POJ-2663-Tri Tiling

解题报告：

1、subproblems:

2、A(n) = A(n-2) + 2 * B(n-1)

3、B(n) = A(n-1) + D(n-1)

4、D(n) = B(n-1)

4、由此可得：

A(n) = A(n-2) + 2 * B(n-1)

A(n) = 3*A(n-2) + 2 * B(n-3)

A(n) = 3*A(n-2) + 2 * A(n-4) + 2 * B(n-5)

...

AC代码：

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
	int i, j, n, tmp, buf[31];
	memset(buf, 0, sizeof(buf));
	buf[0] = 1;
	buf[2] = 3;

	for(i = 4; i <= 30; i += 2)
	{
		tmp = 3 * buf[i-2];
		for(j = 4; j <= i; j += 2)
			tmp += 2 * buf[i-j];
		buf[i] = tmp;
	}

	while(cin >> n && n != -1)
		cout << buf[n] << endl;

	return 0;
}