请补充以下代码：def fib(n): a = 0 b = 1 dic = _____ for i in _____: dic[i]=_____ a,b=b,a+b return ______ fibonac = ______ for key in fibonac: print(fibonac[key], end=",")

答案：

def fib(n): a = 0 b = 1 dic = {} for i in range(n): dic[i]=a a,b=b,a+b return dic fibonac = fib(10) for key in fibonac: print(fibonac[key], end=",")

输出结果为：0,1,1,2,3,5,8,13,21,34,

相关问题

请补充一下代码：def fib(n): a = 0 b = 1 dic = ___(1)___ for i in ___(2)___: dic[i]=___(3)___ a,b=b,a+b return ___(4)___ fibonac = ___(5)___ for key in fibonac: print(fibonac[key], end=",")

def fib(n):
a = 0
b = 1
dic = {1: a, 2: b}
for i in range(3, n + 1):
dic[i] = a + b
a, b = b, a + b
return dic

fibonac = fib(10)
for key in fibonac:
print(fibonac[key], end=",")

请补充下面的代码def fib(n): a = 0 b = 1 dic = ___(1)___ #定义字典 for i in ___(2)___: dic[i]=___(3)___ #把添加斐波那契数到字典 a,b=b,a+b return ___(4)___ #调用函数生成斐波那契数列中的前20个斐波那契数 fibonac = ___(5)___ for key in fibonac: print(fibonac[key], end=",")

请补充下面的代码：

def fib(n):
a = 0
b = 1
dic = {0:0, 1:1} # 表示斐波那契数列的前两项
...

代码中应该补充出现超过前两项的斐波那契数列项数的字典值，可以优化斐波那契数列运算效率，避免重复计算。