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  • 递归

    1)python斐波那切数列1,1,2,3,5,8,13,21,…,编程求第n个斐波那契数列的数:

    def fbql(x):
        if x == 1 or x ==2:
            return 1
        else:
            return fbql(x-1)+fbql(x-2)

    2)n的阶乘(递归实现):

    def fact(n):
        if n == 0:
            return 1
        else:
            return n*fact(n-1)

    number = eval(input("请输入正整数:"))
    result = fact(number)
    print("{}的阶乘是{}".format(number,result))
    ##print("%d的阶乘是%d"%(number,result))

    3)n的阶乘(迭代实现):

    def fact1(n):
        if n == 0:
            return 1
        else:
            result = n
            for i in range(1,n):
                result = n*i
            return result

    number = eval(input("请输入正整数:"))
    result = fact(number)
    print("{}的阶乘是{}".format(number,result))

    4)n的阶乘和(用递归实现):

    def fact(n):
        if n == 0:
            return 1
        else:
            return n*fact(n-1)

    number = eval(input("请输入正整数:"))
    sum = 0
    for i in range (0,number+1):
        sum = sum + fact(i)
        print(i,fact(i),sum)
    print(sum)

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  • 原文地址:https://www.cnblogs.com/beiyue/p/10766605.html
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