Python3: Fizz Buzz (paizaランク C 相当)やってみた

分岐の問題なんですが、単純な繰り返しでもある。

def fizzbuzz_gen():
    F = "Fizz"
    B = "Buzz"
    FB = "Fizz Buzz"
    i = 0
    while True:
        yield from (
            i+1, i+2, F, i+4, B
            , F, i+7, i+8, F, B
            , i+11, F, i+13, i+14, FB
            )
        i += 15

fb = fizzbuzz_gen()

N = int(input())

for _ in range(N):
    print(next(fb))

分岐なしでやってみました。無駄な演算をなるべく減らしてみます。

unbox = lambda *xs: xs[-1]

def count_gen(i):
    while True:
        yield i
        i += 1

def fizzbuzz_gen():
    F = "Fizz"
    B = "Buzz"
    FB = "Fizz Buzz"
    c = count_gen(1)
    while True:
        yield from (
            next(c), next(c), unbox(next(c), F)
            , next(c), unbox(next(c), B), unbox(next(c), F)
            , next(c), next(c), unbox(next(c), F)
            , unbox(next(c), B), next(c), unbox(next(c), F)
            , next(c), next(c), unbox(next(c), FB)
            )
        
fb = fizzbuzz_gen()

N = int(input())

for _ in range(N):
    print(next(fb))

あるいは：

def count_gen(i):
    while True:
        yield i
        i += 1

def fizz_gen():
    c = count_gen(1)
    while True:
        for _ in range(1, 3):
            yield next(c)
        next(c)
        yield "Fizz"

def fizzbuzz_gen():
    f = fizz_gen()
    
    while True:
        for _ in range(1, 3):
            for _ in range(1, 5):
                yield next(f)
                
            next(f)
            yield "Buzz"
            
        for _ in range(1, 5):
            yield next(f)
            
        next(f)
        yield "Fizz Buzz"


fb = fizzbuzz_gen()

N = int(input())

for _ in range(N):
    print(next(fb))

現物を並べただけだとおばかっぽく見えるので。
普段はFizzのジェネレーターの値を採用して、Buzz、FizzBuzzのときだけ "Buzz"、"Fizz Buzz" を返す、って感じです。