Links: http://acm.hdu.edu.cn/showproblem.php?pid=2046
Domino Shop Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) total submission (s): 35779 Accepted S Ubmission (s): 17390
problem Description
In a rectangular square of 2xn, fill the squares with a 1x2 domino, enter N, and output the total number of paving schemes.
For example, when n=3, for 2x3 squares, the Domino placement scheme has three kinds, such as:
Input
The input data consists of multiple lines, each containing an integer n, indicating that the size of the rectangular square of the test instance is 2xn (0<n<=50).
Output
For each test instance, output the total number of placement scenarios, one row for each instance output.
Sample Input
1
3
2
Sample Output
1
3
2
Author
Lcy
Source
Recursive Solution topic exercise (for beginner)
Recommend
Lcy
The rectangular squares of--2*n are covered with 1*2 dominoes. Q: Given an n, the total number of scenarios for the solution placement.
Idea--apparently f (1) =1,f (2) =2,f (3) = 3, now we examine the last column: If the last column is vertical, then the number of schemes is F (n-1), and if the last column is placed sideways, it must be two horizontally, at which point the number of schemes is f (n-2). Therefore, f (n) =f ( n-1) +f (n-2), n>2, is the number of Fibonacci. And because n maximum is 50, so the direct recursion can.
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;int Main () {Ios::sync_with_stdio (false); LL num[51] = {1, 1, 2};for (int i=3; i<51; i++) num[i] = Num[i-1]+num[i-2];int n;while (cin >> N) {cout << nu M[n] << Endl;} return 0;}
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HDU 2046 Dominoes paved squares