输出求解akm（2，1）的递推和求值过程

/*exp6-1.cpp*/

#include<stdio.h>

int count=1;

int akm(int m,int n)

{

int g,f;

printf("\t akm(%d,%d)",m,n);

if(count++ % 3==0) printf("\n");/*每行输入3项*/

if(m==0)

{

printf("\t akm(%d,%d)=%d",0,n,n+1);

if(count++ %3==0) printf("\n");/*每行输出3项*/

return (n+1);

}

else if(n==0)

{

f=akm(m-1,1);

printf("\t akm(%d,%d)=%d",m-1,1,f);

if(count++ %3==0) printf("\n");/*每行输出3项*/

return f;

}

else

{

g=akm(m,n-1);

f=akm(m-1,g);

printf("\t akm(%d,%d)=%d",m-1,g,f);

if(count++ %3==0) printf("\n");/*每行输出3项*/

return f;

}

}

void main()

{

int m=2,n=1;

printf("求akm(%d,%d)的过程如下：\n",m,n);

printf("\t akm(%d,%d)=%d\n",m,n,akm(2,1));

}