中山大学上机实践试题及题目答案

1、 计算机及应用专业 (b080702) 8347 计算机及应用课程实验(二) 1901032019464. 面向对象程序设计 (上机考试)样题. 下列shape类是一个表示形状的抽象类，area()为求图形面积的函数，total()则是一个通用的用以求不同形状的图形面积总和函数。请从shape类派生三角形类(triangle) 、矩形类(rectangle),并给出具体的求面积函数。编写程序验证求面积函数的正确性。shape、 total的定义如下所示。class shapepubilc:virtual float area()=0;float total (shape *s , int n)

2、float sum=0.0;for(int i=0; iarea ();return sum;解答:#include class shape public: virtual float area()=0; ;float total(shape *s, int n) float sum=0; for(int i=0; i area(); return sum;class triangle : public shape protected: float h, w; public: triangle(float h, float w) h=h; w=w; float area() return h*

3、w*0.5;class rectangle : public triangle public: rectangle(float h, float w) : triangle(h, w) float area() return h*w;void main() shape *s4; s0 = new triangle( 3.0, 4.0 ); s1 = new rectangle( 2.0, 4.0 ); s2 = new triangle( 5.0, 8.0 ); s3 = new rectangle( 6.0, 8.0 ); float sum = total(s,4); cout the t

4、otal area is: sum endl;样题. 以面向对象的概念设计一个类，此类包括个私有数据，unlead(无铅汽油), lead有铅汽油, total (当天总收入)。其中，无铅汽油价格是17/升，有铅汽油价格是16/升，请以构造函数的方式建立此值，并编写程序，该程序能够根据加油量，自动计算出油站当天的总收入。解答:#include class income private: float unlead, lead, total; public: income(float ul, float l)unlead=ul; lead=l; total=0.0; float calculate

5、(float, float);float income : calculate(float unleadcontent, float leadcontent) total = unlead*unleadcontent + lead*leadcontent; return total;void main() float unleadcontent, leadcontent, total; income account(17, 16); cout please input unlead content: unleadcontent; cout please input lead content:

6、leadcontent; total = account.calculate(unleadcontent, leadcontent); cout the total income is: total endl;样题. 编写一个计算两个给定长方形的面积的程序，要求长方形用一个类(rectangle)来表示，在该类中增加定义一个成员函数 add_area(), 该成员函数使用对象作为参数，用来计算两个给定长方形的面积。解答:#include class rectangle private: float h, w; public: rectangle(float h, float w)h=h; w=

7、w; float area() return h*w; float add_area(rectangle &);float rectangle : add_area(rectangle &rec) return area() + rec.area();void main() float h1, w1, h2, w2, totalarea; cout please input the 1st rectangle h & w: h1 w1; rectangle rec1(h1,w1); cout please input the 2st rectangle h & w: h2 w2; rectan

8、gle rec2(h2,w2); totalarea = rec1.add_area(rec2); cout the total area of the two rectangle is: totalarea endl;样题4. 定义一个复数类complex,该类至少提供加、减、赋值、输出等操作，所有操作均以友元形式实现。编写程序验证其功能。解答:#include class complex private: float real; float imag; public: complex(); complex(int r, int i) real=r; imag=i; friend compl

9、ex operator + (complex &, complex &); friend complex operator - (complex &, complex &); void show();complex operator + (complex & a, complex & b) float r, i; r=a.real+b.real; i=a.imag+b.imag; return complex(r,i);complex operator - (complex & a, complex & b) float r, i; r=a.real-b.real; i=a.imag-b.im

10、ag; return complex(r,i);void complex : show() if (imag0) if (imag = 1)cout real +i; else cout real + imag i; else if (imag0) if (imag = -1)cout real -i; else cout real imag i; else cout real;void main() complex a(4,5), b(2,3), x, y; x=a+b; y=a-b; a.show(); cout +; b.show(); cout =; x.show(); cout en

11、dl; a.show(); cout -; b.show(); cout =; y.show(); cout endl;样题5.定义一个平面几何中点的位置类position, 它应该包含有移动、计算两点间的距离(包括到原点的距离)，求x坐标、y坐标等操作，其中计算两点间的距离以友元函数形式实现。编写程序验证其功能。解答:#include #include class position private: float x, y; public: position(float xi, float yi) x=xi, y=yi; void move(float xo, float yo) x+=xo,

12、 y+=yo; float getx()return x; float gety()return y; float distancetoorigin(); friend float distance(position &, position &);float position : distancetoorigin() return sqrt( x*x + y*y );float distance(position & a, position & b) float dx = a.x - b.x; float dy = a.y - b.y; return sqrt( dx*dx + dy*dy);

13、void main() position p1(1.5, 3.5), p2(4.5, 6.5); p1.move(3.5, 5.5); float dis0 = p1.distancetoorigin(); float dis = distance( p1, p2 ); cout the distance p1( p1.getx() , p1.gety() ) to origin is: dis0 endl; cout the distance between p1( p1.getx() , p1.gety() ) and p2( p2.getx() , p2.gety() ) is dis

14、endl;样题6.利用类和对象，编制出一个卖瓜的程序。每卖一个瓜要计出该瓜的重量，还要计算所卖出瓜的总重量及总个数，同时卖瓜时还允许退瓜。(提示：将每个瓜设为对象; 用静态成员变量分别统计卖出瓜的总重量和总个数; 卖瓜行为用构造函数模拟，退瓜行为用析构函数模拟。)解答:#include class watermelon private: static int n; static float totalweight; float weight; public: watermelon(float w) n+; weight=w; totalweight += w; watermelon(water

15、melon & wa) n+; weight=wa.weight; totalweight += weight; watermelon() n-; totalweight -= weight; int getweight() return weight; static int getnum() return n; static int gettotal() return totalweight; ;int watermelon : n = 0;float watermelon : totalweight = 0;void main() float w; cout the initial wei

16、ght of watermelon: watermelon : gettotal() endl; cout please input weight of watermelon: w; watermelon wa1(w); cout please input weight of watermelon: w; watermelon wa2(w); cout please input weight of watermelon: w; watermelon wa3(w); cout watermelon : getnum() nos. watermelons were sold. endl; cout

17、 the total weight is: watermelon : gettotal() endl; wa3.watermelon:watermelon(); cout now one watermelon was withdrawed! endl; cout watermelon : getnum() nos. watermelons were sold. endl; cout the total weight is: watermelon : gettotal() endl;样题6.编写一个程序，用于计算三角形、矩形和圆的总面积。(提示：由于尚不能确定该程序计算的具体形状，可以先定义一个

18、抽象的类shape, 对于具体种类的形状，通过从shape派生一个类来对其进行描述。)解答:#include class shape public: virtual float area()=0; ;float total(shape *s, int n) float sum=0; for(int i=0; i area(); return sum;class triangle : public shape protected: float h, w; public: triangle(float h, float w) h=h; w=w; float area() return h*w*0.

19、5;class rectangle : public triangle public: rectangle(float h, float w) : triangle(h, w) float area() return h*w;class circle : public shape protected: float radius; public: circle(float r) radius=r; float area() return radius*radius*3.14; ;void main() shape *s4; s0 = new triangle( 3.0, 4.0 ); s1 =

20、new rectangle( 2.0, 4.0 ); s2 = new circle( 5.0 ); s3 = new circle( 8.0 ); float sum = total(s,4); cout the total area is: sum endl;题样7.编写一个程序，可以输入个学生的总分，并按总分从高到低排序，要求设计一个学生类 student, 并编写其所有成员函数的代码，类student的定义如下：class student int total; / 总分成绩 pubic: void get_score(); /获取一个学生的成绩 void display(); /显示一

21、个学生的成绩 void sort (student *); /将若干学生成绩按总分从高到低排序解答:#include #include class student int total; public: student(int m)total=m; int get_score(); void display(); void sort(student *s3);int student : get_score() return total;void student : display() cout setw(6) total endl;void student : sort(student * s3

22、) student * t; for (int i=0; ii; j-) if (sj-get_score() sj-1-get_score() t=sj;sj=sj-1;sj-1=t; cout the sorted score is: ; for (int k=0; k3; k+) cout setw(6) get_score(); cout endl;void main() int s0, s1, s2; cout please input 3 score: s0s1s2; student *s3; s0=new student(s0); s1=new student(s1); s2=n

23、ew student(s2); s0-sort(s);样题8.设计一个栈操作类，该类包含入和出栈成员函数，编写程序，入栈一组数据：(5, 2, 6, 7, 3)，然后屏幕显示出栈结果。解答:#include #include const int size=20;class stack private: int datasize; int top; public: stack() top=-1; void push(int); int pop();void stack : push( int c) if (top=19) cout stack overflow! endl; else data+

24、top=c;int stack : pop() if (top=-1) cout stack underflow! endl; return null; else return datatop-;void main() stack s; s.push(5); s.push(2); s.push(6); s.push(7); s.push(3); for (int i=0; i5; i+) cout setw(6) s.pop(); cout endl;样题9.编写一个输入、显示学生(用类student表示)和教师(用类teacher表示)数据的程序，学生的数据包括：编号(no)、 姓名(nam

25、e)和班号 (class)， 教师的数据包括：编号(no)、 姓名(name)和部门(dept)。要求将编号、姓名的输入和显示两项操作设计成一个类person，并将它作为student类和teacher类的基类。解答:#include #include class person protected: char no5; char name10; public: virtual void input(); virtual void show();class student : public person private: char class3; public: virtual void inp

26、ut(); virtual void show();class teacher : public person private: char dept3; public: virtual void input(); virtual void show();void person : input() cout please input no.: no; cout please input name: name;void person : show() cout.setf(ios:left); cout.width(8); cout no.: no endl; cout.width(8); cout

27、 name: name endl;void student : input() person : input(); cout please input class: class;void student : show() person : show(); cout.setf(ios:left); cout.width(8); cout class: class endl;void teacher : input() person : input(); cout please input dept: dept;void teacher : show() person : show(); cout

28、.setf(ios:left); cout.width(8); cout dept: dept endl;void main() student stu1, stu2; teacher tea1; cout please input 1st students information: endl; stu1.input(); cout please input 2st students information: endl; stu2.input(); cout please input teachers information: endl; tea1.input(); cout endl stu

29、dent: endl; stu1.show(); stu2.show(); cout endl teacher: endl; tea1.show();样题10.编写一个可以删除文本文件中所有以“/”开头字符串的c+程序。要求必须使用c+的i/o流成员函数来完成。解答:#include #include void main(int argc, char * argv) char ch; if (argc!=3) cout error, you shall use this programs as: nt purgefile filename1 filename2 endl; return; if

30、stream myin(argv1); if (!myin) cout cant open file argv1 endl; return; ofstream myout(argv2); if(!myout) cout cant create file argv2 endl; return; while (myout & myin.get(ch) if ( ch = / ) myin.get(ch); if ( ch = /)do while (myin.get(ch) & (ch!= )&(ch!=n); else myout.put(/);myout.put(ch);myin.get(ch

31、);domyout.put(ch); while (myin.get(ch) & (ch != )&(ch!=n);if (ch = )|(ch = n) myout.put(ch); else if (ch= )|(ch=n) myout.put(ch); else myout.put(ch); myin.get(ch); do myout.put(ch); while (myin.get(ch) & (ch!= )&(ch!=n); if (ch = )|(ch = n) myout.put(ch); myin.close(); myout.close(); cout purge comp

32、leted. endl;样题11.己知类test含有整型私有数据成员num，编写主程序，要求：1) 写出test的完整定义，并含有必要的函数成员;2) 建立长度为9、元素类型为test的动态数组且初值为0;3) 将各元素的值均设置为0x7ff;4) 显示各元素的值;5) 删除动态数组;解答:#include #include class test private: int num; public: test() num=0; test(int n) num=n; void setnum(int n) num=n; int getnum() return num; ;void main() te

33、st *ptr=new test9; int i; for (i=0; i9; i+) ptri.setnum(0x7ff); for (i=0; i9; i+) cout setw(8) hex ptri.getnum(); cout 0 ) return add( succ(a), pred(b) ); if ( b0 ) return sub( pred(a), pred(b) ); if ( b0 ) return sub( succ(a), succ(b) );void main() int a, b, x, y; printf(please input a,b:n); scanf(

34、%d,%d, &a, &b); x=add(a,b); y=sub(a,b); printf(%d+%d=%dn, a, b, x); printf(%d-%d=%dn, a, b, y);样题2. 试编写一个求解josephus问题的函数。用整数序列1, 2, 3, , n 表示顺序围坐在圆桌周围的人，并采用数组表示作为求解过程中使用的数据结构。然后使用n=9, s=1, m=5，以及n=9, s=1, m=0，或者n=9, s=1, m=10作为输入数据，验证程序的正确性。解答:#include “stdio.h”void josephus( int a, int n, int s, in

35、t m) int i, j, k, tmp; if ( m = 0 ) printf(error! m cant set 0!n); return; for ( i=0; i1; k-) if ( i = k ) i=0; i = ( i+m-1 ) % k; if ( i != k-1) tmp = ai; for ( j=i; jk-1; j+ ) aj = aj+1; ak-1 = tmp; for ( k=0; kn/2; k+ ) tmp=ak; ak=an-k-1; an-k-1=tmp; void main() int a20, n, s, m, i; printf(please

36、 input n, s, m:n); scanf(%d,%d,%d, &n, &s, &m); josephus(a, n, s, m); if (m!=0) printf(the score is:); for (i=0; in; i+)printf(%4d, ai); printf(n); 样题3. 依次输入10个整数，分别用顺序表与单链表存储，并实现其就地逆置。解答1 (顺序表):#include “stdio.h”typedef struct int data20; int length; seqlist;void main() seqlist *l; int tmp; int i,

37、j; printf(please input 10 integer:n); for (i=0; idatai); l-length=10; for (i=0, j=l-length-1; idatai; l-datai=l-dataj; l-dataj=tmp; printf(the reversed order:n); for (i=0; ilength; i+) printf(%6d, l-datai); printf(n);解答2 (单链表):#include stdio.h#include malloc.htypedef struct node int data; struct nod

38、e *next; listnode;typedef listnode * linklist;void main() int i, n; linklist l; listnode *s, *r; l=null; r=null; printf(please input 10 integer:n); for(i=0; idata = n; if (l=null) l=s; else r-next = s; r=s; if(r!=null) r-next = null; s=null; while(l) r = l; l = l-next; r-next = s; s = r; l = s; prin

39、tf(the reversed order:n); r=l; while(r) printf(%6d, r-data); r=r-next; printf(n);样题4. 设循环队列类型如下 typedef struct datatype datamaxsize; int front, rear; cycqueue;cyqueue sq;编写入队操作int enqueue(cycqueue sq, datatype x)的代码，并调试通过。(提示：将x入队列sq，成功返回1，否则，返回0)解答:#include stdio.h#define maxsize 100typedef char da

40、tatype;typedef struct datatype datamaxsize; int front, rear; cycqueue;int enqueue(cycqueue *sq, datatype x) if ( (sq-rear+1)%maxsize = sq-front ) printf(queue overflow!n); return 0; sq-rear = (sq-rear+1) % maxsize; sq-datasq-rear = x; return 1;datatype dequeue(cycqueue *sq) if ( sq-rear = sq-front ) printf(queue underflow!n); return null; sq-front = (sq-front+1) % maxsize; return ( sq-datasq-front );void main() int i, n; char c; cycqueue sq; sq.rear=0; sq.front=0; printf(please input cha