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43-liner: the other name for googol (10**100) is ten duotrigintillion solution in Creative category for Speech Module by przemyslaw.daniel
NUMBERS = {
0: 'zero',
1: "one", 10**12: "trillion",
2: "two", 10**15: "quadrillion",
3: "three", 10**18: "quintillion",
4: "four", 10**21: "sextillion",
5: "five", 10**24: "septillion",
6: "six", 10**27: "octillion",
7: "seven", 10**30: "nonillion",
8: "eight", 10**33: "decillion",
9: "nine", 10**36: "undecillion",
10: "ten", 10**39: "duodecillion",
11: "eleven", 10**42: "tredecillion",
12: "twelve", 10**45: "quattuordecillion",
13: "thirteen", 10**48: "quinquadecillion",
14: "fourteen", 10**51: "sedecillion",
15: "fifteen", 10**54: "septendecillion",
16: "sixteen", 10**57: "octodecillion",
17: "seventeen", 10**60: "novendecillion",
18: "eighteen", 10**63: "vigintillion",
19: "nineteen", 10**66: "unvigintillion",
20: "twenty", 10**69: "duovigintillion",
30: "thirty", 10**72: "tresvigintillion",
40: "forty", 10**75: "quattuorvigintillion",
50: "fifty", 10**78: "quinquavigintillion",
60: "sixty", 10**81: "sesvigintillion",
70: "seventy", 10**84: "septemvigintillion",
80: "eighty", 10**87: "octovigintillion",
90: "ninety", 10**90: "novemvigintillion",
100: "hundred", 10**93: "trigintillion",
10**3: "thousand", 10**96: "untrigintillion",
10**6: "million", 10**99: "duotrigintillion",
10**9: "billion", 10**100: "googol"
}
def checkio(number):
if number in NUMBERS and number < 100:
return NUMBERS[number]
k = next(k for k in sorted(NUMBERS)[::-1] if number//k)
a, b = divmod(number, k)
result = [checkio(a)]*(number >= 100)
result += [NUMBERS[k]]+[checkio(b)]*(b > 0)
return ' '.join(result)
Dec. 3, 2017
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