使用量子和产品表进行循环调度
我想进行循环调度。我已经设置了数量= 2.它将生成'时间'和'额外'
时间和额外时间取决于突发时间。量子= 2.突发时间= 10.因此，时间为2，额外为8.如果突发时间= 1，则时间为1，额外为0.如果突发时间= 2，则时间= 2，额外= 0。 / p>
这是我的作品，我有点卡在那里。 Idk如何为时间和额外做出生成。
package roundrobinscheduling;
import java.util.Scanner;
public class RoundRobinScheduling {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int noOfProcess = 0;
int quantum = 2;
System.out.print("Number of Process: ");
noOfProcess = input.nextInt();
int[] burstTime = new int[noOfProcess];
int[] priority = new int[noOfProcess];
String[] process = new String[noOfProcess];
int[] waitingTime = new int[noOfProcess];
for (int i = 0; i < noOfProcess; i++) {
System.out.print("Name Process: ");
process[i] = input.next();
System.out.print("Burst Time: ");
burstTime[i] = input.nextInt();
System.out.print("Priority: ");
priority[i] = input.nextInt();
}
int[] temp = new int[0];
for (int k = 0; k < noOfProcess; k++) {
}
int[] baki = new int[0];
int bakitemp = 0;
System.out.println("|--Process--|--Burst Time--|--Priority--|--Time--|--Extra--|");
for (int j = 0; j < noOfProcess; j++) {
if (burstTime[j] == quantum) {
baki[j] = bakitemp;
time[j] = burstTime[j];
} else if (burstTime[j] > quantum) {
//burstTime[j] = burstTime[j] - quantum;
baki[j] = burstTime[j] - quantum;
time[j] = quantum;
} else {
baki[j] = bakitemp;
time[j] = burstTime[j];
}
if (j > 0) {
waitingTime[j] = waitingTime[j - 1] + burstTime[j-1];
}
System.out.println("| " + process[j] + " | " + burstTime[j] + " | " + priority[j] + " | " + time[j] + " | " + baki[j] + " | " + " A " + waitingTime[j]);
}
}}
新任务： 我有一个我需要在本周日提交的作业。如果你想要完整的问题，那就好了。但是如果我能做到的话，我想逐一弄明白。 “如果用户输入了有效的帐号和正确的PIN码 帐户，屏幕显示主菜单。如果用户输入无效 帐号或PIN码不正确，屏幕显示相应的 消息，然后ATM返回步骤1重新启动身份验证 处理。 “ 那么，如何使数组存储系统中的帐号和密码。当我们保存时输入正确的帐号和密码，我们可以继续。我想制作至少3个帐号。我希望你不介意帮助我。
帐户类：将其添加到与其他ATM类相同的包中
class Account {
private final int accountNumber; // A 5 digit number
private final int pinNumber; // A 5 digit number
private final String accountName; // The first Name
private double accountBalance; // The money in the account
public Account(int accountNumber, int pinNumber, String accountName) {
super();
this.accountNumber = accountNumber;
this.pinNumber = pinNumber;
this.accountName = accountName;
accountBalance = 0.0;
}
public double getAccountBalance() {
return accountBalance;
}
public void setAccountBalance(double accountBalance) {
this.accountBalance = accountBalance;
}
public int getAccountNumber() {
return accountNumber;
}
public int getPinNumber() {
return pinNumber;
}
public String getAccountName() {
return accountName;
}
public void debitAccount(double amount) {
accountBalance -= amount;
}
public void creditAccount(double amount) {
accountBalance += amount;
}
}
AccountService类：
class AccountService {
Account[] accounts;
public AccountService() {
accounts = new Account[3];
accounts[0] = new Account(12345, 123, "Eisom");
accounts[1] = new Account(98765, 321, "Hazman");
accounts[2] = new Account(14123, 456, "Aina");
accounts[0].setAccountBalance(800.00);
accounts[1].setAccountBalance(650.00);
accounts[2].setAccountBalance(500.00);
}
public boolean isValidAccount(int accountNumber, int pinNumber) {
for (int j = 0; j < accounts.length; j++) {
if ((accountNumber == accounts[j].getAccountNumber()) && (pinNumber == accounts[j].getPinNumber())) {
return true;
}
}
return false;
}
public Account getAccount(int accountNumber, int pinNumber) {
for (int j = 0; j < accounts.length; j++) {
if ((accountNumber == accounts[j].getAccountNumber()) && (pinNumber == accounts[j].getPinNumber())) {
return accounts[j];
}
}
return null;
}
}
现在您必须在自动柜员机程序中使用AccountService类：
// Now the ATM has an account
private Account account;
// The service to lookup the account based on account number and pin
AccountService service = new AccountService();
// New loop using account service
while (true) {
MenuDisplay();
InputAccountNumber();
boolean isValid = service.isValidAccount(accountNumber, pinNumber);
if (isValid) {
account = service.getAccount(accountNumber, pinNumber);
Menu();
} else
System.out.println("Wrong Account Number/PIN Number. Please Try Again.");
}
现在，您的其余代码必须使用account。帐户获取者和设置者可以访问此人的姓名，余额。
例如：
public void BalanceEnquiries() {
System.out.println("");
System.out.println("-------BALANCE ENQUIRIES-------");
System.out.println("User Account Balance: RM" + account.getAccountBalance());
System.out.println("-------------------------------");
System.out.println("");
System.out.println("Proceeding back to Main Menu...");
System.out.println("");
}
让我知道到目前为止这是否有意义。
以下是取款示例：
public void attemptWithdrawal(double amount) {
if (account.getAccountBalance() < amount) {
System.out.println("Not Insuffient...");
System.out.println("You put exceed from your account balance. Please try again");
System.out.println("Please put lower amount.");
return;
}
System.out.printf("Withdrawal RM%.2f?\n", amount);
moneyAfterWithdrawal = account.getAccountBalance()-amount;
System.out.printf("psst: your remaining account will be RM%.2f\n", moneyAfterWithdrawal);
System.out.println("[Y]Yes [N]No");
confirmWithdrawal = input.next().charAt(0);
if ((confirmWithdrawal == 'Y') || (confirmWithdrawal == 'y')) {
System.out.println("Congrate! You successfully withdrawal the money");
realMoney -= moneyAfterWithdrawal;
account.debitAccount(amount); // Take the money out
} else if ((confirmWithdrawal == 'N') || (confirmWithdrawal == 'n')) {
// Going back to this method
System.out.println("");
System.out.println("Proceeding to Others again...");
} else {
System.out.println("");
System.out.println("Wrong Character...");
}
}
public void Withdrawal() {
System.out.println("");
System.out.println("-------WITHDRAWAL-------");
System.out.println(" [1]RM50.00 [2]RM100.00");
System.out.println(" [3]Others [4]Exit");
System.out.println("------------------------");
System.out.print("Enter the Code: ");
chooseOptionWithdrawalString = input.next();
chooseOptionWithdrawal = Integer.parseInt(chooseOptionWithdrawalString);
switch (chooseOptionWithdrawal) {
case 1:
attemptWithdrawal(50.0);
break;
case 2:
attemptWithdrawal(100.0);
break;
case 3:
System.out.println("How many money you want to withdrawal: ");
moneyWithdrawal = input.nextDouble();
attemptWithdrawal(moneyWithdrawal);
break;
case 4:
break;
default:
System.out.println("");
System.out.println("You enter wrong input. Please try again.");
System.out.println("");
break;
}
}
备注： 1.帐号和密码。
a）使用字符串会更容易。但是整数都很好。 b）密码必须是5位数（您使用的是3位） c）目前允许使用帐号100000（但不应该是6位数） d）否则本节很好！
- 提款：
- 存款
a）如果您创建attemptWithdraw(double amount)之类的新方法，它会对您有所帮助。你有很多重复的代码（比如50美元，100美元，当用户选择其他数量时）。无论用户是想要50,100还是200，都应该调用完全相同的方法（attemptWithdraw）。在这种方法中，你可以：
a）确保用户有足够的钱（如果他们没有，则发出警告）
b）确保自动柜员机有足够的钱（如果自动柜员机没有，则发出警告）
c）借助amount借记/减少人员帐户（如果a和b是好的）
d）提醒用户拿钱
创建此额外方法后，将其调用为50,100和其他方法。您会发现它更容易，因为您只编写一次相同的代码。
a）我不认为你检查0退出。
b）此人的余额不会被扣除，因为您使用新的变量moneyAfterDeposit。您需要像accountBalanceTemp一样使用realMoney。
c）我认为您需要告诉他们插入存款单。
d）有两分钟计时器;如果他们不这样做，你必须中止交易。
我会在那里停下来。你有很多很好的代码。只需再做一些更改，你就会在那里。
1 个答案:
答案 0 :(得分：1)
执行循环调度而没有任何初始优先级（即A，然后是B）。处理继续，直到所有进程都没有剩余突发。
public class RoundRobinScheduling {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int noOfProcess = 0;
int quantum = 2;
System.out.print("Number of Process: ");
noOfProcess = input.nextInt();
int[] burstTime = new int[noOfProcess];
int[] priority = new int[noOfProcess];
String[] process = new String[noOfProcess];
int waitingTime = 0;
int processesComplete = 0;
for (int i = 0; i < noOfProcess; i++) {
System.out.print("Name Process: ");
process[i] = input.next();
System.out.print("Burst Time: ");
burstTime[i] = input.nextInt();
if (burstTime[i] == 0)
processesComplete++;
System.out.print("Priority: ");
priority[i] = input.nextInt();
}
input.close();
int roundRobinIndex = 0;
System.out.println(" | Process | CPU Burst | Priority | Time | Waiting Time | ");
while (processesComplete < noOfProcess) {
if (burstTime[roundRobinIndex] > 0) {
// Here we want to subtract the full quantum (2)
// But what if burst time left = 1? Then we can't subtract 2
// Math.min takes 2 arguments, and returns the smaller of the two
// Math.min(2, 1) = 1
// Math.min(2, 2) = 2
// Math.min(2, 3) = 2
int time = Math.min(quantum, burstTime[roundRobinIndex]);
burstTime[roundRobinIndex] -= time;
// Determine if this process has finished
// It is finished when bursttime has reduced to zero
// processComplete = processComplete + 1
if (burstTime[roundRobinIndex] == 0)
processesComplete++;
System.out.println(" | " + process[roundRobinIndex] + " | " + burstTime[roundRobinIndex]
+ " | " + priority[roundRobinIndex] + " | " + time + " | " + waitingTime
+ " | ");
// waitingTime += time;
waitingTime += quantum; // I think this is correct (CPU will take full quantum each time
}
roundRobinIndex = (roundRobinIndex + 1) % noOfProcess;
}
}
}
对于提供的数据（流程A-E），产生以下结果。
| Process | CPU Burst | Priority | Time | Waiting Time |
| A | 8 | 3 | 2 | 0 |
| B | 0 | 1 | 1 | 2 |
| C | 0 | 3 | 2 | 4 |
| D | 0 | 4 | 1 | 6 |
| E | 3 | 2 | 2 | 8 |
| A | 6 | 3 | 2 | 10 |
| E | 1 | 2 | 2 | 12 |
| A | 4 | 3 | 2 | 14 |
| E | 0 | 2 | 1 | 16 |
| A | 2 | 3 | 2 | 18 |
| A | 0 | 3 | 2 | 20 |
编辑：这是Round的修复：
public class Round {
int process = 0;
// DON'T create the array yet!
// You don't know how big to make it (yet).
// Until you ask how many processes there are
int[] processvalue;
Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
Round r = new Round();
r.input();
r.loop();
r.print();
}
void input() {
System.out.print("how many process");
// n=Integer.parseInt(br.readLine());
process = sc.nextInt();
// HERE is where you now must create the array
// Now that you know how large to make it
processvalue = new int[process];
}
void loop() {
for (int i = 0; i < process; i++) {
System.out.println("please input the process " + (i + 1));
processvalue[i] = sc.nextInt();
}
}
void print() {
for (int k = 0; k < process; k++) {
System.out.println("value" + processvalue[k]);
}
}
}
编辑正确的等待时间
public class RoundRobinScheduling {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int noOfProcess = 0;
int quantum = 2;
System.out.print("Number of Process: ");
noOfProcess = input.nextInt();
int[] burstTime = new int[noOfProcess];
int[] priority = new int[noOfProcess];
String[] process = new String[noOfProcess];
int[] nextTime = new int[noOfProcess];
int clockTime = 0;
double totalWaitTime = 0;
int processesComplete = 0;
for (int i = 0; i < noOfProcess; i++) {
System.out.print("Name Process: ");
process[i] = input.next();
System.out.print("Burst Time: ");
burstTime[i] = input.nextInt();
if (burstTime[i] == 0)
processesComplete++;
System.out.print("Priority: ");
priority[i] = input.nextInt();
nextTime[i] = 0;
}
input.close();
int roundRobinIndex = 0;
System.out.println(" | Process | CPU Burst | Priority | Time | Clock Time | Wait Time |");
while (processesComplete < noOfProcess) {
if (burstTime[roundRobinIndex] > 0) {
int time = Math.min(quantum, burstTime[roundRobinIndex]);
burstTime[roundRobinIndex] -= time;
if (burstTime[roundRobinIndex] == 0)
processesComplete++;
int waitTime = clockTime - nextTime[roundRobinIndex];
totalWaitTime += waitTime;
System.out.println(" | " + process[roundRobinIndex] + " | " + burstTime[roundRobinIndex]
+ " | " + priority[roundRobinIndex] + " | " + time + " | " + clockTime
+ " | " + waitTime + " |");
//clockTime += quantum;
clockTime += time;
nextTime[roundRobinIndex] = clockTime;
}
roundRobinIndex = (roundRobinIndex + 1) % noOfProcess;
}
System.out.println("Average wait time="+totalWaitTime/noOfProcess);
}
}
创建输出
| Process | CPU Burst | Priority | Time | Clock Time | Wait Time |
| A | 8 | 3 | 2 | 0 | 0 |
| B | 0 | 1 | 1 | 2 | 2 |
| C | 0 | 3 | 2 | 3 | 3 |
| D | 0 | 4 | 1 | 5 | 5 |
| E | 3 | 2 | 2 | 6 | 6 |
| A | 6 | 3 | 2 | 8 | 6 |
| E | 1 | 2 | 2 | 10 | 2 |
| A | 4 | 3 | 2 | 12 | 2 |
| E | 0 | 2 | 1 | 14 | 2 |
| A | 2 | 3 | 2 | 15 | 1 |
| A | 0 | 3 | 2 | 17 | 0 |
Average wait time=5.8
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