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First power of 2 that has leading decimal digits of 12: Difference between revisions
First power of 2 that has leading decimal digits of 12 (view source)
Revision as of 11:54, 4 March 2020
, 4 years agoAdd Swift
(Add Swift) |
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Line 872:
p(12345, 10000) = 284166722
</pre>
=={{header|Swift}}==
{{trans|Go}}
===Slower Version===
<lang swift>let ld10 = log(2.0) / log(10.0)
func p(L: Int, n: Int) -> Int {
var l = L
var digits = 1
while l >= 10 {
digits *= 10
l /= 10
}
var count = 0
var i = 0
while count < n {
let rhs = (Double(i) * ld10).truncatingRemainder(dividingBy: 1)
let e = exp(log(10.0) * rhs)
if Int(e * Double(digits)) == L {
count += 1
}
i += 1
}
return i - 1
}
let cases = [
(12, 1),
(12, 2),
(123, 45),
(123, 12345),
(123, 678910)
]
for (l, n) in cases {
print("p(\(l), \(n)) = \(p(L: l, n: n))")
}</lang>
{{out}}
<pre>p(12, 1) = 7
p(12, 2) = 80
p(123, 45) = 12710
p(123, 12345) = 3510491
p(123, 678910) = 193060223</pre>
===Faster Version===
<lang swift>import Foundation
func p2(L: Int, n: Int) -> Int {
let asString = String(L)
var digits = 1
for _ in 1...18-asString.count {
digits *= 10
}
let ten18 = Int(1e18)
var count = 0, i = 0, probe = 1
while true {
probe += probe
i += 1
if probe >= ten18 {
while true {
if probe >= ten18 {
probe /= 10
}
if probe / digits == L {
count += 1
if count >= n {
count -= 1
break
}
}
probe += probe
i += 1
}
}
let probeString = String(probe)
var len = asString.count
if asString.count > probeString.count {
len = probeString.count
}
if probeString.prefix(len) == asString {
count += 1
if count >= n {
break
}
}
}
return i
}
let cases = [
(12, 1),
(12, 2),
(123, 45),
(123, 12345),
(123, 678910)
]
for (l, n) in cases {
print("p(\(l), \(n)) = \(p2(L: l, n: n))")
}</lang>
{{out}}
Same as before.
=={{header|zkl}}==
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