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+<h1><center>课程作业</center></h1>
+
+<div style="text-align: center;">
+    <div><span style="display: inline-block; width: 65px; text-align: center;">课程名称</span><span style="display: inline-block; width: 25px;">:</span><span style="display: inline-block; width: 210px; font-weight: bold; text-align: left;">操作系统</span></div>
+    <div><span style="display: inline-block; width: 65px; text-align: center;">作业名称</span><span style="display: inline-block; width: 25px;">:</span><span style="display: inline-block; width: 210px; font-weight: bold; text-align: left;">课后习题1</span></div>
+    <div><span style="display: inline-block; width: 65px; text-align: center;">学号</span><span style="display: inline-block; width: 25px;">:</span><span style="display: inline-block; width: 210px; font-weight: bold; text-align: left;">21281280</span></div>
+    <div><span style="display: inline-block; width: 65px; text-align: center;">姓名</span><span style="display: inline-block; width: 25px;">:</span><span style="display: inline-block; width: 210px; font-weight: bold; text-align: left;">柯劲帆</span></div>
+    <div><span style="display: inline-block; width: 65px; text-align: center;">班级</span><span style="display: inline-block; width: 25px;">:</span><span style="display: inline-block; width: 210px; font-weight: bold; text-align: left;">物联网2101班</span></div>
+</div>
+---
+
+## 1. 问题1
+
+**哲学家进餐问题的变形。假设5支筷子都放在圆桌的中间，哲学家进餐时拿起任意两支筷子就可以进餐。其他条件不变。请用信号量实现5个哲学家进程的同步。**
+
+```pascal
+Var 
+	chopsticks, eating_philosophers: Semaphore := 5, 4;
+	
+Procedure Philosopher(i: integer);
+Begin
+    Repeat
+    	wait(eating_philosophers);
+        wait(chopsticks);
+        wait(chopsticks);
+        
+        Eat;
+        
+        signal(chopsticks);
+        signal(chopsticks);
+        signal(eating_philosophers);
+        
+        Think;
+        
+	Until False;
+End;
+```
+
+## 2. 问题2
+
+**四个进程P0,P1,P2,P3和四个信箱M0,M1,M2,M3进程间借助相邻的信箱传递消息：Pi每次从Mi中取出一条消息，经加工送入Mi+1(mod 4)中。其中M0,M1,M2,M3分别设有3,3,2,2个格子，每个格子放一条消息，初始时，M0装满了三条消息，其余为空。写出使用信号量实现进程 (i=0,1,2,3)同步的算法。**
+
+```pascal
+Var 
+	full, empty, mutex: array[0..3] of Semaphore := (3, 0, 0, 0), (0, 3, 2, 2), (1, 1, 1, 1);
+
+Procedure ProcessPi(i: integer);
+Var
+    message: Message;
+Begin
+    Repeat
+        wait(full[i]);
+        wait(mutex[i]);
+        message := Get_Message_From_Mailbox(i);
+        signal(mutex[i]);
+        signal(empty[i]);
+        
+        ProcessMessage;
+        
+        wait(empty[(i + 1) mod 4]);
+        wait(mutex[(i + 1) mod 4]);
+        Send_Message_To_Mail_box((i + 1) mod 4, message);
+        signal(mutex[(i + 1) mod 4]);
+        signal(full[(i + 1) mod 4]);
+        
+    Until False;
+End;
+```
+
+## 3. 问题3
+
+**汽车司机与售票员之间必须协同工作，汽车每到一个站后，一方面，只有当司机已经停下，售票员才能开门上下客，另一方面，只有售票员把车门关好了司机才能开车。假定某辆公共汽车上有前后门各有一名售票员以及一名司机，汽车初始状态是正在始发站停车上客。试设必要的信号量及赋初值，实现售票员进程和司机进程的同步。**
+
+```pascal
+Var
+	driver, ticket_seller: Semaphore := 1, 2;
+
+Procedure Driver;
+Begin
+    Repeat
+		wait(driver);
+        
+        Move;
+        
+        Stop;
+        
+        signal(ticket_seller);
+        signal(ticket_seller);
+
+    Until False;
+End;
+
+Procedure Ticket_seller(i: integer);
+Begin
+    Repeat
+        wait(ticket_seller);
+		wait(ticket_seller);
+		
+        Open;
+        
+        signal(driver);
+        
+        Close;
+        
+    Until False;
+End;
+```