POJ 2506 Tiling

Tiling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6444 Accepted: 3149
Description

In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles?
Here is a sample tiling of a 2x17 rectangle.

Input

Input is a sequence of lines, each line containing an integer number 0 <= n <= 250.
Output www.zzzyk.com

For each line of input, output one integer number in a separate line giving the number of possible tilings of a 2xn rectangle.
Sample Input

2
8
12
100
200
Sample Output

3
171
2731
845100400152152934331135470251
1071292029505993517027974728227441735014801995855195223534251
Source

The UofA Local 2000.10.14
考察点：递推公式 f(n)=f(n-1)+2*(n-2);
                大数相加
注意：   当输入0的时候输出1而不是0
[cpp]
#include <iostream>
#include <iomanip>
#include <string.h>
#include <math.h>
using namespace std;
struct num
{
    int s[1000];
    int top;
}a[260];
int main()
{
    void deal(int x,int y,int z);
    int i,j,n,m,s,t;
    (a[1].s)[0]=1; (a[2].s)[0]=3;
    a[1].top=1;     a[2].top=1;
    for(i=3;i<=250;i++)
    {
        deal(i-1,i-2,i);
    }
    while(cin>>n)
    {
        if(n==0)
        {
            cout<<1<<endl;
        }else
        {
            for(i=a[n].top-1;i>=0;i--)
            {
                cout<<(a[n].s)[i];
            }
            cout<<endl;
        }
    }
    return 0;
}
void deal(int x,int y,int z)
{
    int i,j,l1,l2,max;
    int b[1000],c[1000],d[1000];
    l1=a[x].top; l2=a[y].top;
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    memset(d,0,sizeof(d));
    for(i=0;i<=l1-1;i++)
    {
        b[i]=(a[x].s)[i];
    }
    for(i=0;i<=l2-1;i++)
    {
        c[i]=2*((a[y].s)[i]);
    }
    if(l1>l2)
    {
        max=l1;
    }else
    {
        max=l2;
    }
    for(i=0;i<=max-1;i++)
    {
        d[i]=b[i]+c[i];
    }
    for(i=0;i<=max-1;i++)
    {
        if(d[i]>=10)
        {
            d[i+1]+=d[i]/10;
            d[i]=d[i]%10;
            if(i==max-1)
            {
                max+=1;
            }
        }
    }
    a[z].top=max;
    for(i=0;i<=max-1;i++)
    {
        (a[z].s)[i]=d[i];
    }
}

补充：软件开发 , C++ ,