Merge pull request #200 from AlgorithmWithGod/ShinHeeEul

[20250304] BOJ / P4 / 순열 / 신희을

Author: ShinHeeEul

ShinHeeEul/202503/04 BOJ P4 순열.md

@@ -0,0 +1,109 @@
+```java
+import java.util.*;
+import java.io.*;
+
+public class Main {
+	
+	static int[] segments;
+	static int size;
+	static int N;
+	public static void main(String[] args) throws Exception {
+		N = read();
+		
+		size = 1;
+		
+		while(size < N) {
+			size <<= 1;
+		}
+		
+		segments = new int[(size << 1) + 1];
+		
+		
+		// 빈공간들의 합을 올리고
+		
+		int segmentSize = size << 1;
+		while(segmentSize > 1) {
+			if(segmentSize > size) {
+				segments[segmentSize >> 1] = 2;
+				segmentSize-=2;
+			}
+			else {
+				segments[(segmentSize >> 1)] = segments[segmentSize] + segments[segmentSize - 1];
+				segmentSize -= 2;
+			}
+		}
+
+		for(int i = 1;  i <= N; i++) {
+			int A = read();
+			update(A, i);
+		}
+		
+		StringBuilder sb = new StringBuilder();
+		
+		for(int i = size + 1; i < size + N + 1; i++) {
+			sb.append(segments[i]).append("\n");
+		}
+		
+		System.out.println(sb);
+		
+		
+	}
+	
+	public static void update(int A, int idx) {
+		
+		// 2에서부터 내려가면서
+		int index = 2;
+		
+		while(true) {
+
+			// 거치는 애들 -1
+			segments[index]--;
+			int doubleIndex = index << 1;
+			int doubleIndexMinusOne = doubleIndex - 1;
+			// size보다 커지면? 해당 위치에 idx를 넣고 종료
+			if(index > (size >> 1)) {
+				if(A == 1) {
+					segments[doubleIndex] = idx;
+				} else {
+					if(segments[doubleIndexMinusOne] != 0) {
+						segments[doubleIndex] = idx;
+					} else {
+					segments[doubleIndexMinusOne] = idx;
+					}
+				}
+				return;
+			}
+			
+			// 좌측보다 크거나 같다?
+			int left = segments[doubleIndexMinusOne];
+			
+			if(A >= left) {
+				// 좌측만큼 빼주고 
+				A -= left;
+				// 우측으로 이동
+				index = doubleIndex;
+			}
+
+			// 나머지 경우?
+				// 좌측으로 이동
+			else {
+				index = doubleIndexMinusOne;
+			}
+		}
+	}
+	
+    private static int read() throws Exception {
+        int d, o;
+        d = System.in.read();
+
+        o = d & 15;
+
+        while ((d = System.in.read()) > 32)
+            o = (o << 3) + (o << 1) + (d & 15);
+
+        return o;
+    }
+	
+}
+
+```