HDU_1143_tri tiling
Tri Tiling
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2834 Accepted Submission(s): 1603
Problem Description
In how many ways can you tile a 3xn rectangle with 2x1 dominoes? Here is a sample tiling of a 3x12 rectangle.
![](http://acm.hdu.edu.cn/data/images/1143-1.jpg)
![](http://acm.hdu.edu.cn/data/images/1143-1.jpg)
Input
Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 ≤ n ≤ 30.
Output
For each test case, output one integer number giving the number of possible tilings.
Sample Input
2
8
12
-1
Sample Output
3
153
2131
参考:http://blog.csdn.net/chaoojie/article/details/8860935
当dp[i] 划分成 2 和 i-2 后,再划分成 4和i-4时,4的部分,不能再划分成2 , 2 组合,这样就会和前面的2 和i-2 重复,同样,划分过2 ,i-2 以及 4, i-4 后,再划分 6, i-6时,也不能划分成 2,2,2 以及 2, 4或 4, 2,同理,划分成8, i-8....
把 4, 6, 8.... 看成一整块,就有下图两种情况(正着,倒着)
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int main() { int dp[35]; int t; dp[0]=1; dp[1]=0; dp[2]=3; for(int i=3;i<=30;i++) { if(i%2) { dp[i]=0; continue; } dp[i]=dp[2]*dp[i-2]; for(int j=i-4;j>=0;j-=2) dp[i]+=2*dp[j]; } while(scanf("%d",&t)&&t!=-1) { cout<<dp[t]<<endl; } return 0; }