Baekjoon #18436

Baekjoon #18436

1개의 풀이 · Java

문제 원문 보기 ↗

SOLUTION INFO

Java · Main.java

Main.java
import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) {
        FastReader rd = new FastReader();
        int N = rd.nextInt();
        int[] a = new int[N + 1];
        for (int i = 1; i <= N; i++) {
            a[i] = rd.nextInt();
        }
        int h = (int) Math.ceil(Math.log(N) / Math.log(2));
        int size = 1 << (h + 1);
        int[] tree_odd = new int[size];
        int[] tree_even = new int[size];

        init(1, 1, N, a, tree_odd, tree_even);

        int M = rd.nextInt();

        StringBuffer sb = new StringBuffer();
        for (int i = 0; i < M; i++) {
            int cmd = rd.nextInt();
            int n1 = rd.nextInt();
            int n2 = rd.nextInt();

            if (cmd == 1) {
                update(1, 1, N, n1, n2, tree_odd, tree_even);
            } else if (cmd == 2) {
                sb.append(query(1, 1, N, n1, n2, tree_odd, tree_even, true)).append("\n");
            } else {
                sb.append(query(1, 1, N, n1, n2, tree_odd, tree_even, false)).append("\n");
            }
        }
        System.out.print(sb.toString());
    }


    private static void init(int curr, int st, int ed, int[] a, int[] tree_odd, int[] tree_even) {
        if (st == ed) {
            if (a[st] % 2 == 1) {
                tree_odd[curr] = 1;
            } else {
                tree_even[curr] = 1;
            }
            return;
        }

        int mid = (st + ed) / 2;
        init(curr * 2, st, mid, a, tree_odd, tree_even);
        init(curr * 2 + 1, mid + 1, ed, a, tree_odd, tree_even);
        tree_odd[curr] = tree_odd[curr * 2] + tree_odd[curr * 2 + 1];
        tree_even[curr] = tree_even[curr * 2] + tree_even[curr * 2 + 1];
    }

    private static void update(int curr, int st, int ed, int idx, int newVal, int[] tree_odd, int[] tree_even) {
        if (idx < st || ed < idx) {
            return;
        }

        if (st == ed) {
            tree_even[curr] = newVal % 2 == 0 ? 1 : 0;
            tree_odd[curr] = newVal % 2 == 1 ? 1 : 0;
            return;
        }

        int mid = (st + ed) / 2;
        update(curr * 2, st, mid, idx, newVal, tree_odd, tree_even);
        update(curr * 2 + 1, mid + 1, ed, idx, newVal, tree_odd, tree_even);
        tree_odd[curr] = tree_odd[curr * 2] + tree_odd[curr * 2 + 1];
        tree_even[curr] = tree_even[curr * 2] + tree_even[curr * 2 + 1];
    }

    private static int query(int curr, int st, int ed, int l, int r, int[] tree_odd, int[] tree_even, boolean check) {
        if (r < st || ed < l) {
            return 0;
        }


        if (l <= st && ed <= r) {
            if (check) return tree_even[curr];
            else return tree_odd[curr];
        }


        int mid = (st + ed) / 2;
        int ret1 = query(curr * 2, st, mid, l, r, tree_odd, tree_even, check);
        int ret2 = query(curr * 2 + 1, mid + 1, ed, l, r, tree_odd, tree_even, check);
        return ret1 + ret2;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

SOLUTION DESCRIPTION

풀이 설명

1. 세그먼트 트리 기본 문제. `tree[]` 를 2개 만들어, 임의의 범위에 대한 짝수, 홀수인 경우의 갯수를 따로 구한다.