POJ 2506 Tiling (递推 + 大数加法模拟 )

Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7965		Accepted: 3866

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


算法分析：递推公式：f[i]=f[i-1]+f[i-2]*2; 此公式也不是我自己推导出来的，我也没推导出来，我从ACM之家上的java代码看出的
公式。
代码：

#include <stdio.h>

#include <string.h>

int a[1001][501]={0};

int main()

{

    int n;

    int i, j, h, e;

    a[0][500] = 1;

    a[1][500] = 1;

    a[2][500] = 3;

    for(i=3; i<=250; i++)

    {

        h = 0;

        for(j=500; j>=0; j--)

        {

            e=a[i-2][j]*2+a[i-1][j]+h;

            a[i][j]=e%10;

            h=e/10;

        }

    }

    while(scanf("%d", &n)!=EOF)

    {

        if(n==0)

        {

            printf("1\n"); continue;

        }

        i = 0;

        while(a[n][i]==0)

        {

            i++;

        }

        for(i; i<=500; i++)

        {

            printf("%d", a[n][i] );

        }

        printf("\n");

    }

    return 0;

}

 这还有一份java代码，正确的！

import java.util.*;

import java.math.*;

public class Main{

	static BigInteger[] ans; //

	public static void main(String[] args){

		Scanner reader=new Scanner(System.in);

		ans = new BigInteger[251];

		ans[0]=BigInteger.valueOf(1);

		ans[1]=BigInteger.valueOf(1);

		ans[2]=BigInteger.valueOf(3);

		for(int i=3; i<=250; i++)

		{

			ans[i] = ans[i-1].add(ans[i-2].multiply(BigInteger.valueOf(2)));

		}

		int n;

		while(reader.hasNextInt()){

			n=reader.nextInt();

			System.out.println(ans[n]);

		}

	}

}