题解 | #跳台阶扩展问题#
跳台阶扩展问题
https://www.nowcoder.com/practice/953b74ca5c4d44bb91f39ac4ddea0fee
""" f(n) = f(n-1) + f(n-2)+......+f(0) f(n-1) = f(n-2) + f(n-3)+......+f(0) 所以可得： f(n) = 2 * f(n-1) """ n = int(input()) dp = [0 for i in range(n + 1)] dp[1] = 1 for i in range(2, n + 1): dp[i] = 2 * dp[i - 1] print(dp[n])
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