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

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）。

巴特西

poj_2506_Tiling_201407211555

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