Как склонять по числу?

Question

[RandomNick228]

нужно типо такого
https://pastebin.com/d6G9BiCt

Answer 1

[soremix]

ok google, python склонение числительных, первая ссылка

def conv(n): 
    es = ['а', 'ы', '']
    n = n % 100
    if n>=11 and n<=19:
        s=es[2] 
    else:
        i = n % 10
        if i == 1:
            s = es[0] 
        elif i in [2,3,4]:
            s = es[1] 
        else:
            s = es[2] 
    return s 

for i in range(1,150):
    print('{} минут{}'.format(i, conv(i)))

Answer 2

[Василий Банников]

def pluralize(count: int, one: str, two: str, many: str) -> str:
	significant = count % 100
	if 4 < significant < 21:
		return many
	last = significant % 10
	if last == 1:
		return one
	elif last == 2 or last == 3 or last == 4:
		return two
	else:
		return many