[BOJ][2447] 별 찍기 - 10

1. 문제 링크

https://www.acmicpc.net/problem/2447

2. 사용 알고리즘

분할정복

3. 풀이

i와 j 모두 1, 4, 7 일 때 공백이므로 일반화하면, (i / 3) % 3 == 1 && (j / 3) % 3 == 1

마찬가지로 (i / 3) % 3 == 1 && (j / 3) % 3 == 1을 통해 계산 가능

4. 소스 코드

4-1. C++

https://github.com/dev-aiden/problem-solving/blob/main/boj/2447.cpp

#include <iostream>

using namespace std;

void solve(int r, int c, int size) {
    if(r / size % 3 == 1 && c / size % 3 == 1) cout << " ";
    else if(size / 3 == 0) cout << "*";
    else solve(r, c, size / 3);
}

int main(void) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    int n; cin >> n;
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) solve(i, j, n);
        cout << "\n";
    }
    cout << "\n";
    return 0;
}

4-2. JAVA

https://github.com/dev-aiden/problem-solving/blob/main/boj/2447.java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    private static StringBuilder sb = new StringBuilder();

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) solve(i, j, n);
            sb.append("\n");
        }
        sb.append("\n");
        System.out.println(sb);
    }

    public static void solve(int r, int c, int size) {
        if(r / size % 3 == 1 && c / size % 3 == 1) sb.append(" ");
        else if(size / 3 == 0) sb.append("*");
        else solve(r, c, size / 3);
    }
}