阿里巴巴2021秋招笔试题20211119

Posted 2021-12-17 Starzkg

源代码：https://gitee.com/shentuzhigang/algorithm/tree/master/exam-alibaba/exam-alibaba-20211119

第一题

题目大意：
有长度为n的数组a
有k次机会在连续长度不超过m的区间每个元素+1使得数组全部元素变成偶数

import java.util.LinkedList;
import java.util.Scanner;

/**
 * @author ShenTuZhiGang
 * @version 1.0.0
 * @email [e-mail]
 * @date 2021-11-19 19:16
 */
public class Exam2021111901 
    public static void main(String[] args) 
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) 
            int n =  scanner.nextInt();
            int k = scanner.nextInt();
            int m = scanner.nextInt();
            int[] a = new int[n];
            int add = 0;
            LinkedList<Integer> q= new LinkedList<>();
            for (int j = 0; j < n; j++) 
                a[j] = scanner.nextInt();
                if ((a[j]+add)%2==1)
                    if(q.isEmpty())
                        k--;
                        add++;
                        q.push(j+m-1);
                    else
                        add--;
                        q.pop();
                    
                
                if(q.peek()!=null&&q.peek()==j)
                    q.pop();
                    add--;
                
            
            System.out.println(k>=0?"Yes":"No");
        

    

第二题

（未验证正确性）
题目大意：
有 单词组 strs
人数n，长度b
每个人按单词组依次报【0，i】
例：
king
king hello
king hello wrold
king hello
king
king hello
king hello wrold
king hello

求单词word第X次出现在第几行

import java.util.Scanner;

/**
 * @author ShenTuZhiGang
 * @version 1.0.0
 * @email [e-mail]
 * @date 2021-11-19 20:36
 */
public class Exam2021111902 

    public static void main(String[] args) 
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int b = scanner.nextInt();
        scanner.nextLine();
        String[] s = scanner.nextLine().split(" ");
        String word = s[0];
        int c = Integer.parseInt(s[1]);
        String[] strs = scanner.nextLine().split(" ");
        if(b==1)
            System.out.println(b);
            System.exit(0);
        
        int t = 0;
        for (int i = 0; i < 2*b-2; i++) 
            for (int j = 0; j <= i; j++) 
                if(word.equals(strs[j%n]))
                    t++;
                
            
        
        int k = c/t;
        int t0 = c%t;
        if(t0!=0)
            for (int i = 0; i < 2*b-2; i++) 
                for (int j = 0; j <= i; j++) 
                    if(word.equals(strs[j%n]))
                        t0--;
                        if(t0==0)
                            System.out.println(k*(2*b-2)+i);
                        
                    
                
            
        else 
            System.out.println(k*(2*b-2));