斐波那契数列

f(3) =f(1)+f(2)
f(n+2)=f(n)+f(n+1)
f(1)=f(2)=1
当n=100时，求f(n)?
由题可知，这是一个斐波那契?

# -*- coding:utf-8 -*-


# 第一种
def fab(max):
    n, a, b = 0, 0, 1
    while n < max:
        yield b
        a, b = b, a+b
        n += 1

for i in fab(5):
    print i


# 第二种
def Fibonacci_Recursion_tool(n):
    if n <= 0:
        return 0
    elif n == 1:
        return 1
    else:
        return Fibonacci_Recursion_tool(n-1)+Fibonacci_Recursion_tool(n-2)


def Fibonacci_Recursion_list(n):
    lis = []
    for i in xrange(1, n+1):lis.append(Fibonacci_Recursion_tool(i))
    return lis
print Fibonacci_Recursion_list(5)

#第三种
resu_list = []
def fibo(n):
    for i in xrange(n):
        if i == 0 or i == 1:
            resu_list.append(i)
        else:
            resu_list.append(resu_list[i-1]+resu_list[i-2])
    return resu_list

print fibo(20)

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

#第四种
a = 0
b = 1
while b < 1000:
    print(b,end=',')
    a, b = b, a+b

#第五种
# -*- coding:utf-8 -*-

class Fib(object):
    def __init__(self):
        pass

    def __call__(self, num):
        a, b = 0, 1;
        self.l = []

        for i in range(num):
            self.l.append(a)
            a, b = b, a + b
        return self.l

    def __str__(self):
        return str(self.l)

    __repr__ = __str__


f = Fib()
print f(3)

[0, 1, 1]