记忆化搜索
兔子的数量
https://www.nowcoder.com/practice/8783056676de4396b0bf816a3561d62f
m = {}
def f(x):
if x == 1:
return 2
if x == 2:
return 3
if m.get(x, 0) != 0:
return m[x]
m[x] = f(x - 1) + f(x - 2)
return m[x]
print(f(int(input())))
斐波那契数列更快的解法：矩阵快速幂（逃
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