斐波那契数列

 #include 
#include
#include
using namespace std;
const int N=200;    
int s1[N]={1},s2[N]={1},n;
int i=0,j=3,count=2;
int Fab(int s1[],int s2[],int i,int n,int j)
{
    if(n==1)    
    {
        cout<<"1";
        return 0;
    }
    else if(n==2)      
    {
        cout<<"2";
        return 0;
    }
    else       
    {
            int s3[N],k;
            for(k=0;k<=i;k++)       
            {
                s3[k]=s1[k];
                s1[k]=s1[k] + s2[k];
                s2[k]=s3[k];
            }
            int l=0,b;
            for(l=0;l<=i;l++)     
            	while(s1[l]>=10)
                {
                    b=s1[l]/10;
                    s1[l+1]=s1[l+1]+b;
                    s1[l]=s1[l]-b*10;
                    l++;
                    i++;
                }
            k=i; 
            while(s1[k]==0)     
            {
                k--;
            }
            i=k;
            j++;
        	if(j>n)    
        	{
        	for(;k>=0;k--)    
            {
                cout << s1[k];
            }
            cout<<"n";
            return 0;
			}
        	return Fab(s1,s2,i,n,j);    
    }
}
int main()
{
	clock_t start_time,end_time;
	start_time=clock(); 
    cin>>n;
    Fab(s1,s2,i,n,j);
    end_time=clock();
	printf("ê±??:%f ",(double)(end_time-start_time)/CLOCKS_PER_SEC);
    return 0;
}