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Sum multiples of 3 and 5: Difference between revisions
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(COBOL: Added triangular number solution) |
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NB. continue...
</lang>
We cheat (or not), stealing the simplification from the python implementation,
<lang J>
first =: 0&{
last =: first + skip * <.@:(skip %~ <:@:(1&{) - first)
skip =: 2&{
terms =: >:@:<.@:(skip %~ last - first)
sum_arithmetic_series =: -:@:(terms * first + last) NB. sum_arithmetic_series FIRST LAST SKIP
(0,.10 10000 10000000000000000000x)(,"1 0"1 _)3 5 15x NB. demonstration: form input vectors for 10, ten thousand, and 1*10^(many)
0 10 3
0 10 5
0 10 15
0 10000 3
0 10000 5
0 10000 15
0 10000000000000000000 3
0 10000000000000000000 5
0 10000000000000000000 15
(0,.10 10000 10000000000000000000x)+`-/"1@:(sum_arithmetic_series"1@:(,"1 0"1 _))3 5 15x
23 23331668 23333333333333333331666666666666666668
</lang>
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