POJ 2506 Tiling

Tiling
Time Limit: 1000 MS Memory Limit: 65536 K
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.2cto.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
Research point: recursive formula f (n) = f (n-1) + 2*(n-2 );
Add large numbers
Note: When the input is 0, 1 is output instead of 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];
}
}