C++程序設(shè)計(jì)習(xí)題

/C++程序設(shè)計(jì)練習(xí)題學(xué)院：計(jì)算機(jī)學(xué)院專(zhuān)業(yè)班級(jí)：物聯(lián)網(wǎng)1002學(xué)號(hào)：姓名：徐遠(yuǎn)志第八章1.下面是一個(gè)類(lèi)的測(cè)試程序,試設(shè)計(jì)出能是用如下測(cè)試程序的類(lèi).Intmain<>{Testx;x.initx<30,20>;x.printx<>;return0;}解：#include<iostream>usingnamespacestd;classTest//類(lèi)的開(kāi)始{public: voidinitx<inti,intj>; voidprintx<>;private: intx,y;};voidTest::initx<inti,intj>{ x=i; y=j;}voidTest::printx<>{ cout<<x<<"*"<<y<<"="<<x*y<<endl;}//類(lèi)的結(jié)束intmain<>//測(cè)試函數(shù){ Testx; x.initx<30,20>; x.printx<>; return0;}得到的測(cè)試結(jié)果：4.定義并實(shí)現(xiàn)一個(gè)矩形類(lèi)Crectangle。該類(lèi)包含了下列成員函數(shù)。Crectangle<>:累的構(gòu)造函數(shù).根據(jù)需要可以定義多個(gè)構(gòu)造函數(shù)SetTop<>,SetLeft<>:設(shè)置矩形的左上角坐標(biāo)SetLength<>,SetWidth<>:設(shè)置矩形的長(zhǎng)和寬Perimeter<>:求矩形的周長(zhǎng)Area<>:求矩形的面積GetWidth<>:返回矩形的寬度Getlength<>:返回矩形的長(zhǎng)度IsSquare<>:判斷矩形是否為正方形Move<>:將矩形從一個(gè)位置移動(dòng)到另一個(gè)位置Size<>:改變矩形的大小Where<>:返回矩形的左上角的坐標(biāo)PrintRectangle<>:輸出矩形的四個(gè)頂點(diǎn)的坐標(biāo)數(shù)據(jù)成員inttop,left;intlength,width;解：#include<iostream>usingnamespacestd;classCrectangle//類(lèi)的開(kāi)始{ inttop,left; intlength,width;public: Crectangle<intt=0,intl=0,intlen=1,intw=1> { top=t;left=l; if<len>0>length=len; elselength=0; if<w>0>width=w; elsewidth=0; } voidSetTop<intt> { top=t; } voidSetLeft<intl> { left=l; } voidSetLength<intlen> { length=len; } voidSetWidth<intw> { width=w; } intPerimeter<> { return2*<width+length>; } intArea<> { returnwidth*length; } intGetWidth<> { returnwidth; } intGetLength<> { returnlength; } intIsSquare<> { if<width==length> { cout<<"該矩形是正方形"<<endl; return1; } else { cout<<"該矩形不是正方形"<<endl; return0; } } voidMove<intx,inty> { left+=x; top+=y; } voidSize<intl,intw> { length=l; width=w; } voidPrintRectangle<> { cout<<"左上方的點(diǎn)為:<"<<left<<","<<top<<">"<<endl; cout<<"右上方的點(diǎn)為:<"<<left+length<<","<<top<<">"<<endl; cout<<"左下方的點(diǎn)為:<"<<left<<","<<top+width<<">"<<endl; cout<<"右下方的點(diǎn)為:<"<<left+length<<","<<top+width<<">"<<endl; }};//類(lèi)的結(jié)束intmain<>{ Crectanglea<1,1,5,5>; a.PrintRectangle<>; a.SetTop<2>;a.SetLeft<3>; a.SetLength<4>; a.SetWidth<5>;a.IsSquare<>; cout<<"周長(zhǎng)為:"<<a.Perimeter<><<endl; cout<<"面積為:"<<a.Area<><<endl; a.PrintRectangle<>;cout<<"用getwidth等函數(shù)得到的值為："<<endl; cout<<"長(zhǎng)為:"<<a.GetLength<<endl; cout<<"寬為:"<<a.GetWidth<<endl; return0;}得到的測(cè)試結(jié)果：6.某次歌手大賽中.有JudgeNum個(gè)評(píng)委給選手打分.參加比賽的選手有PlayerNum名.現(xiàn)為比賽積分編寫(xiě)一個(gè)CompetitionResult類(lèi).類(lèi)的定義如下：〔定義略寫(xiě)出所有的成員函數(shù)和實(shí)現(xiàn)代碼。編寫(xiě)main〔函數(shù)對(duì)該類(lèi)進(jìn)行測(cè)試。在函數(shù)體中.定義了一個(gè)competitionResult類(lèi)的對(duì)象數(shù)組r[PlauerNum],他的每個(gè)元素記錄了每個(gè)選手的所有信息.個(gè)評(píng)委的打分通過(guò)鍵盤(pán)的輸入.在屏幕上應(yīng)有提示信息進(jìn)行交互式操作.比賽結(jié)果按選手得分從高到底排列輸出。解：〔1CompetitionResult::CompetitionResult<>{ num=0;strcpy<name,"">; for<inti=0;i<JudgeNum;i++> score[i]=0.0; average=0;}CompetitionResult::CompetitionResult<shortn,char*ps>{ num=n;strcpy<name,ps>; for<inti=0;i<JudgeNum;i++> score[i]=0.0; average=0;}floatCompetitionResult::MaxScore<>{intmax=score[0]; for<inti=0;i<JudgeNum;i++> if<max<score[i]>max=score[i]; returnmax;}floatCompetitionResult::MinScore<>{intmin=score[0];for<inti=0;i<JudgeNum;i++> if<min>score[i]>min=score[i]; returnmin;}voidCompetitionResult::SetAvg<>{inti;floatsum=0.0;for<i=0;i<JudgeNum;i++>sum+=score[i];sum=sum-MaxScore<>-MinScore<>;average=sum/<JudgeNum-2>;}floatCompetitionResult::GetAvg<>{returnaverage;}shortCompetitionResult::GetNo<inti>{returnnum;}voidCompetitionResult::SetNo<inti>{num=i;}charCompetitionResult::*GetName<>{returnname;}floatCompetitionResult::GetScore<intj>{returnscore[j];}voidCompetitionResult::Setscore<intk,floatav>{score[k]=av;}voidSort<CompetitionResult*pr,intn>{inti,j,k; CompetitionResulttemp; for<i=0;i<n-1;i++> { k=i; for<j=i+1;j<n;j++> if<<pr[j].average>><pr[k].average>> k=j; if<k!=j> { temp=pr[i]; pr[i]=pr[k]; pr[k]=temp; } }}<2>intmain<>{ CompetitionResultplayer[PlayerNum]={CompetitionResult<1,"one">, CompetitionResult<2,"two">,CompetitionResult<3,"there">, CompetitionResult<4,"four">,CompetitionResult<5,"five">, CompetitionResult<6,"six">,CompetitionResult<7,"seven">, CompetitionResult<8,"eight">,CompetitionResult<9,"nine">, CompetitionResult<10,"ten">}; floata; for<inti=0;i<PlayerNum;i++> { cout<<"請(qǐng)?jiān)u委給第"<<i+1<<"個(gè)選手打分:"<<endl; for<intj=0;j<JudgeNum;j++> {cin>>a; player[i].Setscore<j,a>; } player[i].SetAvg<>; cout<<endl; } Sort<player,PlayerNum>; cout<<"最后的得分情況為："<<endl; for<intj=0;j<PlayerNum;j++> {cout<<j+1<<""; cout<<setw<6><<player[j].GetAvg<>; cout<<""<<player[j].GetName<><<endl; } return0;}程序運(yùn)行的結(jié)果為：第九章1.在下面的程序中.派生類(lèi)Derived的成員函數(shù)sets〔能否訪問(wèn)基類(lèi)base中得變量a和b？為什么？如果在基類(lèi)base中將數(shù)據(jù)成員a和b定義為私有成員.下面的程序能否通過(guò)編譯？為什么？#include<iostream>usingnamespacestd;classBase{public: voidset<inti,intj>{a=i;b=j;} voidshow<>{cout<<"a="<<a<<",b="<<b<<endl;}protected: inta,b;};classDerived:publicBase{public: voidsets<>{s=a*b;} voidshows<>{cout<<"s="<<s<<endl;}private: ints;};intmain<>{ Derivedobj; obj.set<2,3>; obj.show<>; obj.sets<>; obj.shows<>; return0;}解：可以訪問(wèn)a和b；因?yàn)閍.b為保護(hù)成員.經(jīng)公有繼承.成為派生類(lèi)中得保護(hù)成員.所以派生類(lèi)Derived的成員函數(shù)sets〔能訪問(wèn)基類(lèi)base中得變量a和b。如果定義為私有.則不能通過(guò)編譯。因?yàn)榕缮?lèi)中的成員函數(shù)不能訪問(wèn)基類(lèi)中得私有成員2．聲明一個(gè)圖形基類(lèi)Shape.在它的基礎(chǔ)上派生出矩形類(lèi)Rectangle和圓形類(lèi)Circle.它們都有計(jì)算面積的和周長(zhǎng)、輸出圖形信息的成員函數(shù).再在Rectangle類(lèi)的基礎(chǔ)上派生方形類(lèi)Square。編寫(xiě)程序和各類(lèi)的定義和實(shí)現(xiàn).以及類(lèi)的使用。解：#include<iostream>usingnamespacestd;classShape{public: doublegetArea<>{} doublegetPerimeter<>{}};classRectangle:publicShape{protected: doubleheight; doublewidth;public: Rectangle<>{} Rectangle<doublea,doubleb> { height=a; width=b; } doublegetArea<> {returnheight*width; } doublegetPerimeter<> {return2*<height+width>; }};classCircle:publicShape{public: Circle<doublex>:r<x>{} doublegetPerimeter<> { return2*3.1416*r; } doublegetArea<> { returnr*r*3.1416; }private: doubler;};classSquare:publicRectangle{public: Square<doublex> { height=x; width=x; }};intmain<>{ Rectanglea1<2.2,3.2>; Circlea2<4.2>; Squarea3<5.3>; cout<<"Area="<<a1.getArea<><<endl; cout<<"Perimeter="<<a1.getPerimeter<><<endl; cout<<"Area="<<a2.getArea<><<endl; cout<<"Perimeter="<<a2.getPerimeter<><<endl; cout<<"Area="<<a3.getArea<><<endl; cout<<"Perimeter="<<a3.getPerimeter<><<endl; return0;}5.某銷(xiāo)售公司有銷(xiāo)售經(jīng)理和銷(xiāo)售員工.月工資的計(jì)算辦法為：銷(xiāo)售激勵(lì)的底薪為4000元.并將銷(xiāo)售額的2/1000做提成.銷(xiāo)售員工無(wú)底薪.只提取5/1000為工資。編寫(xiě)程序：定義一個(gè)積累Employee.它包含三個(gè)數(shù)據(jù)成員和salary.以及編號(hào)和姓名的構(gòu)造函數(shù)。有Employee類(lèi)派生Salesman類(lèi)。Salesman類(lèi)包含兩個(gè)新數(shù)據(jù)成員commrate和sales.還包括了用于輸入銷(xiāo)售額并計(jì)算銷(xiāo)售員工工資的成員函數(shù)pay〔和用于輸出打印的print〔。由Salesman派生Salemanager,Salesmmanage類(lèi)包含新數(shù)據(jù)成員monthlypay.以及用于輸入銷(xiāo)售額并計(jì)算銷(xiāo)售經(jīng)理工資的成員函數(shù)pay〔和用于輸出的print〔。編寫(xiě)main〔函數(shù)中測(cè)試所設(shè)計(jì)的類(lèi)結(jié)構(gòu)。解：〔1classEmployee{public: Employee<longnum,char*n> { number=num; strcpy<name,n>; }protected: longnumber; charname[20]; doublesalary;};〔2classSalesman:publicEmployee{public: Salesman<longnum,char*n,floatc=0.005>:Employee<num,n> {commrate=c;} voidpay<> { cout<<"??ê?è?"<<name<<"±???íê3éμ??úê???:"; cin>>sales; salary=commrate*sales; } voidprint<>{ cout<<name<<"μ?1¤×ê?a￡o"<<salary<<endl; }protected: floatcommrate; doublesales;};<3>classSalesmanager:publicSalesman{public: Salesmanager<longnum,char*n,floatc=0.002,doublem=4000>:Salesman<num,n,c>{ monthlypay=m;} voidpay<>{ cout<<"??ê?è?"<<name<<"±???íê3éμ??úê???￡o"; cin>>sales; salary=monthlypay+commrate*sales; } voidprint<>{ cout<<name<<"μ?1¤×ê?a￡o"<<salary<<endl; }private: doublemonthlypay;};<4>intmain<>{ Salesmans1<1234,"à?èe">; Salesmanagers2<1357,"????">; s1.pay<>; s1.print<>; s2.pay<>; s2.print<>; return0;}第十章1.定義一個(gè)輔助類(lèi)Complex,重載運(yùn)算符+、-、*、/使之能用于復(fù)數(shù)的加減乘除運(yùn)算。運(yùn)算符重載函數(shù)為Complex類(lèi)的成員函數(shù)。編寫(xiě)程序.分別求出兩個(gè)復(fù)數(shù)之和、差、積、商。解：#include<iostream>usingnamespacestd;classComplex{public: Complex<doubler=0.0,doublei=0.0>{real=r;imag=i;} Complexoperator+<Complex>; friendComplexoperator-<Complex,Complex>; Complexoperator*<Complex>; friendComplexoperator/<Complex,Complex>; voiddisplay<>;private: doublereal,imag;};ComplexComplex::operator+<Complexc>{ returnComplex<real+c.real,imag+c.imag>;}Complexoperator-<Complexc1,Complexc2>{ returnComplex<c1.real-c2.real,c1.imag-c2.imag>;}ComplexComplex::operator*<Complexc>{ returnComplex<real*c.real-imag+c.imag,imag*c.imag+real*c.real>;}Complexoperator/<Complexc1,Complexc2>{ doubler,i; r=<c1.real*c2.real+c1.imag*c2.imag>/<c2.real*c2.real+c2.imag*c2.imag>; i=<c1.imag*c2.real-c1.real*c2.imag>/<c2.real*c2.real+c2.imag*c2.imag>; return<r,i>;}voidComplex::display<>{ cout<<"<"<<real<<","<<imag<<">"<<endl;}intmain<>{ Complexc1<7.1,8.4>,c2<1.4,5>,c3; cout<<"c1=";c1.display<>; cout<<"c2=";c2.display<>; c3=c1+c2; cout<<"c1+c2=";c3.display<>; c3=c1-c2; cout<<"c1-c2=";c3.display<>; c3=c1*c2; cout<<"c1*c2=";c3.display<>; c3=c1/c2; cout<<"c1/c2=";c3.display<>; return0;}2.定義一個(gè)日期類(lèi)Date.包括年.月.日等私有數(shù)據(jù)成員。要求為所定義的Date類(lèi)設(shè)計(jì)如下重載運(yùn)算符函數(shù)：Dateoperator+<intdays>;返回一日起加天數(shù)days后得到的日期Dateoperator-〔intdays>;返回一日起減天數(shù)days后得到的日期Intoperator-〔Date&b>;返回兩日期相差的天數(shù)解：#include<iostream>usingnamespacestd;intday_tab[12]={31,28,31,30,31,30,31,31,30,31,30,31};classDate{public: Date<>{} Date<inty,intm,intd>{year=y;mon=m;day=d;} voidsetday<intd>{day=d;} voidsetmon<intm>{mon=m;} voidsetyear<inty>{year=y;} intgetday<>{returnday;} intgetmon<>{returnmon;} intgetyear<>{returnyear;} Dateoperator+<intdays> { staticDatedate; intnumber=dton<*this>; number-=days; date=ntod<number>; returndate; } Dateoperator-<intdays> { staticDatedate; intnumber=dton<*this>; number-=days; date=ntod<number>; returndate; }intoperator-<Date&b> { intdays=dton<*this>-dton<b>-1; returndays; } voiddisp<> {cout<<year<<"."<<mon<<"."<<day<<endl; }private: intyear,mon,day; intleap<int>; intdton<Date&>; Datentod<int>;};intDate::leap<intyear>{ if<year%4==0&&year&100!=0||year%400==0> return1; elsereturn0;}intDate::dton<Date&d>{ inty,m,days=0; for<y=1;y<=d.year;y++> if<leap<y>> days+=366; else days+=365; day_tab[1]=leap<d.year>?29:28; for<m=0;m<d.mon-1;m++> days+=day_tab[m]; days+=d.day; returndays;}DateDate::ntod<intn>{ inty=1,m=1,d,rest=n,days_year; while<1> { days_year=leap<y>?366:365; if<rest<days_year> break; else rest-=days_year; y++; } y--; day_tab[1]=leap<y>?29:28; while<1> { if<rest>day_tab[m-1]> rest-=day_tab[m-1]; else break; m++; } d=rest; returnDate<y,m,d>;}intmain<>{ Datenow<2011,12,22>,Nationday<1949,10,1>; cout<<"now:";now.disp<>; cout<<"Nationday:";Nationday.disp<>; cout<<"?à2?ììêy:"<<<now-Nationday><<endl; Dated1=now+365,d2=now-365; cout<<"now+365:";d1.disp<>; cout<<"now-365:";d2.disp<>; return0;}3.編寫(xiě)一個(gè)程序尖酸正方體圓柱體的表面積和體積。要求抽象出一個(gè)公共的基類(lèi)Body.吧他當(dāng)做抽象類(lèi).在該類(lèi)中定義表面積和體積的純虛函數(shù)。抽象類(lèi)中定義一個(gè)數(shù)據(jù)成員data.他可以作為求的半徑.正方形的邊長(zhǎng).或者圓柱體的地面半徑。在三個(gè)類(lèi)中都有計(jì)算白面積和體積的函數(shù)的具體實(shí)現(xiàn)。解：#include<iostream>usingnamespacestd;classBody{protected: doubledata;public: Body<doubledata> { Body::data=data; } virtualdoubles<>=0; virtualdoublev<>=0;};classCube:publicBody{public: Cube<doubledata>:Body<data>{}; doubles<> {returndata*data*6; } doublev<> {returndata*data*data; }};classSphere:publicBody{public: Sphere<doubler>:Body<r>{}; doubles<> { returndata*data*3.1416*4; } doublev<> { return4*data*data*data/3; }};classCylinder:publicBody{ doubleheight;public: Cylinder<doubler,doubleheight>:Body<r>{Cylinder::height=height;} doubles<> { return<2*3.1416*data*data+2*data*3.1416*height>; } doublev<> { returndata*data*3.1416*height; }};intmain<>{ Body*p; Cubeobj1<5>; Sphereobj2<5>; Cylinderobj3<5,5>; p=&obj1; cout<<"正方體的表面積："<<p->s<><<endl; cout<<"正方體的體積："<<p->v<><<endl; p=&obj2; cout<<"球的表面積："<<p->s<><<endl; cout<<"球的體積："<<p->v<><<endl; p=&obj3; cout<<"球的表面積："<<p->s<><<endl; cout<<"球的體積："<<p->v<><<endl; return0;}第十一章已知VoidSort<inta[],intsize>;VoidSort<doublea[],intsize>;是一個(gè)函數(shù)模板的兩個(gè)例子.起功能是將數(shù)組a中得前size個(gè)元素按從小到大的順序依次排列。是設(shè)計(jì)函數(shù)模板。解：#include<iostream>usingnamespacestd;template<classT>voidSort<Tset[],intn>{ inti,j; Ttemp; for<i=1;i<n;i++> for<j=n-1;j>i;j--> if<set[j-1]>set[j]> { temp=set[j-1]; set[j-1]=set[j]; set[j]=temp; }}intmain<>{ inta[]={4,5,8,9,3}; doubleb[]={3,5,6.7,2,5.2,9.2,10.3}; Sort<a,6>; Sort<b,6>; for<inti=0;i<6;i++> cout<<a[i]<<""; cout<<endl; for<i=0;i<6;i++> cout<<b[i]<<""; cout<<endl; return0;}設(shè)有如下聲明classTest{public：VoidSetData1<intval>{data1=val;}VoidSetData2<douleval>{data2=val;}intGetData1<>{returndata1;}doubleGetData2<>{returndata2;}private:intdata1;doubledata2;}試將此類(lèi)聲明為類(lèi)模板的聲明.使得數(shù)據(jù)成員data1和data2可以使任何類(lèi)型#include<iostream>#include<string>usingnamespacestd;template<classT1,classT2>classTest{public:voidSetData1<T1val>{data1=val;}voidSetData2<T2val>{data2=val;}T1GetData1<>{returndata1;}T2GetData2<>{returndata2;}private:intdata1;doubledata2;};intmain<>{ Test<int,double>t1; t1.SetData1<10>; t1.SetData2<5.4>;cout<<t1.GetData1<><<""<<t1.GetData2<><<endl; Test<char,string>t2; t2.SetData1<'a'>; t2.SetData2<"China">; cout<<t2.GetData1<><<""<<t2.GetData2<><<endl; return0;}3棧是一種重要的數(shù)據(jù)結(jié)構(gòu).它是一種只允許在彪的一端進(jìn)行插入或刪除操作的線性表。表中允許進(jìn)行插入、刪除操作的一端稱(chēng)為棧頂。表的令一端稱(chēng)為棧底。下面是一個(gè)整形棧類(lèi)的定義：ConstintMaxSize=100;ClassIstack{Public:Istack<>;voidPush<int&n>;voidpop<>;intGetTop<>;boolEmpty<>;intsize<>;boidClearStack<>;~Istack<>;Private:Intelem[MaxSize];Inttop;}編寫(xiě)一個(gè)棧的類(lèi)模板.以便為任何類(lèi)型的對(duì)象提供棧結(jié)構(gòu)的數(shù)據(jù)操作。解：#include<iostream>#include<string>usingnamespacestd;constintMaxSize=100;template<classT>classStack{ public: Stack<>; voidPush<constT&a>; voidPop<>; TGetTop<>; voidClearStack<>; boolEmpty<>; intSize<>; ~Stack<>{} private: Telem[MaxSize]; intTop;};template<classT>Stack<T>::Stack<>{ Top=-1;}template<classT>voidStack<T>::Push<constT&a>{ if<Top==MaxSize-1> return; Top++; elem[Top]=a;}template<classT>voidStack<T>::Pop<>{ if<Top==-1> return; Top--;}template<classT>voidStack<T>::ClearStack<>{ Top=-1;}template<classT>TStack<T>::GetTop<>{ if<Top==-1> return0; returnelem[Top];}template<classT>boolStack<T>::Empty<>{ if<Top==-1> returntrue; else returnfalse;}template<classT>intStack<T>::Size<>{ returnTop+1;}intmian<>{ Stack<int>s1; Stack<string>s2; s1.Push<123>; s1.Push<456>; cout<<s1.GetTop<><<endl; s1.Pop<>; cout<<s1.Size<><<""<<s1.Empty<><<endl; s2.Push<"beijing">; s2.Push<"wuhan">; cout<<s2.Size<><<""<<s2.Empty<><<endl; return0;}第十三章1.編寫(xiě)程序.重載運(yùn)算符<<和>>.使用戶能直接輸入和輸出復(fù)數(shù)。解：#include<iostream.h>classComplex{public: Complex<doubler=0,doublei=0> { real=r;imag=i;} friendostream&operator<<<ostream&,Complex&>; friendistream&operator>><istream&,Complex&>;private: doublereal,imag;};ostream&operator<<<ostream&output,Complex&obj>{ output<<obj.real; if<obj.imag>0> output<<"+"; if<obj.imag!=0> output<<obj.imag<<"i"; returnoutput;}istream&operator>><istream&input,Complex&obj>{ cout<<"Inputtherealandimgofthecomplex:"<<endl; input>>obj.real; input>>obj.imag; returninput;}intmain<>{ Complexc1,c2,c3; cin>>c1>>c2>>c3; cout<<"c1="<<c1<<endl; cout<<"c2="<<c2<<endl;cout<<"c3="<<c3<<endl; return;}編寫(xiě)一個(gè)程序.統(tǒng)計(jì)文本文件abc.txt中的字符.并給文件所有行加上行號(hào).然后寫(xiě)到newabc.txt中去解：#include<iostream>#include<fstream>usingnamespacestd;intmain<>{ fstreamfile; file.open<"abc.txt",ios::in>; if<!file> { cout<<"abc.txtcan'topen."<<endl; abort<>; } charch; inti=0; while<!file.eof<>> { file.get<ch>; i++; } cout<<"Charater:"<<i<<endl; file.close<>; return0;}6．#include<iostream>#include<fstream>usingnamespacestd;structstudent{