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牛客周赛Round 1-2-11-12-13
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‎jacoco-aggregate-report/pom.xml‎

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<artifactId>leetcode-29</artifactId>
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<version>1.0-SNAPSHOT</version>
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</dependency>
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<dependency>
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<groupId>com.devyy</groupId>
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<artifactId>leetcode-30</artifactId>
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<version>1.0-SNAPSHOT</version>
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</dependency>
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<dependency>
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<groupId>com.devyy</groupId>
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<artifactId>leetcode-offer</artifactId>
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<version>1.0-SNAPSHOT</version>
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</dependency>
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<dependency>
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<groupId>com.devyy</groupId>
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<artifactId>nowcoder</artifactId>
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<version>1.0-SNAPSHOT</version>
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</dependency>
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</dependencies>
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<build>

‎leetcode/leetcode-03/src/main/java/Solution238.java‎

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public class Solution238 {
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public int[] productExceptSelf(int[] nums) {
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int len = nums.length;
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int[] leftProduct = new int[len];
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int[] rightProduct = new int[len];
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leftProduct[0] = 1;
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for (int i = 1; i < len; i++) {
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leftProduct[i] = leftProduct[i - 1] * nums[i - 1];
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int n = nums.length;
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int[] L = new int[n];
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int[] R = new int[n];
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L[0] = 1;
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for (int i = 1; i < n; i++) {
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L[i] = L[i - 1] * nums[i - 1];
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}
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rightProduct[len - 1] = 1;
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for (int i = len - 2; i >= 0; i--) {
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rightProduct[i] = rightProduct[i + 1] * nums[i + 1];
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R[n - 1] = 1;
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for (int i = n - 2; i >= 0; i--) {
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R[i] = R[i + 1] * nums[i + 1];
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}
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int[] res = new int[len];
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for (int i = 0; i < len; i++) {
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res[i] = leftProduct[i] * rightProduct[i];
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int[] ans = new int[n];
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for (int i = 0; i < n; i++) {
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ans[i] = L[i] * R[i];
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}
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return res;
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return ans;
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}
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}
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/*

‎leetcode/leetcode-29/src/main/java/Solution2873.java‎

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3 <= nums.length <= 100
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1 <= nums[i] <= 10^6
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同 2874. 有序三元组中的最大值 II
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同: 2874. 有序三元组中的最大值 II
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https://leetcode.cn/problems/maximum-value-of-an-ordered-triplet-ii/description/
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*/

‎leetcode/leetcode-30/pom.xml‎

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<?xml version="1.0" encoding="UTF-8"?>
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<project xmlns="http://maven.apache.org/POM/4.0.0"
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xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
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xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
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<modelVersion>4.0.0</modelVersion>
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<parent>
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<groupId>com.devyy</groupId>
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<artifactId>leetcode</artifactId>
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<version>1.0-SNAPSHOT</version>
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</parent>
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<artifactId>leetcode-30</artifactId>
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<properties>
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<maven.compiler.source>17</maven.compiler.source>
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<maven.compiler.target>17</maven.compiler.target>
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<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
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</properties>
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<dependencies>
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<dependency>
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<groupId>com.devyy</groupId>
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<artifactId>leetcode-core</artifactId>
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<version>1.0-SNAPSHOT</version>
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</dependency>
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</dependencies>
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</project>

‎leetcode/pom.xml‎

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<module>leetcode-27</module>
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<module>leetcode-28</module>
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<module>leetcode-29</module>
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<module>leetcode-30</module>
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<module>leetcode-core</module>
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<module>leetcode-extends</module>

‎nowcoder/pom.xml‎

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<?xml version="1.0" encoding="UTF-8"?>
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<project xmlns="http://maven.apache.org/POM/4.0.0"
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xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
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xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
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<modelVersion>4.0.0</modelVersion>
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<parent>
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<groupId>com.devyy</groupId>
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<artifactId>leetcode-hub-java</artifactId>
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<version>1.0-SNAPSHOT</version>
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</parent>
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<artifactId>nowcoder</artifactId>
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<properties>
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<maven.compiler.source>8</maven.compiler.source>
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<maven.compiler.target>8</maven.compiler.target>
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<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
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</properties>
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<dependencies>
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<dependency>
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<groupId>com.devyy</groupId>
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<artifactId>algo-core</artifactId>
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<version>1.0-SNAPSHOT</version>
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</dependency>
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</dependencies>
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</project>

‎nowcoder/src/main/java/nc60245/NC60245A.java‎

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package nc60245;
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import java.util.Scanner;
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public class NC60245A {
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in);
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int n = scanner.nextInt();
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System.out.println(solve(n));
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}
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private static String solve(int n) {
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String[] ans = new String[n * 4];
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for (int i = 0; i < n * 3; i++) {
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// String s = "*".repeat(n) + ".".repeat(n * 2) + "*".repeat(n);
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String s = repeat("*", n) + repeat(".", n * 2) + repeat("*", n);
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ans[i] = s;
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}
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int j = 1;
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for (int i = n * 3; i < n * 4; i++) {
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// String s = ".".repeat(j) + "*".repeat(n) + ".".repeat(n - j)
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// + ".".repeat(n - j) + "*".repeat(n) + ".".repeat(j);
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String s = repeat(".", j) + repeat("*", n) + repeat(".", n - j)
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+ repeat(".", n - j) + repeat("*", n) + repeat(".", j);
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ans[i] = s;
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j++;
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}
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return String.join(System.lineSeparator(), ans);
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}
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static String repeat(String s, int count) {
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StringBuilder stringBuilder = new StringBuilder();
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for (int i = 0; i < count; i++) {
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stringBuilder.append(s);
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}
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return stringBuilder.toString();
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}
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}
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/*
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牛客周赛 Round 1
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游游画U
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https://ac.nowcoder.com/acm/contest/60245/A
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题目描述
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游游想让你画一个大小为 n 的 "U" ，你能帮帮她吗？
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具体的画法见样例说明。
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输入描述:
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一个正整数 n。
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1≤n≤50
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输出描述:
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共输出4n 行，每行输出一个长度为4n 的，仅包含 '*' 和 '.' 两种字符的字符串。
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找规律
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======
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示例1
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输入
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1
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输出
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*..*
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*..*
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*..*
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.**.
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示例2
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输入
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2
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输出
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**....**
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**....**
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**....**
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**....**
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**....**
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**....**
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.**..**.
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..****..
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示例3
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输入
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3
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输出
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***......***
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***......***
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***......***
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***......***
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***......***
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***......***
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***......***
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***......***
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***......***
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.***....***.
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..***..***..
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...******...
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*/

‎nowcoder/src/main/java/nc60245/NC60245B.java‎

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package nc60245;
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import java.util.HashMap;
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import java.util.Map;
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import java.util.Scanner;
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public class NC60245B {
8+
static int n;
9+
static int[] a;
10+
static String s;
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12+
public static void main(String[] args) {
13+
Scanner scanner = new Scanner(System.in);
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n = scanner.nextInt();
15+
a = new int[n];
16+
for (int i = 0; i < n; i++) {
17+
a[i] = scanner.nextInt();
18+
}
19+
s = scanner.next();
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System.out.println(solve());
21+
}
22+
23+
private static String solve() {
24+
Map<Integer, Integer> red = new HashMap<>();
25+
Map<Integer, Integer> blue = new HashMap<>();
26+
for (int i = 0; i < n; i++) {
27+
if (s.charAt(i) == 'R') {
28+
red.put(a[i], red.getOrDefault(a[i], 0) + 1);
29+
} else {
30+
blue.put(a[i], blue.getOrDefault(a[i], 0) + 1);
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}
32+
}
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long ans = 0;
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for (Map.Entry<Integer, Integer> entry : red.entrySet()) {
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long rcnt = entry.getValue();
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int bcnt = blue.getOrDefault(entry.getKey(), 0);
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ans += rcnt * bcnt;
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}
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return String.valueOf(ans);
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}
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}
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/*
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游游的数组染色
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https://ac.nowcoder.com/acm/contest/60245/B
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题目描述
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游游拿到了一个数组，其中一些数被染成红色，一些数被染成蓝色。
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游游想知道，取两个不同颜色的数，且它们的数值相等，有多少种不同的取法？
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我们定义，两种取法如果取的某个数在原数组的位置不同，则定义为不同的取法。
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输入描述:
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第一行输入一个正整数n，代表数组的长度。
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第二行输入n 个正整数ai，代表数组中的元素。
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第三行输入一个长度为n ，仅包含 'R' 和 'B' 两种字符的字符串，第i 个字符为 'R' 代表数组第i 个元素被染成红色，为 'B' 代表被染成蓝色。
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1≤n≤200000
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1≤ai≤10^9
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输出描述:
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输出一个整数，代表取数的方案数量。
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乘法原理
61+
======
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示例1
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输入
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5
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1 2 1 2 2
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BRRBB
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输出
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3
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说明
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第一种取法，取第一个数和第三个数，颜色不同且都是1。
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第二种取法，取第二个数和第四个数，颜色不同且都是2。
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第三种取法，取第二个数和第五个数，颜色不同且都是2。
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示例2
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输入
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3
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2 3 3
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BBB
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输出
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0
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说明
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所有数都是蓝色，显然取不出两个颜色不同的数。
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*/

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