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

8
12
-1
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;
}
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