POJ 2506

  
  
  
  

   
   
   
   Tiling
  
  
  
  
  
  
  
  

   
   
   
   
 
    
 
     
 
      
Time Limit: 1000MS
 
      
 
 
      
Memory Limit: 65536K
 
     
 
     
 
      
Total Submissions: 8097
 
      
 
 
      
Accepted: 3921
 
     
 
    
 
   
  
  
  
  
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

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
char s[260][250];
int main()
{
    int n,m,i,j,k;
    memset(s,'0',sizeof(s));
    s[0][0]='1';s[1][0]='1',s[2][0]='3',s[3][0]='5';
    for(i=4;i<=250;i++)
    {
        for(j=0;j<200;j++)
        {
            k=(2*(s[i-2][j]-'0')+s[i-1][j]-'0'+s[i][j]-'0');//之所以还要加上s[i][j]，这是前一位向后一位的进位。
            s[i][j]=k%10+'0';
            s[i][j+1]=k/10+'0';
        }
    }

    while(cin>>n)
    {
        int x=200;
        for(i=200;i>=0;i--)
        {
            if(s[n][i]!='0')
            {
                x=i;
                break;
            }
        }
        for(i=x;i>=0;i--)
        {
            printf("%c",s[n][i]);
        }
        cout<<endl;
    }
    return 0;
}

方法二：

   JAVA 过大数feel excellent!

import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
	static BigInteger[] ans;
	public static void main(String srga[])
	{
		Scanner cin=new Scanner(System.in);
		ans=new BigInteger[260];
		ans[0]=BigInteger.valueOf(1);
		ans[1]=BigInteger.valueOf(1);
		ans[2]=BigInteger.valueOf(3);
		int i;
		for(i=3;i<=255;i++)
		{
			ans[i]=ans[i-1].add( ans[i-2].multiply(BigInteger.valueOf(2))   );
		}
		while(cin.hasNext())
		{
			int a;
			a=cin.nextInt();
			System.out.println(ans[a]);
		}
		cin.close();
	}
}