题解 | #牛牛的Ackmann#
牛牛的Ackmann
https://www.nowcoder.com/practice/3a7a4c26420c4358a1a5cda3da2fa1c8
#include <stdio.h>
int Ackmann(int m, int n) {//这里最蠢的就是n和m交换了位置
if (m == 0)
return n + 1;
if (m > 0 && n == 0)
return Ackmann(m - 1, 1);
else
return Ackmann(m - 1, Ackmann(m, n - 1));
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
printf("%d", Ackmann(m, n));
return 0;
}
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