[BOJ][11727] 2×n 타일링 2
Updated:
1. 문제 링크
https://www.acmicpc.net/problem/11727
2. 사용 알고리즘
DP
3. 풀이
d[n] : 2 * n 크기의 직사각형을 채우는 방법의 수
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2 * (n - 2) 크기의 직사각형에 1 * 2 타일 또는 2 * 2 타일을 붙이면 2 * n 크기의 직사각형
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2 * (n - 1) 크기의 직사각형에 2 * 1 타일을 붙이면 2 * n 크기의 직사각형
∴ d[n] = d[n - 2] * 2 + d[n - 1]
4. 소스 코드
4-1. C++
4-1-1. Top-Down
https://github.com/dev-aiden/problem-solving/blob/main/boj/11727.cpp
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#include <iostream>
using namespace std;
int d[1003];
int solve(int num) {
if (num < 2) return 1;
if (d[num] != 0) return d[num];
return d[num] = (solve(num - 2) * 2 + solve(num - 1)) % 10007;
}
int main(void) {
ios_base::sync_with_stdio(false);
int n; cin >> n;
cout << solve(n) << "\n";
return 0;
}
4-1-2. Bottom-Up
https://github.com/dev-aiden/problem-solving/blob/main/boj/11727_2.cpp
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#include <iostream>
using namespace std;
int d[1003];
int main(void) {
ios_base::sync_with_stdio(false);
int n; cin >> n;
d[0] = d[1] = 1;
for (int i = 2; i <= n; ++i) d[i] = (d[i - 2] * 2 + d[i - 1]) % 10007;
cout << d[n] << "\n";
return 0;
}
4-2. JAVA
4-2-1. Top-Down
https://github.com/dev-aiden/problem-solving/blob/main/boj/11727.java
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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static int[] d = new int[1003];
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
System.out.println(solve(n));
}
public static int solve(int num) {
if (num < 2) return 1;
if (d[num] != 0) return d[num];
return d[num] = (solve(num - 2) * 2 + solve(num - 1)) % 10007;
}
}
4-2-2. Bottom-Up
https://github.com/dev-aiden/problem-solving/blob/main/boj/11727_2.java
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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static int[] d = new int[1003];
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
d[0] = d[1] = 1;
for(int i = 2; i <= n; ++i) d[i] = (d[i - 2] * 2 + d[i - 1]) % 10007;
System.out.println(d[n]);
}
}
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