poj_2506_Tiling_201407211555

Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7509		Accepted: 3672

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

Source

The UofA Local 2000.10.14

 

 #include <stdio.h>
 #include <string.h>
 #define MAX 1000
 int s[][MAX];
 int main()
 {
     int i,j,k,t,n,len=;
     memset(s,,sizeof(s));
     s[][]=;
     s[][]=;
     s[][]=;
     for(i=;i<=;i++)
     {
         t = ;
         for(j=;j<=len;j++)
         {
             s[i][j]=s[i-][j]+*s[i-][j]+t;
             if(s[i][j]>)
             {
                 t=s[i][j]/;
                 s[i][j]%=;
             }
             else
             t = ;
             if(j==len&&t==)
             len++;
         }
     }
     while(scanf("%d",&n)!=EOF)
     {
         for(i=MAX-;i>&&(s[n][i]==);i--);
         for(;i>=;i--)
         printf("%d",s[n][i]);
         printf("\n");
     }
     return ;
 }

首先，无论任何一种方案，最左边的要么是一根竖的，要么是两根横的，要么是一个方的，对吧？所以：
当是一根竖时剩下的方案数是f[i-1]
当是两根横时剩下的方案数是f[i-2]
当是一个方时剩下的方案数是f[i-2]
故f[i]=f[i-1]+2*f[i-2]

//递推+大数

注意：0的时候输出1（暂时没搞明白为什么，请大神指教）。