题解 | 牛牛的Ackmann

#include <stdio.h>
int Ackman (int m, int n) {

    if (m == 0) {
        return n + 1;
    } else if (m > 0 && n == 0) {
        return Ackman(m - 1, 1);
    } else {
        return Ackman(m - 1, Ackman(m, n - 1));
    }
}


int main() {
    /* code */
    int m, n;
    scanf("%d %d", &n, &m);

    printf("%d", Ackman(m,n));

    return 0;
}