题解 | 牛牛的Ackmann
牛牛的Ackmann
https://www.nowcoder.com/practice/3a7a4c26420c4358a1a5cda3da2fa1c8
import sys
sys.setrecursionlimit(1000000) # python得限制递归深度
n, m = map(int, input().split(' '))
def ackmann(m,n):
if m == 0:
return n+1
elif m > 0 and n == 0:
return ackmann(m-1,1)
elif m > 0 and n > 0:
return ackmann(m-1,ackmann(m,n-1))
print(ackmann(m,n))
