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 case
int 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_);
else
System.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];
else
clo[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
