Abundant odd numbers: Difference between revisions
no edit summary
No edit summary |
|||
Line 6,395:
First abundant odd number > 1 000 000 000:
1000000575 proper divisor sum: 1083561009
</pre>
=={{header|Vlang}}==
{{trans|go}}
<lang vlang>fn divisors(n i64) []i64 {
mut divs := [i64(1)]
mut divs2 := []i64{}
for i := 2; i*i <= n; i++ {
if n%i == 0 {
j := n / i
divs << i
if i != j {
divs2 << j
}
}
}
for i := divs2.len - 1; i >= 0; i-- {
divs << divs2[i]
}
return divs
}
fn sum(divs []i64) i64 {
mut tot := i64(0)
for div in divs {
tot += div
}
return tot
}
fn sum_str(divs []i64) string {
mut s := ""
for div in divs {
s += "${u8(div)} + "
}
return s[0..s.len-3]
}
fn abundant_odd(search_from i64, count_from int, count_to int, print_one bool) i64 {
mut count := count_from
mut n := search_from
for ; count < count_to; n += 2 {
divs := divisors(n)
tot := sum(divs)
if tot > n {
count++
if print_one && count < count_to {
continue
}
s := sum_str(divs)
if !print_one {
println("${count:2}. ${n:5} < $s = $tot")
} else {
println("$n < $s = $tot")
}
}
}
return n
}
const max = 25
fn main() {
println("The first $max abundant odd numbers are:")
n := abundant_odd(1, 0, 25, false)
println("\nThe one thousandth abundant odd number is:")
abundant_odd(n, 25, 1000, true)
println("\nThe first abundant odd number above one billion is:")
abundant_odd(1_000_000_001, 0, 1, true)
}</lang>
{{out}}
<pre>
Same as Go entry
</pre>
| |||