Huawei machine test questions-20190417

Introduce

This machine test question , Time of occurrence ：2019 year 04 month 17 Japan ,19：00.

Last exam question , Time of occurrence ：2019 year 04 month 03 Japan ,19：00.

This machine test 3 topic , all AC.

1

Title Description ：
Given an array , There are 6 It's an integer , Find the maximum that this array can represent 24 What's the base time , Output this time , Cannot represent output invalid.

Input description ：
The input is an array of integers , There are six integers in the array .
The length of the input integer array is 6, There is no need to consider other lengths , The element value is 0 Or a positive integer ,6 Each number can only be used once .

Output description ：
Output as a 24 Time in hexadecimal format , Or string “invalid”.

Example 1
Input
[0,2,3,0,5,6]
Output
23:56:00
Example 2
Input
[9,9,9,9,9,9]
Output
invalid

remarks ：
The output time format is xx:xx:xx Format .

Code

// Direct violence solution .
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		final String s = "invalid";
		final String c = ":";
		Scanner in = new Scanner(System.in);
		String tempString = in.nextLine();
		in.close();
		tempString = tempString.substring(1, tempString.length() - 1);
		String [] tempInt = tempString.split(",");
		tempString = null;
		int [] input = new int [6];
        for(int i = 0; i < 6; i ++) {
            input[i] = Integer.parseInt(tempInt[i]);
        }
        tempInt = null;
        Arrays.sort(input);
        ArrayList<Integer> list = new ArrayList<>();
        for(int i = 0; i < 6; i ++) {
        	list.add(input[i]);
        }
        input = null;
        StringBuilder output = new StringBuilder();
        int i = getSpeci(list, 3); // first place 
        if(i != -1 ) {
        	output.append(list.remove(i));
        } else {
        	System.out.print(s);
        	return ;
        }
        if(output.charAt(0) == '2') {
        	i = getSpeci(list, 4); // Second 
        	if(i != -1 ) {
        		output.append(list.remove(i));
        		output.append(c);
        	}else {
            	System.out.print(s);
            	return ;
            }
        } else {
        	i = getSpeci(list, 10); // Second 
        	if(i != -1 ) {
        		output.append(list.remove(i));
        		output.append(c);
        	}else {
            	System.out.print(s);
            	return ;
            }
        }
        i = getSpeci(list, 6); // Third 
        if(i != -1 ) {
    		output.append(list.remove(i));
    	}else {
        	System.out.print(s);
        	return ;
        }
        i = getSpeci(list, 10); // Fourth place 
    	if(i != -1 ) {
    		output.append(list.remove(i));
    		output.append(c);
    	}else {
        	System.out.print(s);
        	return ;
        }
    	i = getSpeci(list, 6); // Fifth 
        if(i != -1 ) {
    		output.append(list.remove(i));
    	}else {
        	System.out.print(s);
        	return ;
        }
        i = getSpeci(list, 10); // Sixth place 
    	if(i != -1 ) {
    		output.append(list.remove(i));
    	}else {
        	System.out.print(s);
        	return ;
        }
        System.out.print(output);
	}
	static int getSpeci(ArrayList<Integer> list, int speci) {
		for(int i = list.size() - 1; i > -1; i --) {
			if(list.get(i) < speci) {
				return i;
			}
		}
		return -1;
	}
}

2

Title Description ：
Xiao Wang has a little spare money , I want to do some small business selling fruits . Give two arrays m、n, use m[i] On behalf of the i The cost price of a fruit ,n[i] On behalf of the i The price at which fruit can be sold , If you have capital now k, How much money can you earn at most in the end ？

explain ：
1 You only need to buy each kind of fruit once , It can only be sold once
2 Array m、n Not larger than 50
3 Array elements are positive integers , No more than 1000

Input description ：
1 Array m、n
2 capital k
remarks ：
1 Enter a comma separated array on the first line m The element value of
2 In the second line, enter a comma separated array n The element value of
3 On the third line, enter the capital

Output description ：
How much money can you earn at most .

Example 1
Input
4,2,6,4
5,3,8,7
15
Output
22

explain
Example calculation process ：
Before buying first 3 Kind of fruit , Sell all , Buy again 4 Kind of fruit , Resell , Finally, the principal becomes 22.

Code

import java.util.Arrays;
import java.util.Scanner;
public class Main {
	static class B implements Comparable<B> {
		int m;
		int n;
		B(int m, int n) {
			this.m = m;
			this.n = n;
		}
		@Override
		public int compareTo(B o) {
			if(m > o.m) {
				return 1;
			}
			if(m < o.m) {
				return -1;
			}
			if(n > o.n) {
				return -1;
			}
			if(n < o.n) {
				return 1;
			}
			return 0;
		}
	}
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String [] mm = in.nextLine().split(",");
        String [] nn = in.nextLine().split(",");
        int k = in.nextInt();
        B [] b = new B [mm.length];
        for(int i = 0; i < b.length; i ++) {
        	b[i] = new B(Integer.parseInt(mm[i]), Integer.parseInt(nn[i]));
        }
        mm = null;
        nn = null;
        Arrays.sort(b);
        for(int i = 0; i < b.length; i ++) {
        	if(k >= b[i].m && b[i].n > b[i].m) {
        		k += (b[i].n - b[i].m);
        	}
        }
        System.out.print(k);
        in.close();
        
    }
}

3

Title Description
A multiprocessor multiprocessor batch processing system allows all jobs to be transferred into memory at one time , And can execute in parallel , Its parallel number is equal to the number of processors . The system adopts SJF The scheduling method of （ The shortest work is preferred , When the system is scheduling , Always give priority to the jobs with the shortest processing time ）.
Now the number of processors is given m, Number of assignments n, The processing time of each job is t1, t2, … tn.
When n>m when , First, deal with those with short time m Jobs enter the processor for processing , Others enter and wait , When a job is processed , Take out the jobs with the shortest processing time from the waiting queue in turn and enter the processing .
Find out how much time it takes the system to process all jobs ？
notes ： Do not consider the consumption of job switching .

Input description ：
Input 2 That's ok , First act 2 It's an integer （ Space separated ）, Respectively identify the number of processors m Number of jobs n; Second line input n It's an integer （ Space separated ）, Identify the processing time of each job t1, t2,… tn.0<m, n<100,0<t1, t2,… tn<100.

Output description ：
Total output processing time

Example 1
Input
3 5
8 4 3 1 10
Output
13

remarks ：
notes ： Do not consider the legitimacy of input .

Code ：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
public class Main{
	static class CPU{
		int time;
		CPU() {
			time = 0;
		}
	}
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int m = in.nextInt();
		int n = in.nextInt();
		int [] task = new int [n];
		for(int i = 0; i < n; i ++) {
			task[i] = in.nextInt();
		}
		in.close();
		ArrayList<CPU> cpu = new ArrayList<>();
		for(int i = 0; i < m; i ++) {
			cpu.add(new CPU());
		}
		Arrays.sort(task);
		int total = 0;
		int index = 0;
		while(true) {
			int i = getZero(cpu);
			if(i == 101) {
				break;
			}
			if(i >= 0) {
				if(index == n) {
					cpu.remove(i);
				} else {
					cpu.get(i).time = task[index];
					index ++;
				}
			} else {
				total -= i;
			}
		}
		System.out.print(total);
	}
	static int getZero(ArrayList<CPU> cpu) {
		if(cpu.size() == 0) {
			return 101;
		}
		int min = 100;
		for(int i = 0; i < cpu.size(); i ++) {
			if(cpu.get(i).time == 0) {
				return i;
			}
			min = min > cpu.get(i).time ? cpu.get(i).time : min;
		}
		for(int i = 0; i < cpu.size(); i ++) {
			cpu.get(i).time -= min;
		}
		return -min;
	}
}