题解 | #跳台阶扩展问题#
跳台阶扩展问题
https://www.nowcoder.com/practice/953b74ca5c4d44bb91f39ac4ddea0fee
import sys # for line in sys.stdin: # a = line.split() # print(int(a[0]) + int(a[1])) n = int(input()) dp =[0 for i in range(n+1)] # dp[0]=1 dp[0]=1 #递推公式是i个台阶的跳法是i-1的2倍数,初始值i=0为第一个台阶dp为1,所以n-1表示第n个台阶 for i in range(1,n): dp[i] = 2*dp[i-1] print(dp[n-1])