直接用 Python 自带的 int 就好，这玩意自带高精。
by SlaineTroyard @ 2024-07-24 01:24:21
@[ironwaste97213](/user/777207) 你这代码全RE啊
by Maisie586_ @ 2024-07-24 07:32:20
@[Maisie586_](/user/1362512) 贴错代码了ww这个能过60
```python
import numpy as np
n = (int)(input());
# print(n)
# n = long(n)
res = np.eye(2);
# print(res)
n -= 1
a = np.array([[1,1],[1,0]])
# print(a)
while(n) :
if (n & 1) : res = np.dot(res,a)
# print(a)
# print(n)
a = np.dot(a,a)
n>>=1
# print(res)
print("%d"%(res[0][0] + res[1][0]))
```
by ironwaste97213 @ 2024-07-24 12:17:58
@[SlaineTroyard](/user/450246) 确实但是我想用矩阵快速幂来搞这个，但是好像numpy的库里面最高支支持uint32或者float32，就用矩阵的dot去写就写不出来了，我想知道矩阵库能不能有这种直接高精的方式？
by ironwaste97213 @ 2024-07-24 12:19:38
@[ironwaste97213](/user/777207) 求关（DP+快速幂
```python
def climb_stairs(n):
if n <= 2:
return n
dp = [0] * (n + 1)
dp[0], dp[1] = 1, 1
for i in range(2, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]
return dp[n]
n = int(input())
print(climb_stairs(n))
```
by Maisie586_ @ 2024-07-24 12:21:38
@[Maisie586_](/user/1362512)
啊？ 这也没有dp和矩阵快速幂啊
by ironwaste97213 @ 2024-07-24 20:26:17