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[300][MAX];
int main()
{
    int i,j,k,t,n,len=0;
    memset(s,0,sizeof(s));
    s[0][0]=1;
    s[1][0]=1;
    s[2][0]=3;
    for(i=3;i<=300;i++)
    {
        t = 0;
        for(j=0;j<=len;j++)
        {
            s[i][j]=s[i-1][j]+2*s[i-2][j]+t;
            if(s[i][j]>9)
            {
                t=s[i][j]/10;
                s[i][j]%=10;                
            }
            else
            t = 0;
            if(j==len&&t==1)
            len++;
        }
    }
    while(scanf("%d",&n)!=EOF)
    {
        for(i=MAX-1;i>0&&(s[n][i]==0);i--);
        for(;i>=0;i--)
        printf("%d",s[n][i]);
        printf("
");                
    }
    return 0;
}

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）。