HOJ 2124 &POJ 2663Tri Tiling(动态规划)

Tri Tiling Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9016 Accepted: 4684 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

2 8 12 -1 Sample Output

3 153 2131 Source

之前也写过这种堆方块，就是找递推的方程，还是比较简单的题目。但是这道题目却是一个升级，因为他的递推方程需要变形，所以以后遇到这类题目就又多了一点见识。 dp[n]=3*dp[n-2]+2*dp[n-4]+2*dp[n-6]+……2*dp[0]; 如果只拿这个方程去解肯定麻烦或者还可能超时 变形 dp[n-2]=3*dp[n-4]+2*(dp[n-6]+dp[n-8]+…..dp[0]); dp[n-2]-dp[n-4]=2*(dp[n-4]+dp[n-6]+dp[n-8]+….dp[0]); dp[n]=4*dp[n-2]-dp[n-4]; 就搞定了，

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;
int dp[35];
int n;
int main()
{
    dp[0]=1;
    dp[1]=0;
    dp[2]=3;
    for(int i=3;i<=30;i++)
    {
         if(i&1)
            dp[i]=0;
        else
            dp[i]=dp[i-2]*4-dp[i-4];
    }
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1)
            break;
        printf("%d\n",dp[n]);
    }
    return 0;
}