the first time and then I enlarge the array of std::string and got AC.
So I think the really tricky part of this problem is that the blank line above and under the grid. They are not counted within the 30 lines. So an array of 30 rows could result in RE.
By ...