递归小题中的空间换时间思想

#include <stdio.h>

int bian(int num);
//static int shu[100]={1,1};

int main()
{
    int num;
    while ( printf("请输入编号数:"), scanf("%d",&num)==1 )
    {
        printf("序号%d的数字:%d\n",num, bian(num));
    }
    return 0;
}


/*重复计算了太多步骤，当序号较大时耗时太多*/
/*
int bian( int num )
{
    if (num <= 2)
    {
        return 1;    
    }
    else
    {
        return bian(num-1)+bian(num-2);
    }
    return 0;
}
*/

/*优化后，将计算过的数值保存起来，用空间换时间的思想*/
int bian( int num )
{
      static int shu[100]={1,1};//注意static
    if (!shu[num])
    {
        shu[num]= bian(num-1)+bian(num-2);
        return shu[num];
    }
    else
    {
        return shu[num];
    }
    return 0;
}