斐波那契数列问题-求余数-递归-分治python

n=int(input())
#算n的余数就是算n-1与n-2的余数之和
def f(n):
    a=1
    b=1
    for i in range(n-1):
        a,b=b,(a+b)%10007
    print(a)
f(n)

求字符串在fn出现的次数

n=int(input())
s=input()

def f(num):
    if num == 0 or num ==1:
        return str(num)
    else:
        return f(num-1)+f(num-2)

f1=f(n)
print(f1.count(s))