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牛客周赛 Round 92 题解 Java

news 来源：原创 2025/6/2 21:23:38

A 小红的签到题

B 小红的模拟题

C 小红的方神题

D 小红的数学题

E 小红的 ds 题

F 小红的小苯题

A 小红的签到题

直接构造类似于 a_aaaa，a_aaaaaaaa 这种即可

// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;/*** 题目地址**/// xixi♡西
public class Main {static IoScanner sc = new IoScanner();static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);static int n;static int arr[];static boolean visited[];static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();/*** @throws IOException*/private static void solve() throws IOException {// todoint n=sc.nextInt();dduo("a_");for(int i=0;i<n-2;i++) {dduo("a");}}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}static <T> void dduo(T t) {System.out.print(t);}static <T> void dduoln() {System.out.println("");}static <T> void dduoln(T t) {System.out.println(t);}
}/*** IoScanner类** @author Dduo* @version 1.0* @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。*/
class IoScanner {BufferedReader bf;StringTokenizer st;BufferedWriter bw;public IoScanner() {bf = new BufferedReader(new InputStreamReader(System.in));st = new StringTokenizer("");bw = new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException {return bf.readLine();}public String next() throws IOException {while (!st.hasMoreTokens()) {st = new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException {return next().charAt(0);}public int nextInt() throws IOException {return Integer.parseInt(next());}public long nextLong() throws IOException {return Long.parseLong(next());}public double nextDouble() throws IOException {return Double.parseDouble(next());}public float nextFloat() throws IOException {return Float.parseFloat(next());}public BigInteger nextBigInteger() throws IOException {return new BigInteger(next());}public BigDecimal nextDecimal() throws IOException {return new BigDecimal(next());}
}

B 小红的模拟题

第一眼以为是 dfs

但是看到了

只有一个陷阱格子

所以只要规定一条到终点的路线先一直往右走再一直往下走

如果陷阱格子在这条线路上

就一直往下走再一直往右走以到达终点

// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;/*** 题目地址**/// xixi♡西
public class Main {static IoScanner sc = new IoScanner();static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);static int n;static int arr[];static boolean visited[];static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();/*** @throws IOException*/private static void solve() throws IOException {// todoint n=sc.nextInt();int m=sc.nextInt();char arr[][]=new char[n][m];for(int i=0;i<n;i++) {String str=sc.next();arr[i]=str.toCharArray();}int x=0;int y=0;for(int i=0;i<n;i++) {for(int j=0;j<m;j++) {if(arr[i][j]=='#') {x=i;y=j;}}}if(x==0||y==m-1) {for(int i=0;i<n-1;i++) {dduo("S");}for(int i=0;i<m-1;i++) {dduo("D");}}else if(y==0||x==n-1){for(int i=0;i<m-1;i++) {dduo("D");}for(int i=0;i<n-1;i++) {dduo("S");}}else {for(int i=0;i<n-1;i++) {dduo("S");}for(int i=0;i<m-1;i++) {dduo("D");}}}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}static <T> void dduo(T t) {System.out.print(t);}static <T> void dduoln() {System.out.println("");}static <T> void dduoln(T t) {System.out.println(t);}
}/*** IoScanner类** @author Dduo* @version 1.0* @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。*/
class IoScanner {BufferedReader bf;StringTokenizer st;BufferedWriter bw;public IoScanner() {bf = new BufferedReader(new InputStreamReader(System.in));st = new StringTokenizer("");bw = new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException {return bf.readLine();}public String next() throws IOException {while (!st.hasMoreTokens()) {st = new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException {return next().charAt(0);}public int nextInt() throws IOException {return Integer.parseInt(next());}public long nextLong() throws IOException {return Long.parseLong(next());}public double nextDouble() throws IOException {return Double.parseDouble(next());}public float nextFloat() throws IOException {return Float.parseFloat(next());}public BigInteger nextBigInteger() throws IOException {return new BigInteger(next());}public BigDecimal nextDecimal() throws IOException {return new BigDecimal(next());}
}

C 小红的方神题

其实我想了一会

然后根据样例试了试

输入 4

1 4 3 2

输入 5

1 5 4 3 2

都是符合的

// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;/*** 题目地址**/// xixi♡西
public class Main {static IoScanner sc = new IoScanner();static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);static int n;static int arr[];static boolean visited[];static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();/*** @throws IOException*/private static void solve() throws IOException {// todoint n=sc.nextInt();if(n==1||n==2) {dduoln("-1");return;}dduo(1+" ");for(int i=n;i>=2;i--) {dduo(i+" ");}}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}static <T> void dduo(T t) {System.out.print(t);}static <T> void dduoln() {System.out.println("");}static <T> void dduoln(T t) {System.out.println(t);}
}/*** IoScanner类** @author Dduo* @version 1.0* @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。*/
class IoScanner {BufferedReader bf;StringTokenizer st;BufferedWriter bw;public IoScanner() {bf = new BufferedReader(new InputStreamReader(System.in));st = new StringTokenizer("");bw = new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException {return bf.readLine();}public String next() throws IOException {while (!st.hasMoreTokens()) {st = new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException {return next().charAt(0);}public int nextInt() throws IOException {return Integer.parseInt(next());}public long nextLong() throws IOException {return Long.parseLong(next());}public double nextDouble() throws IOException {return Double.parseDouble(next());}public float nextFloat() throws IOException {return Float.parseFloat(next());}public BigInteger nextBigInteger() throws IOException {return new BigInteger(next());}public BigDecimal nextDecimal() throws IOException {return new BigDecimal(next());}
}

D 小红的数学题

我的首先想到的是韦达定理

要求是 x1 x2 互不相同而且 x1+x2+x1*x2==k

之后 x1+x2 为 p ，x1*x2 为 q

其次进行了打表

获取了大量数据

发现奇数的话是一定可以的只要构造其中一个数为 1 即可

接着是偶数

我找出的规律是这样的

然后根据规律推

放大循环的次数就过了

// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;/*** 题目地址**/// xixi♡西
public class Main {static IoScanner sc = new IoScanner();static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);static int n;static int arr[];static boolean visited[];static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();/*** @throws IOException*/private static void solve() throws IOException {// todo//    	TreeSet<Integer>hs=new TreeSet<>();
//    	for(int i=1;i<=100;i++) {
//    		for(int j=i+1;j<=100;j++) {
//    			// k
//    			if( (i+j+i*j) %2==0) {
//    				dduoln((i+j)+" "+(i*j)+" "+(i+j+i*j));	
//    			}
//    			hs.add((i+j+i*j));
//    		}
//    	}
//    	
//    	for(Integer i:hs) {
//    		if(i%2==0) {
//    			dduoln(i);
//    		}
//    	}long k=sc.nextLong();if(k%2!=0) {// 奇数if(k==1||k==3) {dduoln("-1");return;}long ans1=k-1;ans1/=2;long ans2=(1+ans1);dduoln((ans1)+" "+(ans2));	}else {// 偶数long zuobian=14;long youbian=6;long zuopbianjia=12;long youbianjia=4;// todofor(int i=0;i<1000000;i++) {if(k-zuobian<0) {dduoln("-1");return;}long youbianshengxialai=k-zuobian;if(youbianshengxialai%youbian==0) {long nn=youbianshengxialai/youbian;
//    				dduoln(nn);
//    				dduoln((zuobian-youbian)+" "+nn*(youbian-2));dduoln((youbian+2*nn)+" "+((zuobian-youbian)+(nn*(youbian-2))));return;}zuopbianjia+=8;zuobian+=zuopbianjia;youbian+=youbianjia;}}}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}static <T> void dduo(T t) {System.out.print(t);}static <T> void dduoln() {System.out.println("");}static <T> void dduoln(T t) {System.out.println(t);}
}/*** IoScanner类** @author Dduo* @version 1.0* @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。*/
class IoScanner {BufferedReader bf;StringTokenizer st;BufferedWriter bw;public IoScanner() {bf = new BufferedReader(new InputStreamReader(System.in));st = new StringTokenizer("");bw = new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException {return bf.readLine();}public String next() throws IOException {while (!st.hasMoreTokens()) {st = new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException {return next().charAt(0);}public int nextInt() throws IOException {return Integer.parseInt(next());}public long nextLong() throws IOException {return Long.parseLong(next());}public double nextDouble() throws IOException {return Double.parseDouble(next());}public float nextFloat() throws IOException {return Float.parseFloat(next());}public BigInteger nextBigInteger() throws IOException {return new BigInteger(next());}public BigDecimal nextDecimal() throws IOException {return new BigDecimal(next());}
}

E 小红的 ds 题

我感觉这题的难度小于 D

给出二叉树每一层的节点

然后构造出二叉树

每个节点有 0 个 1 个 2 个节点

我们直接顺着往下构造就行

直接一层一层推

什么情况下构造不出来呢

当这下层的节点数大于这一层的两倍时无法构造因为这一层的每个节点最多连下一层的两个节点

接着就是一个个构造

// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;/*** 题目地址**/// xixi♡西
public class Main {static IoScanner sc = new IoScanner();static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);static int n;static int arr[];static boolean visited[];static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();/*** @throws IOException*/private static void solve() throws IOException {// todoint n=sc.nextInt();long arr[]=new long[n+1];for(int i=1;i<=n;i++) {arr[i]=sc.nextLong();}for(int i=2;i<=n;i++) {if( (arr[i]) > (arr[i-1]*2) ) {dduoln("-1");return;}}dduoln("1");long cnt=2;for(int i=1;i<=n;i++) {long ans=arr[i]; // 当前层有多少节点if(i==n) {// 最后一层for(int j=0;j<ans;j++) {dduoln("-1 -1");}}else {// 下一层有多少个节点long next=arr[i+1];
//    			cnt+=ans;for(int j=0;j<ans;j++) {if(next>=2) {dduo(cnt);dduo(" ");cnt++;dduoln(cnt);cnt++;next-=2;}else if(next==1){dduo(cnt);dduo(" ");cnt++;dduoln("-1");next-=1;}else {dduoln("-1 -1");}}    			}
//			cnt+=ans;}}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}static <T> void dduo(T t) {System.out.print(t);}static <T> void dduoln() {System.out.println("");}static <T> void dduoln(T t) {System.out.println(t);}
}/*** IoScanner类** @author Dduo* @version 1.0* @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。*/
class IoScanner {BufferedReader bf;StringTokenizer st;BufferedWriter bw;public IoScanner() {bf = new BufferedReader(new InputStreamReader(System.in));st = new StringTokenizer("");bw = new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException {return bf.readLine();}public String next() throws IOException {while (!st.hasMoreTokens()) {st = new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException {return next().charAt(0);}public int nextInt() throws IOException {return Integer.parseInt(next());}public long nextLong() throws IOException {return Long.parseLong(next());}public double nextDouble() throws IOException {return Double.parseDouble(next());}public float nextFloat() throws IOException {return Float.parseFloat(next());}public BigInteger nextBigInteger() throws IOException {return new BigInteger(next());}public BigDecimal nextDecimal() throws IOException {return new BigDecimal(next());}
}

F 小红的小苯题

构造一个矩阵每行每列的异或和构成排列

这样构造然后正好要满足这个情况

就是行加列%4==3

很奇妙！

// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;/*** 题目地址**/// xixi♡西
public class Main {static IoScanner sc = new IoScanner();static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);static int n;static int arr[];static boolean visited[];static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();/*** @throws IOException*/// 2 5// 0 0 0 0 7// 1 2 3 4 2private static void solve() throws IOException {// todoint n = sc.nextInt();int m = sc.nextInt();int k = n + m;if (k % 4 != 3) {dduoln("-1");return;}// 行int[] rows = new int[n];for (int i = 0; i < n; i++) {rows[i] = k - i;}// 列int[] cols = new int[m];for (int i = 0; i < m; i++) {cols[i] = i + 1;}int[][] arr = new int[n][m];for (int i = 0; i < n - 1; i++) {for (int j = 0; j < m - 1; j++) {arr[i][j] = 0;}arr[i][m - 1] = rows[i];}if (m == 0) {dduoln("-1");return;}// 列处理for (int j = 0; j < m - 1; j++) {arr[n - 1][j] = cols[j];}int ans = 0;for (int i = 0; i < n - 1; i++) {ans ^= rows[i];}int x = cols[m - 1] ^ ans;arr[n - 1][m - 1] = x;for (int i = 0; i < n; i++) {int xor = 0;for (int num : arr[i]) {xor ^= num;}assert xor == rows[i] : "Row " + i + " xor error";}for (int j = 0; j < m; j++) {int xor = 0;for (int i = 0; i < n; i++) {xor ^= arr[i][j];}assert xor == cols[j] : "Col " + j + " xor error";}for (int[] row : arr) {StringBuilder sb = new StringBuilder();for (int num : row) {sb.append(num).append(" ");}dduoln(sb.toString());}}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}static <T> void dduo(T t) {System.out.print(t);}static <T> void dduoln() {System.out.println("");}static <T> void dduoln(T t) {System.out.println(t);}
}/*** IoScanner类** @author Dduo* @version 1.0* @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。*/
class IoScanner {BufferedReader bf;StringTokenizer st;BufferedWriter bw;public IoScanner() {bf = new BufferedReader(new InputStreamReader(System.in));st = new StringTokenizer("");bw = new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException {return bf.readLine();}public String next() throws IOException {while (!st.hasMoreTokens()) {st = new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException {return next().charAt(0);}public int nextInt() throws IOException {return Integer.parseInt(next());}public long nextLong() throws IOException {return Long.parseLong(next());}public double nextDouble() throws IOException {return Double.parseDouble(next());}public float nextFloat() throws IOException {return Float.parseFloat(next());}public BigInteger nextBigInteger() throws IOException {return new BigInteger(next());}public BigDecimal nextDecimal() throws IOException {return new BigDecimal(next());}
}