Sum multiples of 3 and 5: Difference between revisions
Odin version
(Add TI SR-56 solution) |
(Odin version) |
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</pre>
=={{header|Odin}}==
Note: 1e19 is the largest sum that can be calculated with 128 bit integers.
<syntaxhighlight lang="odin">
package main
import "core:fmt"
sumdiv :: proc(n, d: i128) -> i128 {
m := n / d
return (m % 2 == 0)? \
m/2 * (m + 1) * d : \
(m + 1)/2 * m * d
}
sum3or5 :: proc(n: i128) -> i128 {
return sumdiv(n, 3) + sumdiv(n, 5) - sumdiv(n, 15)
}
main :: proc() {
sum := 0
for n in 1..=999 {
if n % 3 == 0 || n % 5 == 0 {
sum += n
}
}
fmt.println("The sum of all multiples of 3 and 5 < 1000 is", sum)
fmt.println("The sum of all multiples of 3 and 5 < 1e19 is", sum3or5(1e19 - 1))
}
</syntaxhighlight>
{{Out}}
<pre>
The sum of all multiples of 3 and 5 < 1000 is 233168
The sum of all multiples of 3 and 5 < 1e19 is 23333333333333333331666666666666666668
</pre>
=={{header|OCaml}}==
<syntaxhighlight lang="ocaml">let sum_m3m5 n =
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