Acwing 141 周赛 解题报告 | 珂学家 | 逆序数+奇偶性分析

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前言

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整体评价

很普通的一场比赛，t2思维题，初做时愣了下，幸好反应过来了。t3猜猜乐，感觉和逆序数有关，和奇偶性有关。不过要注意int溢出。

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A. 客人数量

题型: 签到

累加和即可

import java.io.BufferedInputStream;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(new BufferedInputStream(System.in));
        int n = sc.nextInt();
        long res = 0;
        for (int i = 0; i < n; i++) {
            long v = sc.nextLong();
            res += v;
        }
        System.out.println(res);
     }

}

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B. 指针运动

思路: 思维题

其实只要找到最小值，然后模拟，这样时间复杂度就能控制在$O(n)$

值域很大，这是最佳的策略

不过枚举应该也可以

import java.io.BufferedInputStream;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(new BufferedInputStream(System.in));
        int n = sc.nextInt();

        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextLong();
        }

        long minTimes = Arrays.stream(arr).min().getAsLong();
        int now = (int)(minTimes % n);

        long cutoff = minTimes;
        while (arr[now] > cutoff) {
            now = (now + 1) % n;
            cutoff++;
        }
        System.out.println(now + 1);
    }

}

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C. 随机排列

思路: 逆序对数 + 奇偶分析

求逆序数的大概有两种解法

• CDQ, 分治归并排序
• 树状数组

这边采用了树状数组，因为是1~n的排列，所以不用离散化处理

时间复杂度为$O(nlogn)$

当然更好的解法是置换群(置换环)，因为侧重于交换的奇偶次数

这样的话，时间复杂度可以控制在$O(n)$

• 树状数组解法

// package acwing.acw141;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    static class BIT {
        int n;
        int[] arr;
        public BIT(int n) {
            this.n = n;
            this.arr = new int[n + 1];
        }
        void update(int p, int d) {
            while (p <= n) {
                arr[p] += d;
                p += p & -p;
            }
        }
        int query(int p) {
            int r = 0;
            while (p > 0) {
                r += arr[p];
                p -= p & -p;
            }
            return r;
        }
    }

    public static void main(String[] args) {
        AReader sc = new AReader();
        int n = sc.nextInt();
        int[] arr = new int[n + 1];

        for (int i = 1; i <= n; i++) {
            arr[i] = sc.nextInt();
        }

        long res = 0;
        BIT bit = new BIT(n);
        for (int i = 1; i <= n; i++) {
            int v = arr[i];
            int d = bit.query(n) - bit.query(v);
            res += d;
            bit.update(v, 1);
        }

        if (res % 2 != n % 2) {
            System.out.println(2);
        } else {
            System.out.println(1);
        }

    }

    static
    class AReader {
        private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        private StringTokenizer tokenizer = new StringTokenizer("");
        private String innerNextLine() {
            try {
                return reader.readLine();
            } catch (IOException ex) {
                return null;
            }
        }
        public boolean hasNext() {
            while (!tokenizer.hasMoreTokens()) {
                String nextLine = innerNextLine();
                if (nextLine == null) {
                    return false;
                }
                tokenizer = new StringTokenizer(nextLine);
            }
            return true;
        }
        public String nextLine() {
            tokenizer = new StringTokenizer("");
            return innerNextLine();
        }
        public String next() {
            hasNext();
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }

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

//        public BigInteger nextBigInt() {
//            return new BigInteger(next());
//        }
        // 若需要nextDouble等方法，请自行调用Double.parseDouble包装
    }

}

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