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English cardinal anagrams: Difference between revisions

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English cardinal anagrams (view source)

Revision as of 11:17, 5 May 2023

1,271 bytes removed ,  1 year ago
→‎{{header|Wren}}: 'Numbers to names' routines now built into 'Fmt' module.
m (→‎{{header|Phix}}: sort less)
(→‎{{header|Wren}}: 'Numbers to names' routines now built into 'Fmt' module.)
Line 400:
{{libheader|Wren-sort}}
{{libheader|Wren-fmt}}
I've reused the code for determining the number names from the related task.
<syntaxhighlight lang="ecmascript">import "./sort" for Sort
import "./fmt" for Fmt, Name
 
var small = [
"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven",
"twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"
]
 
var tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
 
var illions = ["", " thousand", " million", " billion"," trillion", " quadrillion", " quintillion"]
 
var say
say = Fn.new { |n|
var t = ""
if (n < 0) {
t = "negative "
n = -n
}
if (n < 20) {
t = t + small[n]
} else if (n < 100) {
t = t + tens[(n/10).floor]
var s = n % 10
if (s > 0) t = t + "-" + small[s]
} else if (n < 1000) {
t = t + small[(n/100).floor] + " hundred"
var s = n % 100
if (s > 0) t = t + " " + say.call(s)
} else {
var sx = ""
var i = 0
while (n > 0) {
var p = n % 1000
n = (n/1000).floor
if (p > 0) {
var ix = say.call(p) + illions[i]
if (sx != "") ix = ix + " " + sx
sx = ix
}
i = i + 1
}
t = t + sx
}
return t
}
 
for (limit in [1000, 10000]) {
var ana = {}
for (i in 0..limit) {
var key = Sort.quick(sayName.callfromNum(i).toList).join("")
if (ana.containsKey(key)) {
ana[key].add(i)
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