[BOJ][2446] 별 찍기 - 9
Updated:
1. 문제 링크
https://www.acmicpc.net/problem/2446
2. 사용 알고리즘
구현
3. 풀이
반복문을 돌며 별 출력
4. 소스 코드
4-1. C++
https://github.com/dev-aiden/problem-solving/blob/main/boj/2446.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
#include <iostream>
using namespace std;
int main(void) {
ios_base::sync_with_stdio(false);
int n; cin >> n;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) cout << " ";
for (int j = i * 2; j < (n * 2) - 1; ++j) cout << "*";
cout << "\n";
}
for (int i = 0; i < n - 1; ++i) {
for (int j = i; j < n - 2; ++j) cout << " ";
for (int j = 0; j < (i * 2) + 3; ++j) cout << "*";
cout << "\n";
}
return 0;
}
4-2. JAVA
https://github.com/dev-aiden/problem-solving/blob/main/boj/2446.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder();
int n = Integer.parseInt(br.readLine());
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) sb.append(" ");
for (int j = i * 2; j < (n * 2) - 1; ++j) sb.append("*");;
sb.append("\n");
}
for (int i = 0; i < n - 1; ++i) {
for (int j = i; j < n - 2; ++j) sb.append(" ");
for (int j = 0; j < (i * 2) + 3; ++j) sb.append("*");;
sb.append("\n");
}
System.out.println(sb);
}
}
Leave a comment