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SHENZHEN I/O >  创意工坊 > shigawire 的创意工坊
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DIV 2
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2024 年 9 月 25 日 上午 5:38
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shigawire
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DIV 2

描述
Just divide a number by 2.

I mean how hard could it be?
10 条留言
< >
Gilgamesh 8 月 18 日 下午 3:03 
15/4628/33:
https://psteamcommunity.yuanyoumao.com/sharedfiles/filedetails/?id=3551687440
Moose 6 月 6 日 下午 9:34 
My solution: https://psteamcommunity.yuanyoumao.com/sharedfiles/filedetails/?id=3494555087
Moose 6 月 6 日 下午 9:30 
Got 8/705/19 trivially, I suspect you can use dst to increase throughput but it remains to be seen whether its actually more efficient giventhe pre-processing necessary
NoobDancer100 6 月 2 日 上午 7:56 
I got 8/70.6k/15 not particually good (especially compared to other people) but easy to understand solution.
Verdammte Heinz 2024 年 11 月 6 日 上午 11:34 
Optimized the low power version for costs to 8 / 705 / 19 . As I could not optimize it further checked the solution from Shigawire, nice one! I only considered taking two lowest digits and multiply those at the same time, but did not think about taking the two higher ones.
Verdammte Heinz 2024 年 11 月 5 日 下午 7:57 
Solved in 6 / 70603 / 12 and 10 / 668 / 18
Scoptlie 2024 年 10 月 14 日 上午 8:33 
@shigawire Oh that's a very clever way to do it! Using the simple IO pin basically for extra storage is something I wish I had considered as I was stuck on eliminating the need for the dat register . I much appreciate your sharing, and thanks for the fun puzzle.
shigawire  [作者] 2024 年 10 月 11 日 上午 7:21 
@Scoptlie : Mostly it relies on using simple IO to make sharing intermediate values cheap between 2 MC4000s. and that the instructions:
mov <src> acc
dgt 0
can be replaced with
dst 0 <src>
if you know that acc is less than 10

(Screenshot of the 6/631/17 solution )
Scoptlie 2024 年 10 月 9 日 下午 12:05 
This one is driving me mad. The best I can manage after many attempts is 8/705/19 . I'd be curious to see the 6/631/17 solution.
shigawire  [作者] 2024 年 9 月 26 日 上午 7:45 
So far I have 6/631/17 or 6/50511/13