(Hdu step 3.1.5) Tiling_easy version (calculate the number of grid solutions with 2x1 and 2x2 bone cells filled with 2 x n), hdutiling_easy

(Hdu step 3.1.5) Tiling_easy version (calculate the number of grid solutions with 2x1 and 2x2 bone cells filled with 2 x n), hdutiling_easy

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Question:

Tiling_easy version

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission (s): 515 Accepted Submission (s): 447

Problem Description has a grid of 2 x n, and now we need to fill it with two types of dominoes, which are 2x1 and 2x2, respectively, calculate the total number of laying methods.

The first line of Input contains a positive integer T (T <= 20), indicating a total of T groups of data, followed by T rows of data, each row contains a positive integer N (N <= 30), indicating that the grid size is two rows and N columns.

How many laying methods are Output, and the Output of each group of data occupies one row.

Sample Input 32812

Sample Output 31712731

Source ACM Short Term Exams _ software engineering and other majors

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Question Analysis:

Simple recursion. In fact, this question is quite similar to "hdu step 3.1.2)", but it only adds a type of bone lattice. But the idea is the same.

The Code is as follows:

/** E. cpp ** Created on: February 5, 2015 * Author: Administrator */# include <iostream> # include <cstdio> using namespace std; const int maxn = 33; int dp [maxn]; void prepare () {dp [1] = 1; dp [2] = 3; int I; for (I = 3; I <maxn; ++ I) {dp [I] = dp [I-1] + dp [I-2] * 2; // This time is dp [I-2] * 2. because the last 2*2 bone cells can be solved by two horizontally placed 2*1, or by a 2*2 filled} int main () {prepare (); int t; scanf ("% d", & t); while (t --) {int n; scanf ("% d", & n ); printf ("% d \ n", dp [n]);} return 0 ;}