算法笔记非递归归并排序拓展题

int A[8] = {66,12,33,57,64,27,18};
int n = 6;
for(int step =2;step/2<=n;step *= 2)
{
for(int i=0; i<=n; i+=step)
{
int mid = min(i+step/2-1,n); //这里问题还蛮大的
if(mid<=n)
{
printf(“此时的参数为：%d%d%d\n”,i,mid,min(i+step-1,n));
test8(A,i,mid,mid+1,min(i+step-1,n));
}
}
for(int i = 0;i<=6;i++)
{
printf("%d “,A[i]);
}
printf(”\n");
}
for(int i = 0;i<=6;i++)
printf("%d ",A[i]);
return 0;