首先是劃分為n-2,和2兩部分，有3種。劃分為n-4和4兩部分，不重復的劃分有2種。劃分為n-6和6兩部分，不重復的劃分還是有2種。。。
所以遞推公式為 ?F(n)=3*F(n-2)+2×F(n-4)+2*F(n-6)+。。。+2*F(0);
化簡：
?F(n)=3*F(n-2)+2×F(n-4)+2*F(n-6)+。。。+2*F(0)
與F(n-2)=3*F(n-4)+2×F(n-6)+2*F(n-8)+。。。+2*F(0)?相減得 ? F(n)=4*F(n-2)-F(n-4)

Tri Tiling

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1486????Accepted Submission(s): 850


Problem DescriptionIn how many ways can you tile a 3xn rectangle with 2x1 dominoes? Here is a sample tiling of a 3x12 rectangle.



InputInput consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 ≤ n ≤ 30.?

OutputFor each test case, output one integer number giving the number of possible tilings.?

Sample Input2812-1
Sample Output31532131
SourceUniversity of Waterloo Local Contest 2005.09.24
RecommendEddy
#include <cstdio>#include <cstring>
using namespace std;
int a[32];
int main(){memset(a,0,sizeof(a));a[0]=1; a[2]=3;for(int i=4;i<32;i=i+2){a=4*a[i-2]-a[i-4];}int n;while(scanf("%d",&n)&&n!=-1){printf("%d\n",a[n]);}
return 0;}

轉(zhuǎn)載于:https://www.cnblogs.com/CKboss/archive/2013/05/13/3351065.html

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