1 /* 2     1 r; 2 b; 3 w 3     2不能在最前面，所以dp[1] = 2;    dp[2] = 2: 13 or 31 4  5     dp[i] = dp[i-1] + dp[i-2]; 6     只加1或3时，总数dp[i-1];    只加12或32时，总数dp[i-2]; 7     详细解释：http://www.cnblogs.com/vongang/archive/2011/09/30/2196847.html 8 */ 9 #include 
     
      10 #include 
      
       11 #include 
       
        12 #include 
        
         13 using namespace std;14 15 const int MAXN = 55;16 const int INF = 0x3f3f3f3f;17 long long dp[MAXN];18 19 int main(void)        //URAL 1225 Flags20 {21     //freopen ("A.in", "r", stdin);22 23     int n;24     while (scanf ("%d", &n) == 1)25     {26         memset (dp, 0, sizeof (dp));27 28         dp[1] = 2;    dp[2] = 2;29         for (int i=3; i<=n; ++i)30         {31             dp[i] = dp[i-1] + dp[i-2];32         }33 34         printf ("%I64d\n", dp[n]);35     }36 37     return 0;38 }