【POJ】2506 - Tiling（递推，高精度）

FishWang

发布于 2025-08-26 19:46:11

6400

代码可运行

运行总次数：0

代码可运行

Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8687		Accepted: 4183

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

Sample Output

3
171
2731
845100400152152934331135470251
1071292029505993517027974728227441735014801995855195223534251

Source

The UofA Local 2000.10.14

看到n的取值范围，直接懵了。然后就学了高精度，感觉好棒。

代码如下：

#include <cstdio>
int main()
{
	int f[300][100];
	f[0][0] = 1;		//至于这里为什么是1，本人表示不理解，2 * 0 的格子难道不是 0 种方案吗 
	f[1][0] = 1;
	f[2][0] = 3;
	for (int i = 3 ; i <= 250 ; i++)
	{
		int t = 0;
		for (int j = 0 ; j <= 99 ; j++)
		{
			t += f[i-1][j] + f[i-2][j] * 2;
			f[i][j] = t % 10;
			t /= 10;
		}
	}
	int n;
	while (~scanf ("%d",&n))
	{
		int tt;
		for (tt = 99 ; tt >= 0 ;tt--)
		{
			if (f[n][tt] != 0)
				break;
		}
		for (;tt >= 0 ; tt--)
			printf ("%d",f[n][tt]);
		printf ("\n");
	}
	return 0;
}

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原始发表：2025-08-26，如有侵权请联系 cloudcommunity@tencent.com 删除

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