Acwing 141 周赛解题报告 | 珂学家 | 逆序数+奇偶性分析-编程知识

前言

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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);}}

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);}}

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 $O (n l o g n)$

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);}}staticclass 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}}