计算机二级C++操作题答案-绝版小抄.docx

00:00:0000:01:00(1)if (+seconds) = 60) (2)return *this; (3)show(+sw); My name is Zhang.(1)Person() (2)virtual void (3): Person(s), gpa(g) (4)Person* p s1 = abcdes2 = abcs3 = abdif (s = NULL) return true; int len = strlen(s); if (len size) return false; for (int i = 0; i len; i+) if ( stri != si) return false; return true;12345678910(1)delete array; (2)cout arrayi ; (3)MyClass obj(10); 10,9,8,7,6,5,4,3,2,1，1,2,3,4,5,6,7,8,9,10，(1) anum = e; (2) Array(s) (3) aj = aj - 1; (4) ai = e; 二维向量的类MyVector；程序应当显示(6,8)。但程序中有缺失部分，请按照以下提示，把缺失部分补充完整：(1)在/*1* *found* 的下方是构造函数的定义，它用参数提供的坐标对x和y进行初始化。(1) MyVector:MyVector (2) i.x - j.x, i.y - j.y (3) v1 + v2 u1234(1)const int Size = 4; (2)void MyClass:Print() (3)MyClass obj(intArray); 教 材 名：C+语言程序设计页 数：299作 者：张三相关课程：面向对象的程序设计(1)writer = new charstrlen(the_writer)+1; (2)delete title,writer; (3)Book(the_title,pages,the_writer) (4)a_book.theCourse() endl 01:02:03.00406:04:06.021(1)Number seconds; (2)advanceMinutes(seconds.advance(k) (3) s+; n -= base; 1.41421(1)Point(double m=0.0, double n=0.0):x(m), y(n)(2)double Point:distanceTo(const Point& p)(3)coutf1.distanceTo(f2)switchOff()空货物表的总价：0添加 办公桌/12.00 后：6144添加 书柜/6.00 后：6144添加 椅子/35.00 后：26724char the_nameMAX_NAME_LENGTH;strncpy(the_name, name, MAX_NAME_LENGTH);for(int i=0; icounter; i+) if(strcmp(the_name, namesi)=0) return i;return -1;1.41421(1)Point(double m=0.0, double n=0.0):x(m), y(n)(2)double distanceBetween(const Point& p1, const Point& p2)(3)coutdistanceBetween(f1, f2)switchOn();空货物表的总价：0添加 办公桌/12.00 后：6144添加 书柜/6.00 后：6144添加 椅子/35.00 后：26724for(int i=0; i=0) total+=prices.getValue(index)*goods.getValue(i); 1.41421(1)Point(double m=0.0, double n=0.0):x(m), y(n)(2)double distanceBetween(const Point& p1, const Point& p2)(3)coutdistanceBetween(f1, f2)switchOn();刚建立时：添加8，3后：8 3添加5后：8 3 5删除首元素8后：3 5添加-7后：3 5 -7删除首元素3后：5 -7Node *p;p = head;while(p) osdatanext;1958年3月23日比2001年11月5日更早2001年11月5日比1958年3月23日更晚2001年11月5日与2001年11月5日是同一日期(1)Date(int yyyy, int mm, int dd): year(yyyy),month(mm),day(dd)(2)return day=a1p1 + p2 的结果: 7.3X4+20.6X3-41.2X2-2.4X+5p2 + p3 的结果: -2.3X5+14.6X4+12.8X3+2.8X2+0.2X+1int i;for( i=0; inum_of_terms;i+) ci=coefi;for( i=0; ix.num_of_terms; i+) ci+=x.coefi;1958年3月23日比2001年11月5日更早2001年11月5日比1958年3月23日更晚2001年11月5日与2001年11月5日是同一日期(1)Date(int yyyy, int mm, int dd): year(yyyy),month(mm),day(dd)(2)return daydt.day;(3)void compareDate( Date dt1, Date dt2) c, b, a（1）charmaxSize（2） p（3）c（4）ptop- +7.3X4+20.6X3-41.2X2-2.4X1+5X0-2.3X5+14.6X4+12.8X3+2.8X2+0.2X1+1X0for(int i=0; iSum(); sum=55。(1)public Base (2)Base:sum() (3)Derived: (4)Base(m1, m2) 5 85,4,3,2,10,0,8.4,5.6,4.5,3.4,2.3,1.2for (int i = 0, j = m-1; i j; i+, j-) Type temp = ai; ai = aj; aj = temp; 2 5 8 3 66 3 8 5 2（1）a=new intn;（2）return a;（3）for(i=0; i5; i+) coutbi=a1? (x1=a0，x2=a1): (x1=a1，x2=a0); for (int i = 2; i size; i+) if (x2 ai) if (x1 ai) x2 = x1; x1 = ai; else x2 = ai; return x1+x2;2006-1-12005-12-312005-12-312006-1-1(1): year(y), month(m), day(d) (2)Date(const Date & d) (3)int year, month, day;Point: (0,0)3Point: (4,5)628.2743 113.097(1)cc.PrintP() (2)Circle (3)cc(cen) (4)b.PrintP()829, 20, 33, 12, 18, 66, 25, 1412, 14for (int i = 2; i ai) if (x1 ai) x2 = x1; x1 = ai; else x2 = ai; 32_11(1)int ABC:c=0;(2)ABC:ABC(int aa, int bb)(3)coutx.Sum()_y.Sum()endl;得到的输出应该是：s=28(1)public(2)int Sum()(3)Base:Sum()(4)mem3(m3)8 55 3int i, c=0;for(i=0; i=x) c+;return c;u(4,4)(1)Point(double x, double y) :_x(x),_y(y) 或Point(double x, double y) _x = x; _y = y; (2)void Move(double xOff, double yOff) (3)cout ( pt.GetX() , pt.GetY() ) endl; Im a derived class.(1)virtual void Show() = 0 (2)strcpy( _p, s) (3)public Base1, private Base2 (4)Base2(s)从文本文件in.dat中读取全部整数，将整数序列存放到intArray类的对象myArray中，然后对整数序列按非递减排序，最后由函数writeToFile选择序列中的部分数据输出到文件out.dat中。文件in.dat中的整数个数不大于300个。for (int i = 0; i length; +i) for (int j = i; j arrayj) int temp; temp = arrayi; arrayi = arrayj; arrayj = temp; for (int a = 0; a length; +a) cout arraya ;57（1）ai=aai;（2）d+=x.GetA(i);（3）int f=x.SumA(5);s=144（1）mem1(m1)，mem2(m2)（2）return mem1*mem2（3）mem3(m3)（4）return XBase:product()*mem33 52 2.3 24.5 7 5int i;Type sum= 0，temp=a0;for(i=0; i ai) temp = ai;x =sum;return temp; 890（1）Desk(int kg):weight(kg)，price(0.0)（2）void setPrice(double p) price=p;（3）d.setPrice(890);1.41421（1）x(p.getX()，y(p.getY()（2）double px，double py（3）end1(x1，y1)，end2(x2，y2)（4）sqrt(dx*dx+dy*dy)删除前：5 -3 -4 2 -8 9 1 4 -7 8 -9 3删除后：5 2 9 1 4 8 3while(tocounter)if(arrayto0) for(int i=to; ilink = head (2)p = new node (3)Insert(p) v = 1,2,3,4,5v = 2,3,4,5,1v = 3,4,5,1,2v = 4,5,1,2,3v = 5,1,2,3,4for (int i = 0; i size-1; i+) int temp = vi; vi = vi+1; vi+1 = temp; uConstructorThe value is 10Destructor(1)MyClass() (2)int value; (3)void MyClass:Print() const 013,11,0XB(1)virtual const char* toString()const = 0 (2)HexNumber(int k) : Number(k) (3)cout number.toString() (4) show(HexNumber(11) 小型字符串类，字符串长度不超过100。实现默认构造函数MiniString(const char *s)和析构函数MiniString()。MiniString( const char *s = ) length = strlen(s); sPtr = new charlength+1; strcpy(sPtr, s); uConstructor called.The value is 10Max number is 20Destructor called.(1)MyClass(int i) (2)int Max(int x, int y, int z) (3)cout The value is obj.GetValue()next = top (0,2)(2,2)(2,0)(0,0)4 (1)MyPoint p1, MyPoint p2 (2)up_left.getX(), down_right.getY() (3)double MyRectangle:uConstructor called of 10The value is 10Destructor called of 10(1)void Print() const (2)MyClass() (3)int value; 其中定义了类Bag和用于测试该类的主函数main。类Bag是一个袋子类，用于存放带有数字标号的小球(如台球中用的球，在类中用一个整数值表示一个小球)，其运算符成员函数 = 用来判断两个袋子对象是否相同( (1)InBag(bagi) != b.InBag(bagi) (2)return false (3)ball = bagi (4)count+ 6.828432(1)point1(p1), point2(p2), point3(p3) (2):perimeter()const (3)(this-perimeter()/2 uConstructor called.The value is 10Copy constructor called.The value is 10Destructor called.Destructor called.(1)MyClass(int i = 10) (2)MyClass(const MyClass & p) (3)MyClass() 1 2 3 4 52 3 4 5 63 4 5 6 71 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 0 (1)getElement(i, j) (2)MatrixBase(rows, cols) (3)new doublerows*cols (4)r = c 两个数据表:1,2,3,4,5,63,4,5,6,7,8两个数据表之和:4,6,8,10,12,14for( int i=0; i len; +i) ddi = di + list.di;uEnd point 1=(1,8), End point 2=(5,2), length=7.2111。(1)Line(double x1, double y1, double x2, double y2) (2)cout ),length = length(*this) 。 endl; (3)return sqrt(l.getX1() - l.getX2() * (l.getX1() - l.getX2() + (l.getY1() - l.getY2() * (l.getY1() - l.getY2(); (1,2,3,4,5)(0,0,0,0,0,0)(1)getElement(length()-1)(2)delete val (3)s += vali (4)return 0.0; 插入前:1,2,4,5,7,8,10插入6和3后:1,2,3,4,5,6,7,8,10for( int i = 0; i len; +i) if (data i; k-) ddk = dk-1; ddi = data; for (int j = 0; j i; j+) ddj = dj; delete d; d = new doublelen; for (int index = 0; index name，name)=0;Dog named Lucky speaks WoofCat named Daisy speaks Miaow（1）charstrlen(str)+1（2）return Woof;（3）Animal(str)（4）Animal& a3 6 5 4 16 10 20 153 4 5 6 10 15 16 20int temp; for (int i = 0; i size; +i) for (int j = i; j aj) temp = ai; ai = aj; aj = temp; hello world!（1）record = new char2;（2）int size = strlen(record)+1;（3）strcpy(newrecord，record); bool IsIsosceles() const; /判断该三角形是否是等腰三角形 bool IsEquilateral() const; /判断该三角形是否是等边三角形 double CalculateSide(Point p1, Point p2); /根据顶点p1和p2计算边长 Point A,B,C; /三角形三个顶点double AB,BC,AC; /三角形的三条边长（1）(p1.GetX()-p2.GetX()*(p1.GetX()-p2.GetX()+(p1.GetY()-p2.GetY()*(p1.GetY()-p2.GetY()（2）true（3）false（4）fabs(AB-BC)MINNUM & fabs(AB-AC)MINNUM & fabs(BC-AC)A-#A-#A-#exiting inner blockexiting outer blocksListItem* temp = new sListItem; temp-data = c; temp-next = h; h = temp; Initial times are0 d : 0 h : 0 m : 59 sAfter one second times are0 d : 0 h : 1 m : 0 s(1)void Clock:print() const (2)set(+total_sec); (3)return *this;此图形是一个抽象图形, 周长=0, 面积=0此图形是一个三角形, 周长=6.82843, 面积=2（1）x(x0)，y(y0)（2）Point point1， point2， point3（3）return length(point1，point2)+length(point2，point3)+length(point3，point1)（4）void show(Shape * shape)初始字符串：C+ Primer Plus (5th Edition)转换后：C+ PRIMER PLUS (5TH EDITION)int length = Length();char Tchar;for (int i=0; i=a)&(Tchartype = type; (2)delete name; (3)strcpy(this-name,); 2010(1)virtual int area(void) = 0; (2)area() (3)length * height (4)CPolygon ValArray v1 = 1,2,3,4,5ValArray v2 = 1,2,3,4,5size = other.size; v = new intsize; for (int i = 0; i size; i+) setArray(i,other.vi)uName: SmithAge: 21ID: 99999CourseNum: 12Record: 970(1)char *Name; (2)StudentInfo() (3)StudentInfo:StudentInfo(char *name, int age, int ID, int courseNum, float record) 801501001(1)MaxSpeed(maxspeed), Weight(weight) ; (2)virtual(3)virtual(4)public bicycle, public motorcar 其中声明的CDeepCopy是一个用于表示矩阵的类。请编写这个类的赋值运算符成员函数operator =，以实现深层复制。n = r.n; delete p; p = new intn; for (int i = 0; i n; i+) pi = r.pi; return *this;uName: SmithAge: 21ID: 99999CourseNum: 12Record: 970(1)char *Name; (2)StudentInfo(char *name, int age, int ID, int courseNum, float record); (3)StudentInfo() delete Name; The area of the Circle is 78.5The perimeter of the Circle is 31.4The area of the Rectangle is 24The perimeter of the Rectangle is 20(1)virtual (2)virtual (3)itsLength(len), itsWidth(width) (4)Shape* sp; 其中声明的CDeepCopy是一个用于表示动态数组的类。请编写其中的拷贝构造函数。n = r.n; p = new intn; for (int index = 0; index n; index+) pindex = r.pindex; Name：SmithAge：21ID：99999CourseNum：12Record：970(1)char *Name; (2)StudentInfo(char *name, int age, int ID, int courseNum, float record); (3)void StudentInfo:show() const30,50center=120,89; radius=2.7(1)x(xValue), y(yValue) (2)virtual void Disp() (3): Point(xValue,yValue), radius(radiusValue)v1 =1,2,3,4,5v2 =1,2,3,4v3 =1,2,3,4,6v4 =1,2,3,4,5v1!=v2v1!=v3v1=v4if (size != other.size)/判断数组长度 return false; for (int i = 0; i size; i+)/判断数组元素 if (vi != other.vi) return false; return true;Name: Smith Age: 21 ID: 99999 CourseNum: 12 Record: 970（1）char *Name;（2）StudentInfo(char *name， int age， int ID， int courseNum， float record);（3）void StudentInfo:show()const1 2 3 4 5 6 7（1）!InSet(si)（2）setdatanum=si（3）setdatai=c（4）return true初始字符串：I love C+ Programming Language反转后的字符串：egaugnaL gnimmargorP +C evol Ichar temp = stri; stri = strj; strj = temp;Name: SmithAge: 21ID: 99999CourseNum: 12Record: 970(1)char *Name; (2)StudentInfo(char *name, int age, int ID, int courseNum, float record); (3)StudentInfo() delete Name; The area of the Circle is 78.5The perimeter of the Circle is 31.4The area of the Rectangle is 24The perimeter of the Rectangle is 20(1)virtual (2)virtual (3)itsLength(len), itsWidth(width) (4)Shape* sp;初始的两个字符串：mStr1: I love C+.mStr2: Its amazing.MiniString ms3;strcpy(buf，ms1.str);strcat(buf，ms2.str);ms3.str=new charstrlen(buf)+1;strcpy(ms3.str，buf);return ms3;uNUM = 0Value = 1(1)MyClass(int i=0) : NUM(0) (2)void Increment(MyClass& f) f._i+; (3)Increment(obj); 其中定义了Score类。Score是一个用于管理考试成绩的类。其中，数据成员_s指向存储成绩的数组，_n表示成绩的个数；成员函数Sort使用冒泡排序法将全部成绩按升序进行排列。请在程序中的横线处填写适当的代码，然后删除横线，以实现Score类的成员函数Sort。(1)i+ (2)_n-1 (3)_sj = _sj-1 (4)_sj-1 = t ValArray v1 = 1,2,3,4,5ValArray v2 = 2,2,2,2,2size = other.size; v = new int other.size; for (int i = 0; i x=x,this-y=y; (2)void show() const cout ( x , ynext = top; 求交集前：1 2 3 5 8 102 8 9 11 30 56 67求交集后：1 2 3 5 8 102 8 9 11 30 56 672 8for (int i = 0; i x=x,this-y=y; (2)void show() const cout ( x , y ); (3): p1(pt1), p2(pt2) 或p1=pt1;p2=p210 10 8 6 4 2（1）top(-1)（2）datatop-（3）return datatop（4）top=-1求交集前：set1: 1 2 3 5 8 10set2: 2 8 9 11 30 10 67set3:for (int i = 0; i =set.GetEnd(); i+) if (IsMemberOf(set.GetElement(i) asize+ = set.GetElement(i);usum=60(1)public: (2)for(int i=0; in; i+) ai=aai; (3)for(int i=0; iimagPart + otherComplex.imagPart; sum.realPart = this-realPart + otherComplex.realPart; return sum; usmallersmallersmallerlargest(1)void Judge(MyClass &obj) (2)(ptr+i)-Set (i + 1); (3)delete ptr; Before reverse：abcdefgAfter reverse：cbagfed(1)strlen(s) + 1 (2)delete m_str (3)m_strn-i-1 (4)m_strn-i-1 = tmp 重载 + 运算符。MiniString& operator+ (const MiniString& s) char *pt = new char length+1; strcpy(pt, sPtr); int blength = length; length += s.length; delete sPtr; sPtr = new charlength+1; strcpy(sPtr,pt); delete pt; fo