杭电1143-Tri Tiling

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Tri Tiling

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1575 Accepted Submission(s): 898

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
AC代码+详细解释：
#include<iostream>//当n为奇数时，面积也是奇数，所以不可能用偶数的面积填满，即n为奇数时输出为0#include<cstring>//当n为偶数时仔细观察会发现/*可以看到图形的高都是3，所以要么在最上面放一层或者在最下面放一层，让高变成2，而让图不可分剩下的高为2的部分，必须在两头各竖着放一个要不图形就可以分成两个小图形了这样就不满足我们递推的思路了这样其实可以得到一个规律，只有k为2时g(k)是3其余的情况，g(k)都是2这样我们就可以简单的写出递推式：f(n)=3*f(n-2)+2*f(n-4)+.....+2*f(2)同理f(n-2)=3*f(n-4)+2*f(n-6)+.....+2*f(2)f(n)-4*f(n-2)=-f(n-4)；所以最终的递推式就是f(n)=4*f(n-2)-f(n-4)*/#include<string>#include<cstdio>#include<algorithm>const int MAX=31;int s[MAX];using namespace std;int main(){    int i,n;    s[0]=1;    s[2]=3;    for(i=4;i<MAX;i+=2)    {        s[i]=4*s[i-2]-s[i-4];    }    while(cin>>n,n>=0)    {        if(n&1)        cout<<0<<endl;        else        cout<<s[n]<<endl;    }    return 0;}