面向对象(C++)程序设计 (上机考试)
样题１. 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 shape{ Pubilc:
Virtual float area()=0 };
float total (shape *s[ ], int n) {
float sum=0.0;
for(int i=0; i
解答: #include
class shape{ public:
virtual float area()=0; };
float total(shape *s[], int n) {
float sum=0;
for(int i=0; i
class triangle : public shape{ protected: float H, W; public:
triangle(float h, float w) { H=h;
W=w;}
float area() { return H*W*0.5;} };
class rectangle : public triangle{ public:
rectangle(float h, float w) :
triangle(h, w) {}
float area() { return H*W;} };
void main() {
shape *s[4];
s[0] = new triangle( 3.0, 4.0 ); s[1] = new rectangle( 2.0, 4.0 ); s[2] = new triangle( 5.0, 8.0 ); s[3] = new rectangle( 6.0, 8.0 ); float sum = total(s,4);
cout << \
endl;
}
样题２. 以面向对象的概念设计一个类，此类包括
３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(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 << \<< endl;
cin >> unleadcontent;
cout << \endl;
cin >> leadcontent;
total = account.calculate(unleadcontent, leadcontent);
1
cout << \<< endl; }
样题３. 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=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 << \H & W:\ cin >> h1 >> w1;
Rectangle rec1(h1,w1);
cout << \H & W:\ cin >> h2 >> w2;
Rectangle rec2(h2,w2);
totalarea = rec1.add_area(rec2);
cout << \total area of the two rectangle is:\ }
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 complex 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.imag; return complex(r,i); }
void complex :: show() {
if (imag>0) {
if (imag == 1) cout << real << \
else cout << real << \<< imag << \ }
else if (imag<0) {
if (imag == -1) cout << real << \
else cout << real << imag << \ }
else cout << real; }
2
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 << endl; 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
float X, Y; public:
Position(float xi, float yi){ X=xi, Y=yi;} void Move(float xo, float yo){ X+=xo, 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); }
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 << \distance p1(\<< p1.GetX() << \
<< \to origin is: \<< dis0 << endl;
cout << \p1.GetX() << \
<< \\ << dis << 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(Watermelon & wa) {
n++;
weight=wa.weight; totalWeight += weight; }
3
~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 << \initial weight of watermelon: \
<< Watermelon :: getTotal() << endl;
cout << \input weight of watermelon:\ cin >> w;
Watermelon wa1(w);
cout << \input weight of watermelon:\ cin >> w;
Watermelon wa2(w);
cout << \input weight of watermelon:\ cin >> w;
Watermelon wa3(w);
cout << Watermelon :: getNum() << \ cout << \total weight is: \<< Watermelon :: getTotal() << endl; wa3.Watermelon::~Watermelon();
cout << \one watermelon was withdrawed!\
cout << Watermelon :: getNum() << \ cout << \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
class triangle : public shape{ protected: float H, W; public:
triangle(float h, float w) { H=h; W=w;}
float area() { return H*W*0.5;} };
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; } };
4