diff --git a/LiiNi-coder/202507/.gitkeep b/LiiNi-coder/202507/.gitkeep
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+
diff --git "a/LiiNi-coder/202507/10 BOJ \354\244\221\353\263\265\354\240\234\352\261\260.md" "b/LiiNi-coder/202507/10 BOJ \354\244\221\353\263\265\354\240\234\352\261\260.md"
new file mode 100644
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+++ "b/LiiNi-coder/202507/10 BOJ \354\244\221\353\263\265\354\240\234\352\261\260.md"	
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+# 첫풀이 (hashSet이용)
+## 코드
+```java
+import java.io.BufferedReader;
+import java.io.BufferedWriter;
+import java.io.IOException;
+import java.io.InputStreamReader;
+import java.io.OutputStreamWriter;
+import java.util.HashSet;
+import java.util.StringTokenizer;
+
+public class Main {
+	private static BufferedReader br;
+	private static StringTokenizer st;
+	private static BufferedWriter bw;
+
+	public static void main(String[] args) throws IOException {
+		br = new BufferedReader(new InputStreamReader(System.in));
+		bw = new BufferedWriter(new OutputStreamWriter(System.out));
+		String line = br.readLine();
+		st = new StringTokenizer(line);
+		
+		var sh = new HashSet<Integer>();
+		while(st.hasMoreElements()){
+			int ni = Integer.parseInt(st.nextToken());
+			if(!sh.contains(ni)) {
+				
+				bw.write(ni + " ");
+				sh.add(ni);
+			}	
+		}
+		
+		
+		bw.flush();
+		br.close();
+		bw.close();
+	}
+}
+```
+## 결과
+맞았습니다!!	396560KB	2260ms	Java 11
+# 두번째 풀이(상태값bit집합이용)
+## 코드
+```java
+import java.io.BufferedReader;
+import java.io.BufferedWriter;
+import java.io.IOException;
+import java.io.InputStreamReader;
+import java.io.OutputStreamWriter;
+import java.util.HashSet;
+import java.util.StringTokenizer;
+
+public class Main {
+	private static BufferedReader br;
+	private static StringTokenizer st;
+	private static BufferedWriter bw;
+
+	public static void main(String[] args) throws IOException {
+		br = new BufferedReader(new InputStreamReader(System.in));
+		bw = new BufferedWriter(new OutputStreamWriter(System.out));
+		String line = br.readLine();
+		st = new StringTokenizer(line);
+
+		// 2^25개의 숫자가 등장하였는지 여부를 저장하는데에, int하나는 32개의 숫자를 검사 가능하니, 2^25 / 32 하여서 2^20개의 int필요
+		int[] isAppeardNumber = new int[1<<20];
+		
+		while(st.hasMoreElements()){
+			int ni = Integer.parseInt(st.nextToken());
+			// isAppeardNumber 에 해당 숫자가 존재하면 == isAppeardNumber의 해당 bit가 1이라면
+			int byte_index = ni/32;
+			int bit_index = ni%32;
+			if((isAppeardNumber[byte_index] & (1 << (bit_index))) == 0) {
+				bw.write(ni + " ");
+				// isAppeardNumber에 반영
+				isAppeardNumber[byte_index] |= (1 << (bit_index));
+			}	
+		}
+		
+		
+		bw.flush();
+		br.close();
+		bw.close();
+	}
+}
+```
+## 결과
+맞았습니다!!	332500KB	1244ms	Java 11