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])
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