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Number names: Difference between revisions
→{{header|Python}}
(→{{header|Kotlin}}: Updated example see https://github.com/dkandalov/rosettacode-kotlin for details) |
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Line 4,279:
-2 -> minus two
0 -> zero</pre>
An alternative solution that can name very large numbers.
<lang python>
def int_to_english(n):
if n < 0: return "minus " + int_to_english(-n)
if n < 10:
return ["zero", "one", "two", "three", "four", "five",
"six", "seven", "eight", "nine"][n]
if n < 20:
return ["ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",
"sixteen", "seventeen", "eighteen", "nineteen"][n-10]
if n < 100:
tens = ["twenty", "thirty", "forty", "fifty", "sixty",
"seventy", "eighty", "ninety"][(n // 10 - 2)%10]
if n % 10 != 0:
return tens + "-" + int_to_english(n % 10)
else:
return tens
if n < 1000:
if n % 100 == 0:
return int_to_english(n // 100) + " hundred"
else:
return int_to_english(n // 100) + " hundred and " +\
int_to_english(n % 100)
# http://www.isthe.com/chongo/tech/math/number/tenpower.html
powers = [("thousand", 3), ("million", 6),
("billion", 9), ("trillion", 12), ("quadrillion", 15),
("quintillion", 18), ("sextillion", 21), ("septillion", 24),
("octillion", 27), ("nonillion", 30), ("decillion", 33),
("undecillion", 36), ("duodecillion", 39), ("tredecillion", 42),
("quattuordecillion", 45), ("quindecillion", 48),
("sexdecillion", 51), ("eptendecillion", 54),
("octadecillion", 57), ("novemdecillion", 61),
("vigintillion", 64)]
ns = str(n)
idx = len(powers) - 1
while True:
d = powers[idx][1]
if len(ns) > d:
first = int_to_english(int(ns[:-d]))
second = int_to_english(int(ns[-d:]))
if second == "zero":
return first + " " + powers[idx][0]
else:
return first + " " + powers[idx][0] + " " + second
idx = idx - 1
if __name__ == "__main__":
print(int_to_english(42))
print(int_to_english(3 ** 7))
print(int_to_english(2 ** 100))
print(int_to_english(10 ** (2*64)))
</lang>
{{out}}
<pre>
forty-two
two thousand one hundred and eighty-seven
one nonillion two hundred and sixty-seven octillion six hundred and fifty septillion six hundred sextillion two hundred and twenty-eight quintillion two hundred and twenty-nine quadrillion four hundred and one trillion four hundred and ninety-six billion seven hundred and three million two hundred and five thousand three hundred and seventy-six
one vigintillion vigintillion
</pre>
=={{header|Racket}}==
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