2023年第六届传智杯程序设计挑战赛（个人赛）B组 赛后复盘

传智杯赛后复盘

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2023年第六届传智杯程序设计挑战赛（个人赛）B组 赛后复盘
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1. 字符串拼接

细节：一定要清楚nextLine()和next()的区别
nextLine()是遇到回车会停下来
next是遇到空格会停下来
很明显这里必须得选nextLine()
踩坑实录…

import java.util.Scanner;// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);String str1 = in.nextLine();String str2 = in.nextLine();StringBuffer s1 = new StringBuffer(str1);StringBuffer s2 = new StringBuffer(str2);s1.append(s2);System.out.println(s1);}
}

2. 差值

import java.util.Arrays;
import java.util.Scanner;public class Main {public static void main(String[] args) {// 创建Scanner对象用于接收输入Scanner in = new Scanner(System.in);// 输入战士数量int n = in.nextInt();int[] strengths = new int[n];for (int i = 0; i < n; i++) {strengths[i] = in.nextInt();}// 对战士战斗力进行排序 以便比较相邻两位战士的战力之差Arrays.sort(strengths);// 初始化最小差值为一个较大的值int minDif = 0x3f3f3f3f;// 枚举相邻两名战士战斗力之差的最小值 for (int i = 0; i < n - 1; i++) {int currentDif = strengths[i + 1] - strengths[i];//更新战力之差的最小值if (currentDif < minDif) {minDif = currentDif;}}System.out.println(minDif);in.close();}
}

3. . 红色和紫色

很有趣的一题，奇数和偶数的区别

import java.util.Scanner;// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();int m = in.nextInt();//方格数量为偶数则yukari赢 因为yukari总会染成紫色if(n % 2 == 0 || m % 2 == 0){System.out.println("yukari");}else{//方格数量为奇数则akai赢 因为最后akai不能染成紫色    System.out.println("akai");}}
}

4. abb

dp 举出后面相同的字符就+1

import java.util.*;// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);// 注意 hasNext 和 hasNextLine 的区别while (in.hasNextInt()) { // 注意 while 处理多个 caseint n = in.nextInt();String str = in.next();int[][] dp = new int[n + 1][26];Arrays.fill(dp[n], 0);for (int i = n - 1; i >= 1; i--) {char ch = str.charAt(i);for (int j = 0; j < 26; j++) {if (ch - 'a' == j) {dp[i][j] = dp[i + 1][j] + 1;} else {dp[i][j] = dp[i + 1][j];}}}long res = 0;for (int i = 1; i <= n; i++) {char c = str.charAt(i - 1);for (int j = 0; j < 26; j++) {if (c - 'a' != j && dp[i][j] >= 2) {res += dp[i][j] * (dp[i][j] - 1) / 2;}}}System.out.println(res);}}
}

5. kotorti和素因子

dfs + 质数筛

import java.util.Scanner;
import java.util.ArrayList;public class Main {static final int maxn = 1005;static final int INF = 0x3f3f3f3f;static int n, m, sum, min_, X;static ArrayList<Integer>[] e = new ArrayList[maxn];static boolean[] vis = new boolean[maxn];static boolean flag;static boolean isPrime(int n) {if (n == 1)return false;for (int i = 2; i <= Math.floor(Math.sqrt(n)); i++) {if (n % i == 0)return false;}return true;}static void prime(int n, int x) {for (int i = 1; i <= Math.floor(Math.sqrt(n)); i++) {if (n % i == 0) {if (isPrime(i))e[x].add(i);if (i * i != n && isPrime(n / i))e[x].add(n / i);}}}static void dfs(int y) {if (y == X) {flag = true;min_ = Math.min(min_, sum);return;}for (int i = 0; i < e[y].size(); i++) {if (!vis[e[y].get(i)]) {sum += e[y].get(i);vis[e[y].get(i)] = true;dfs(y + 1);vis[e[y].get(i)] = false;sum -= e[y].get(i);}}}public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();for (int i = 0; i < maxn; i++) {e[i] = new ArrayList<>();}X = 0;for (int i = 0; i < n; i++) {m = scanner.nextInt();prime(m, X++);}min_ = INF;dfs(0);if (flag)System.out.println(min_);elseSystem.out.println(-1);}
}

6. 红和蓝

赛后补题

import java.util.Scanner;
import java.util.Arrays;public class Main {static final int N = 2 * 100000 + 10;static int[] e = new int[N];static int[] h = new int[N];static int[] ne = new int[N];static int idx;static void add(int a, int b) {e[idx] = b;ne[idx] = h[a];h[a] = idx++;e[idx] = a;ne[idx] = h[b];h[b] = idx++;}static int n;static int cnt;static int[] colour = new int[N];static boolean flag;static void dfs1(int x, int fa) {int son = 0;for (int i = h[x]; i != -1; i = ne[i]) {int ver = e[i];if (ver == fa)continue;son++;dfs1(ver, x);}if (son == 0 || colour[x] == 0) {if (colour[fa] != 0 || fa == 0) {flag = true;return;}colour[x] = colour[fa] = ++cnt;}}static int[] clo = new int[N];static void dfs2(int x, int fa) {for (int i = h[x]; i != -1; i = ne[i]) {int ver = e[i];if (ver == fa)continue;if (colour[ver] == colour[x])clo[ver] = clo[x];elseclo[ver] = clo[x] ^ 1;dfs2(ver, x);}}public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();Arrays.fill(h, -1);for (int i = 1; i < n; i++) {int x = scanner.nextInt();int y = scanner.nextInt();add(x, y);}dfs1(1, 0);if (flag) {System.out.println("-1");return;}dfs2(1, 0);for (int i = 1; i <= n; i++)System.out.print(clo[i] != 0 ? "B" : "R");}
}

总结

ACM模式，大部分题目是从牛客题库、寒假训练营抽出来的，Dfs、DP、图论、质数筛的混合考察比较多，平时多练习多debug
ACM注意罚时的重要性，考虑一些细节不对，提交报错则罚时严重。
拿到题目，先把题目全部扫一遍，不要一股脑只是做题，应该先把题目先过一遍，确定考点后，由易入难。
确保会的都写对，不会的尝试一下，多debug