poj 2506 Tiling

Tiling Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8943   Accepted: 4257 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 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

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公式为a
=a[n-1]+a[n-2]*2，但是我当时没推出来～，

我推出来的是：当i为偶数时a[i]=a[i-1]*2+1;当i为奇数时a[i]=a[i-1]*2-1;
然后用大数相加来写～
还有就是 这道题比较坑的地方吧：n可以为0，并且哪位0时 输出1种方式～～～

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f

int a[300][110];
int main()
{
int n,i,index,j;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
{
printf("1\n");
continue;
}
memset(a,0,sizeof(a));
a[1][1]=1;
for(i=2;i<=n;i++)
{
if(i%2==0)
{
for(j=1;j<110;j++)
{
if(j==1)
a[i][j]=a[i][j]+a[i-1][j]*2+1;
else
a[i][j]=a[i][j]+a[i-1][j]*2;
if(a[i][j]>=10)
{
a[i][j+1]=a[i][j+1]+a[i][j]/10;
a[i][j]=a[i][j]%10;
}
}
}
else
{
for(j=1;j<110;j++)
{
if(j==1)
a[i][j]=a[i][j]+a[i-1][j]*2-1;
else
a[i][j]=a[i][j]+a[i-1][j]*2;
if(a[i][j]>=10)
{
a[i][j+1]=a[i][j+1]+a[i][j]/10;
a[i][j]=a[i][j]%10;
}
}
}
}
for(i=110;i>0;i--)
{
if(a[n][i]!=0)
{
index=i;
break;
}
}
for(i=index;i>=1;i--)
printf("%d",a[n][i]);
printf("\n");
}
return 0;
}