Links: http://acm.hdu.edu.cn/showproblem.php?pid=1143
Tri tilingtime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) total submission (s): 2799 Accept Ed Submission (s): 1585
problem Description
In what 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 was a line containing an integer 0≤n≤30.
Output
For each test case, the output one integer number giving the number of possible tilings.
Sample Input
2
8
12
-1
Sample Output
3
153
2131
Source
University of Waterloo Local Contest 2005.09.24
Recommend
Eddy
The rectangular squares of--3*n, covered with 2*1 dominoes. Q: Given an n, the total number of scenarios is solved.
Idea-obviously this is a recursive topic. It is obvious that when n is odd, there is no solution. and f (0) =1,f (2) = 3, now we examine the last two columns: if the last two columns are covered, there are 3 scenarios, so the number of scenarios is 3*f (n-2), and if the last two columns are not covered, then there must be four columns in the back, with 2 schemes, so the number of scenarios is f (n-4). And so on, when N>=4, f (n) =3*f (n-2) +2*f (n-4) +2*f (n-6) + +2*f (0). Make 2*m=n, then f (m) =3*f (m-1) +2*f (m-2) + +2*f (0), F (m-1) =3*f (m-2) +2*f (m-3) + +2*f (0), subtract and collate two, F (m) =4*f (M-1)-F (m-2). Thus, f (n) =4*f (n-2)-F (n-4), n>=4, and N is an even number. And because N is the maximum of 30, the direct recursion can be.
Complexity analysis--time complexity: O (n), spatial complexity: O (n)
Attach the AC code:
#include <iostream> #include <cstdio> #include <string> #include <cmath> #include <iomanip > #include <ctime> #include <climits> #include <cstdlib> #include <cstring> #include < algorithm>using namespace std;typedef unsigned int ui;typedef long long ll;typedef unsigned long long ull;typedef long Double Ld;const double PI = 3.14159265;const double E = 2.71828182846; LL num[31] = {1, 0, 3, 0};void init (); int main () {Ios::sync_with_stdio (false); Init (); Short N;while (CIN >> N &&am P n! =-1) {cout << num[n] << Endl;} return 0;} void Init () {for (int i=4; i<31; i++) num[i] = 4*num[i-2]-num[i-4];}
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HDU 1143 Tri Tiling
