[20251028] BOJ / G5 / 가장 긴 짝수 연속한 수열(large)

0224LJH/202510/28 BOJ 가장 긴 짝수 연속한 수열(large).md

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+```java
+import java.io.BufferedReader;
+import java.io.IOException;
+import java.io.InputStreamReader;
+import java.util.StringTokenizer;
+
+public class Main {
+
+
+	static int len,totalErase,curErase,start,end,ans,curLen;
+	static int[] arr;
+
+    public static void main(String[] args) throws IOException {
+    	init();
+    	process();
+    	print();
+    }
+
+    private static void init() throws IOException{
+    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));;
+    	StringTokenizer st = new StringTokenizer(br.readLine());
+    	len = Integer.parseInt(st.nextToken());
+    	totalErase = Integer.parseInt(st.nextToken());
+    	arr = new int[len];
+    	st = new StringTokenizer(br.readLine());
+    	for (int i = 0; i < len; i++) {
+    		arr[i] = Integer.parseInt(st.nextToken());
+    	}
+
+    	start = 0;
+    	end = 0;
+    	ans = 0;
+    	curErase = 0;
+    	curLen = 0;
+
+    }
+
+    private static void process() {
+    	if (arr[0] %2 == 0) {
+    		curLen++;
+    		ans++;
+    	} else {
+    		curErase++;
+    	}
+
+    	while( end < len) {
+    		if (curErase < totalErase) {
+    			end++;
+    			if (end == len)break;
+    			if (arr[end] %2 == 0) curLen++;
+    			else curErase++;
+
+    		} else if (curErase == totalErase && end + 1 < len && arr[end+1]%2 ==0) {
+    			end++;
+    			curLen++;
+    		} else {
+    			if (arr[start] %2 == 0) {
+    				curLen--;
+    			} else {
+    				curErase--;
+    			}
+    			start++;
+    		}
+
+
+
+    		ans = Math.max(ans, curLen);
+    	}
+    }
+
+
+
+    private static void print() {
+    	System.out.println(ans);
+    }
+}
+
+
+```