变态青蛙跳

一次可以跳一个，也可以跳n个

思考：在dp[n] = dp[n-1] + dp[n-2] + .. + dp[1] + 1(直接跳n)步骤
即dp[n]=∑n−1i=1dp[i]+1
class Solution:
    def find_ways(self,number):
        if number==1 or number ==2:
            return number
        ret=sum_=3
        for i in range(number-2):
            ret=sum_+1
            sum_+ =ret

        return ret