Tiling Java大数

最新推荐文章于 2021-02-25 14:18:12 发布

原创最新推荐文章于 2021-02-25 14:18:12 发布 · 187 阅读

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本文介绍了一种通过递推公式解决2xn矩形使用2x1或2x2瓷砖进行铺砌的问题，提供了完整的Java代码实现，展示了如何高效地计算不同尺寸矩形的所有可能铺砌方式的数量。

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

import java.util.*;
import java.math.*;

public class Main {
	static BigInteger[] a;
	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		Scanner in=new Scanner(System.in);
		a=new BigInteger[251];
		a[0]=BigInteger.valueOf(1);
		a[1]=BigInteger.valueOf(1);
		a[2]=BigInteger.valueOf(3);
		for(int i=3;i<=250;i++) {
			a[i]=a[i-1].add(a[i-2].multiply(BigInteger.valueOf(2)));
		}
		int n;
		while(in.hasNextInt()) {
			n=in.nextInt();
			System.out.println(a[n]);	
		}
	}
}