HDOJ 1143 Tri Tiling

首先是划分为n-2,和2两部分，有3种。划分为n-4和4两部分，不重复的划分有2种。划分为n-6和6两部分，不重复的划分还是有2种。。。
所以递推公式为  F(n)=3*F(n-2)+2×F(n-4)+2*F(n-6)+。。。+2*F(0);
化简：
 F(n)=3*F(n-2)+2×F(n-4)+2*F(n-6)+。。。+2*F(0)
与F(n-2)=3*F(n-4)+2×F(n-6)+2*F(n-8)+。。。+2*F(0) 相减得   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 Description

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

 

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

2812-1

 

Sample Output

31532131

 

Source

University of Waterloo Local Contest 2005.09.24

 

Recommend

Eddy

#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;

}