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
"); continue;
        }
        i = 0;
        while(a[n][i]==0)
        {
            i++;
        }
        for(i; i<=500; i++)
        {
            printf("%d", a[n][i] );
        }
        printf("
");
    }
    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]);
		}
	}
}