Anonymous user
Sum multiples of 3 and 5: Difference between revisions
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→{{header|Tcl}}: fix fencepost error
m (→{{header|Tcl}}: fix fencepost error) |
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=={{header|Tcl}}==
<lang tcl># Fairly simple version; only counts by 3 and 5, skipping intermediates
proc mul35sum {n} {
for {set total [set threes [set fives 0]]} {$threes<
if {$threes<$fives} {
incr total $threes
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return $total
}</lang>
However, that's pretty dumb. We can do much better by observing that the sum of the multiples of <math>k</math> below some <math>n+1</math> is <math>k T_{n/k}</math>, where <math>T_i</math> is the <math>i</math>'th [[wp:Triangular number|triangular number]], for which there exists a trivial formula. Then we simply use an overall formula of <math>3T_{n/3} + 5T_{n/5} - 15T_{n/15}</math> (that is, summing the multiples of three and the multiples of five, and then subtracting the multiples of 15 which were double-counted).
<lang tcl># Smart version; no iteration so very scalable!
proc tcl::mathfunc::triangle {n} {expr {
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}}
# Note that the rounding on integer division is exactly what we need here.
proc sum35 {n}
incr n -1
expr {3*triangle($n/3) + 5*triangle($n/5) - 15*triangle($n/15)}
Demonstrating:
<lang tcl>puts [mul35sum 1000],[sum35 1000]
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