The 2021 Spring Recruitment Algorithm Written Test Questions are organized for your reference!
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top notch
Xiaomei is learning the operating system recently.
Streaming is an important concept in operating systems.
Xiaomei has also implemented a device similar to /dev/random, but it can only provide a pre-set but looping arrangement of lowercase letters.
package MT;
import java.util.Scanner;
public class MT1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String in1 = sc.next();
String in2 = sc.next();
char[] chars1 = in1.toCharArray();
char[] chars2 = in2.toCharArray();
int length1 = chars1.length;
int length2 = chars2.length;
int count = 0;
int i = 0;
while(true) {
for (int j = 0; j < length1; j++) {
if (chars1[j] != chars2[i]) {
count++;
} else {
if (i < length2-1) i++;
else break;
}
}
if (i == length2-1) break;
}
System.out.println(count);
}
}
Two, multiple sets
Xiao Tuan now has two equal-sized multiple sets A and B. She now wants to make A set and B set equal.
package MT;
import java.util.HashMap;
import java.util.Scanner;
public class MT2 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int[] a = new int[n];
int[] b = new int[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextInt();
for (int i = 0; i < n; i++)
b[i] = sc.nextInt();
HashMap<Integer, Integer> mapA = new HashMap<>();
HashMap<Integer, Integer> mapB = new HashMap<>();
for (int i = 0; i < n; i++) {
if (mapA.containsKey(a[i])) mapA.put(a[i], mapA.get(a[i]) + 1);
else mapA.put(a[i], 1);
}
//System.out.println(mapA);
for (int i = 0; i < n; i++) {
if (mapB.containsKey(b[i])) mapB.put(b[i], mapB.get(b[i]) + 1);
else mapB.put(b[i], 1);
}
//System.out.println(mapB);
for (int i = 1; i < m; i++) {
boolean match = true;
for (Integer num : mapA.keySet()) {
int numMod = (num + i) % m;
if (mapB.containsKey(numMod) && mapA.get(num) == mapB.get(numMod))
continue;
else {
match = false;
break;
}
}
if (match) System.out.println(i);
}
}
}
//6 8
//1 1 4 5 1 4
//3 0 4 0 3 0
Three, milk tea
The school is holding a milk tea competition. For this competition, n cups of milk tea were prepared, with serial numbers from 1 to n.
package MT;
import java.util.Scanner;
public class MT3 {
private static int minTime = Integer.MAX_VALUE;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
double c = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
findTime(arr, 0, m, 0, c);
System.out.println(minTime);
}
private static void findTime(int[] arr, int begin, int people, int time, double c) {
if (people == 1) {
double sum = 0;
for (int i = begin; i < arr.length; i++) {
sum += arr[i];
}
time = Math.max(time, (int) Math.ceil(sum / c));
minTime = Math.min(minTime, time);
} else {
for (int i = begin; i < arr.length; i++) {
double sum = 0;
for (int j = begin; j <= i; j++) {
sum += arr[j];
}
findTime(arr, i + 1, people - 1, Math.max(time, (int) Math.ceil(sum / c)), c);
}
}
}
}
Four, on duty
There are n people in Xiaomei's and Xiaotuan's classes, numbered 1 to n respectively. Among them, Xiaomei’s number is 1, and Xiaotuan’s number is 2.
Insert picture description here
package MT;
import java.util.Scanner;
public class MT4 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int[][] duty = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
duty[i][j] = sc.nextInt();
}
}
System.out.println(studentID(m, duty));
}
static int studentID(int m, int[][] dutyTable) {
int id = 0;// 值日的人
// 初值
int a1 = 1, a2 = 2;
// 第一天小美 “1”，第二天小团 “2”，从第三天开始，第三天dutyTable[a2-1][a1-1]
int i = 0;
while ( i < (m - 2)) {
id = dutyTable[a2-1][a1-1];
a1 = a2;
a2 = id;
i++;
}
return id;
}
}
// 输入
//3 7
//0 3 2
//3 0 3
//2 1 0
Five, solution
Xiaomei is doing a chemistry experiment. This experiment needs to configure a variety of solution concentrations, so the solution concentration needs to be calculated in detail before the solution is prepared to avoid misoperation.
package MT;
import java.util.Scanner;
public class MT5 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// 初始溶液质量 m0 和 质量分数 w0
int m0 = sc.nextInt();;
double w0 = sc.nextInt() / 100.0;
int n = sc.nextInt(); // 操作数
// A操作之后的溶液质量和质量分数 mNew、wNew
int mNew = m0; double wNew = w0;
// 记录 A 操作加入溶液质量 m，加入溶液质量分数 w 的缓存值，方便B操作回退
int mTmp = m0; double wTmp = w0;
String B = "B", A = "A";
int i = 0;
while (i < n) {
// 操作 n 次，
String input = sc.next();
if (input == A) {
int m = sc.nextInt(); // 加入溶液质量 m
double w = sc.nextInt() / 100.0; // 加入溶液质量分数 w
// A操作之后的溶液质量和质量分数 mNew、wNew
mNew = m0 + m;
wNew = (m0 * w0 + m * w) / mNew;
mTmp = m; wTmp = w; // 记录 A 操作加入的量
}
if (input == B) {
if (mNew == m0 && wNew == wTmp) continue;
// 回退 上一步的 A，就是 根据公式反求一下
mNew -= mTmp;
wNew = (mNew * wNew - mTmp * wTmp) / (mNew - mTmp);// 反推公式中的变量
}
i++;
System.out.println(mNew);
System.out.println(String.format("%.5f",wNew)); // 保留5位小数
}
}
}