具有stdin和stdout的FizzBu​​zz python代码

如何解决具有stdin和stdout的FizzBu​​zz python代码

我使用FizzBu​​zz程序进行了在线面试测试。编写一个程序，打印给定范围内的数字。但是对于三倍打印“ Fizz”而不是数字，对于五倍打印“ Buzz”。对于三和五的倍数的数字，请打印“ FizzBu​​zz。在每个字符串或数字之后打印新行。 输入格式：第一行将是测试用例的数量T。下一行将有N表示的T个整数。 输出格式：对于每个测试用例，请打印从1到N的数字，但请遵循问题中给出的规则。 约束：1

N= int(input())
for fizzbuzz in range(N+1):
    if fizzbuzz % 3 == 0 and fizzbuzz % 5 == 0:
        print("fizzbuzz")
        continue
    elif fizzbuzz % 3 == 0:
        print("fizz")
        continue
    elif fizzbuzz % 5 == 0:
        print("buzz")
        continue
    print(fizzbuzz)

解决方法

T = input("Enter some values: ")
value_list = str.split(T)

for value in value_list:
    N = int(value)
    for number in range(0,N+1):
        if number % 15 == 0:
            print("FizzBuzz")
        elif number % 5 == 0:
            print("Buzz")
        elif number % 3 == 0:
            print("Fizz")
        else:
            print(number)

输出：

Enter some values: 2 5 17 80
FizzBuzz
1
2
FizzBuzz
1
2
Fizz
4
Buzz
FizzBuzz
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
FizzBuzz
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz
Fizz
22
23
Fizz
Buzz
26
Fizz
28
29
FizzBuzz
31
32
Fizz
34
Buzz
Fizz
37
38
Fizz
Buzz
41
Fizz
43
44
FizzBuzz
46
47
Fizz
49
Buzz
Fizz
52
53
Fizz
Buzz
56
Fizz
58
59
FizzBuzz
61
62
Fizz
64
Buzz
Fizz
67
68
Fizz
Buzz
71
Fizz
73
74
FizzBuzz
76
77
Fizz
79
Buzz

,

最好的方法

c3 = 0
c5 = 0

for i in range(1,101):
    c3 += 1
    c5 += 1
    d = ""
    if (c3 == 3):
        d += "fizz"
        c3 = 0
    if (c5 == 5):
        d+= "buzz"
        c5 = 0
    if (d == ""):
        print(i)
        
    else:
        print(d)