问题
Here is a different approach for the Project Euler #1 solution:
+/~.(3*i.>.1000%3),5*i.>.1000%5
How to refactor it?
回答1:
[:+/@~.@,3 5([*i.@>.@%~)]
usage example:
f =: [:+/@~.@,3 5([*i.@>.@%~)]
f 1000
or
+/~.,3 5([*i.@>.@%~)1000
%~ = 4 : 'y % x'
i.@>.@%~ = 4 : 'i. >. y % x'
[*i.@>.@%~ = 4 : 'x * i. >. y % x'
3 5([*i.@>.@%~)] = 3 : '3 5 * i. >. y % 3 5'
[:+/@~.@,3 5([*i.@>.@%~)] = 3 : '+/ ~. , 3 5 * i. >. y % 3 5'
回答2:
+/(#~ ( (0= 3| ]) +. (0 = 5 |]) )) 1+i.999
0 = ( 3 | ]) uses (twice) the trick of verb train (fork) with n u v (discussed at the end of http://www.jsoftware.com/help/learning/09.htm)
A different way of writing it:
+/(#~ ( ((0&=) @ (3&|)) +. ((0&=) @ (5&|)))) 1+i.999
回答3:
Here is another approach, using a simple, generic verb
multiplesbelow =: 4 : 'I. 0 = x | i.y'
+/ ~. ,3 5 multiplesbelow"0 [ 1000
来源：https://stackoverflow.com/questions/1557522/how-to-refactor-this-in-j