1. 示例代码：

/* 用递归调用求 Ackerman(m, n) 的值 */

#include <stdio.h>

int Ackerman(int m, int n);

int main(void)
{
	int m = 0, n = 0;
	printf("Please input m and n: ");
	scanf_s("%d%d", &m, &n);
	printf("Ackerman(%d, %d) = %d\n", m, n, Ackerman(m, n));

	return 0;
}

int Ackerman(int m, int n)
{
	int z = 0;
	if (m == 0) {
		z = n + 1;
	} else if (n == 0) {
		z = Ackerman(m - 1, 1);
	} else {
		z = Ackerman(m - 1, Ackerman(m, n - 1));
	}

	return z;
}

2. 输出示例：