Python（斐波纳契数的位置）

Question

我已经找到用于查找斐波那契数字的代码。 如果输入的不是斐波那契数字，程序将打印出最接近的数字。 接下来，如何确定该数字在系列中的位置？ 就像n = 8一样，它的位置是7th，所以输出是7。我尝试使用index，但是python指出index不适用于整数

def fibs():
    a,b = 0,1
    yield a
    yield b
    while True:
        a,b = b,a+b
        yield b

def nearest_fib(n):
    # If n is a Fibonacci number return Yes and n
    # Otherwise, return No and the nearest Fibonacci number

    n = int(input("Please enter an integer:"))
    for fib in fibs():
        if fib == n:
            return print("Yes! your integer is an fibonacci number")
        elif fib < n:
            prev = fib
        else:
            # Is n closest to prev or to fib?
            if n - prev < fib - n:
                return print("No! Your answer is not a fibonacci number"), prev
            else:
                return print("No! Your answer is not a fibonacci number"), fib

for i in range(1):
    print(i,nearest_fib(i))

Answer 1

使用enumerate ，它将为您提供索引和值：

 for ind, fib in enumerate(fibs()):

将索引和值都存储在当前存储prev ：

def fibs():
    a,b = 0,1
    yield a
    yield b
    while True:
        a,b = b,a+b
        yield b

def nearest_fib(n):
    #If n is a Fibonacci number return Yes and n
    #Otherwise, return No and the nearest Fibonacci number

    for ind, fib in enumerate(fibs()):
        if fib == n:
            print("Yes! your integer is fibonacci number #%d" % (ind + 1) )
            return n
        elif fib < n:
            prev_index = ind
            prev = fib
        else:
            # Is n closest to prev or to fib?
            if n - prev < fib - n:
                print("No! Your answer is not a fibonacci number. It is closest to number #%d" % (prev_index + 1))
                return prev
            else:
                print("No! Your answer is not a fibonacci number. It is closest to number #%d" % (prev_index + 2))
                return fib

for i in range(10):
    n = int(input("Please enter an integer:"))
    print(n,nearest_fib(n))