C程序设计基础课后习题

C++程序设计基础课后习题

Homeworkproblems

Whataretheelementsofa1-1object-orientedsystem?

Answer:theobject-orientedsystemcontainsthreeelements:

objects,classes,andinheritance.

1-2pleasedetermineifthefollowingstatementsarevalidand

why:

Answer:1)/*Thisis/*aexplanation*/sentence.The

compilerthinkstheannotationstatementisfinishedwhenit

encountersthefirst*/.Theactualannotationisthoughtto

be/*Thisis/*aexplanation*/sentence.*//Thisis

aexplanationsentence.//

Legal./Thisisanexplanationsentence./Thisisa

explanationsentence.Commentsymbol/**/and//nomatch.

//Thisisaexplanationsentence.Unmatchedannotation

symbols*/mightcausecommentstoenderrorsprematurely,

similarto(1).

Whatarethelegalc++variablenamesineachofthefollowing

lexicalsymbols?Italsoexplainswhyotherlexicalsymbolsare

illegal.

Answer:invalidvariablename:3ndcannotbeginwithanumber.

Youcanonlystartwith26lettersorunderscores.The@price

isnotlegalatthebeginning.A*b*isanillegalvariable

namecharacter.PIillegalvariablenamecharacter.Depthis

notlegalatthebeginning.Thevia.numdot"isaninvalid

variablenamecharacter.

1-4pointsouttheerrorsinthefollowingprogram:

Answer:#include<iostream.h>

Voidmain()

Const:x=1;//constshouldnothaveacolon.

Y=6;//yisassignedwithoutdefinition.

X=x+y;//xisconstconstantandcannotbereassigned.

Cout<<'X='<<<<<<<<//stringsshouldbewritten

indoublequotationmarks,"X=".

1-5trytoanalyzethevaluesofthefollowingfourexpressions

todetermineiftheirvaluesarethesame.Andcomparethe

calculationrulesofexpressionsinac++andwhatarethe

similaritiesanddifferencesinmathematicsoperationrules,

soastoexplainthefourexpressionevaluationprocessandthe

calculationofthevariousstepsoftheintermediateresults.

1)2000*5/72)2000/7*53)5/7*7*2)5/7*5)5/7*5)

5/7*5)5/7*1000

Answer:1)2000*5/7=10000/7=14282)2000/7*5=285*

5=14253)5/7*2000=0*2000=0*2000=0*2000=0.714286

*2000=1428.575)1000*5/7=7146)5/7*1000=0

1-6unlessforcedtypeconversionisperformed,thefour

arithmeticresultsbetweenintegersarestillintegers.The

integeranddecimalarithmeticresultsinthehigherprecision

principle.3.2.6pleasepointoutthedifferencesofthefour

variablesvarl〜var4shownbelow:

Answer:1)intvarl=1999;Variistheintegervariable.2)

constvar2=1999;Var2istheintegerconstant.3)volatile

intvar3=1999;Integervar3isthememoryvariable,andthe

programwillnotbestoredintheregister.3)constvolatile

var4=1999;Integervar4ismemoryconstant.

TherearefourmembersofA:A,B,CandDinAdormitory.They

meetthefollowingconditions:1)Aneversingsinthedormitory;

2)onlyAisinthedormitory,BandDsing.3)Cdoesn'tsing

inanysituation.Accordingtotheconditionsgiven,writethe

c++logicalexpressionfor"nobodysinginginthedorm”.

A:Aisinthedormitory,B,DmeansBandDsing.Aslongas

BandDarenotsinginginthedormitory,nooneissingingin

thedormitory,nomatterwhatAorCis.AslongasAisnot

inthedorm,BandDdon'tsing,andnooneinthedormcansing.

Sothefullconditionalexpressioncanbewrittenas:solution

one:!A).(necessaryandsufficientcondition)solutiontwo:!

(B&D);Method:three!B&&!D;(actually,thelogical

transformationofsolutiontwo).A;

Whichstatementshavethesamefunctioninthefollowingfour

ifstatements?1)if(pl)if(p2)statement;2)if(p2)if(pl)

statement;3)thestatement;4)if(pl,|,|p2)statement;

Answer:1),2),3)thesamefunction,bothconditionalpland

p2,whenthestatementisexecuted.

1-9afactoryhasprioritymaintenanceonthefollowing

equipment:productioncapacityisgreaterthan1000unitsper

hour,andhasbeenrepairedlessthan5timesinthepast,or

morethan10yearsofservicelife.Pleasecompileanddebug

thec++programthatdeterminesiftheplant'sequipment

shouldbemaintainedfirst.

Answer:thekeytotheproblemistowritethelogical

expressionofprioritymaintenance:ability>1000&&ers.1:

#include<iostream.H>

Voidmain()

Intability,repair,lifetime;

Cout<<"pleaseentertheproductioncapacityofthedevice

(piece/hour),pastmaintenancetimesandservicelife

(year):";

Cin>,>,>repair>lifetime.

"If”(ability>1000&&)

Theequipmentshouldberepairedfirst.;

Theelse

Cout<<"thedeviceshouldnotberepairedfirst.

)

Method2:

If(ability>1000){

If(repair<5||lifetime>10)

Theequipmentshouldberepairedfirst.;

)

Theelse

Cout<<"thedeviceshouldnotberepairedfirst.

Usethesecondmethodshouldpayattentiontoistheif-else

statementsmatchingambiguity:c++,theelseconnectedtothe

firstandisvisibleifnotmatching,theontologyasthesecond

iftheelsearenotvisible,sotheelseifmatchedwithablity

infront.Butforthereadabilityofaprogram,thesecond

programmingapproachisnotadvocated,andthefirstisbetter

forreadability.

1-10trytocreateac++programthatsimulatesacalculator

forfouroperations.

Answer://sourcecodeisasfollows:

#include<iostream.H>

Voidmain()

DoubleDNumADNumB;

CharCComput;

TheInputnumberandoperator

“('p'toquit,'c,toclean)

Cin>>DNumA;

BoolBComputer=true;

While(BComputer)

CComput=，p，;

Cin>>CComput;

Theswitch(CComput){

Case'+':

Cin>>DNumB;

DNumA=DNumA+DNumB;

Cout<<\t=，z<<<<<<<<<<<<<<<<<<<<<<<

<<<<<<<<<<<<<<<<<<<

Break;

Case:'*'

Cin>>DNumB;

DNumA=DNumA*DNumB;

Cout<<\t=，z<<<<<<<<<<<<<<<<<<<<<<<

<<<<<<<<<<<<<<<<<<<

Break;

Case'-':

Cin>>DNumB;

DNumA=DNumA-DNumB;

Cout<<\t="<<<<<<<<<<<<<<<<<<<<<<<

<<<<<<<<<<<<<<<<<<<

Break;

Case'/':

Cin>>DNumB;

If(DNumB!=0.0)

(

DNumA=DNumA/DNumB;

Cout<<\t=，z<<<<<<<<<<<<<<<<<<<<<<<

<<<<<<<<<<<<<<<<<<<

}

Theelse

Zerodivisor.

Break;

Case'c

TheInputnumberandoperator

“('p'toquit,'c'toclean):“

Cin>>DNumA;

Break;

Default:

BComputer=false;

TheProgramquited.

Break;

)

}

)

//programcompute.CPPrunstheresults:

Inputnumberandoperator('p'toquit,'c'toclean)

6.7*5

=33.5

89.87

=56.37

c

Inputnumberandoperator('p,toquit,'c'toclean):

5.4/7.908=0.682853

1to11.Writeascheduletoprintprogram,accordingtothe

inputoftheweek(1~5)toprintoutthedayofthecourse,

andtheprintingprocesscanbeloop,andsettheendofthe

cycle.

A:thec++languageprovidesthreetypesofloopstructures:

whilestatements,do_whi1estatements,andforstatements.

Thisproblemcanbedoneusingeitherawhileordo_whileloop

controlmethod.Thewhilestatementfirstdeterminesthe

evaluationoftheexpression,

Iftheexpressionresultistruethentheloopbodyisexecuted,

otherwisethewhilestatementisended;Do_whilestatementis

moresuitablefortheloopbodyexecutedatleastonce,its

characteristicistoexecutetheloopbodyfirst,andthenjudge

expressionisevaluatedasaresult,iftheexpressionresults

istruecontinuestoimplementtheloopbody,otherwiseend

do_whilestatements,executingthenextstatement.

Methodone:

//sourcecodeisasfollows:

#include"iostream,h”

Voidmain()

(

Intday=1;

While(day!=0){

"Cout"<<"\nPleaseinputthedayyouwanttobesure

(1~5=daynumber;0=end):\t.

Cin>>day.

Theswitch(day){

Case1:cout<<"Monday:…“<<endl;Break;

Tuesday:"Tuesday".<<endl;Break;

Case3:cout<<"Wednesday，.."<<endl;Break;

Case4:cout<<Thursday:<<endl;Break;

Case5:cout<<"Friday:"<<endl;Break;

Case0:cout.Break;

Default:cout,pleasereinput.

)

Method2:

#include“iostream,h”

Voidmain()

(

Intday;

Do{

"Cout"<<nPleaseinputthedayyouwanttobesure”

(1~5=daynumber;0=end):\t.

Cin>>day.

Theswitch(day){

Case1:cout<<"Monday:…“<<endl;Break;

Tuesday:"Tuesday".<<endl;Break;

Case3:cout<<"Wednesday，.."<<endl;Break;

Case4:cout<<Thursday//<<endl;Break;

Case5:cout<<"Friday:"<<endl;Break;

Case0:cout.Break;

Default:cout<<<<"<<”>

)

}while(day!=0)

}

//programpromptsfor1~5numberqueryMondaytoFriday,

0endquery,inputotherfigureswereup//inputerrorsand

re-enteryourrequirementsnumberquery.

1-12peopleuselettersinsteadofdecimalnumbercalculation,

writethefollowingformula:EGALxL-LAGEpleaseusec++

programming,findouttheselettersinsteadofNumbers.

Answer://sourcecodeisasfollows:

#include<iostream.H>

Voidmain()

Inte,g,a,1,tempal,tempa2,tempb,see;

For(e=0;e<=9;e++){

For(g=0;g<=9;g++){

For(a=0;a<=9;a++){.

For(1=0;1<=9;1++){.

Tempal=(e*1000+g*100+a*10+1);

Tempa2=tempal*1;

Tempb=1*1000+a*100+g*10+e;

See=(e-g)*(e-a)*(g-a)*(g-1)*(a-l).

If(tempa2===tempb&&see!=0)

Thecout<<<<<<<<<<<<<><<<<<<<><<<<

<><<<><<<<<><<<<<><<<<<<<>><<<<

<<><<<<

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

<<<<<<<<<<<<<><<<<<<<<<<<<<<<><

<<<<<<<<<<<<<<<<

)

//theresultoftheprogramis1089*9---9801

1-13trytowriteaprogramthatenterstwopositiveintegers

andasksfortheirgreatestcommonandleastcommonmultiples.

Answer:aandbareintegergreatestcommondivisorandleast

commonmultipleproblemsthekeyistofindthegreatestcommon

divisortfirst,mathematicianEuclid(Euclid)hasgivena

classicalsolution:step1:ifa<b,theexchangeofaandb.

Steptwo:letrbetheremainderofa/b.Step3:ifr=0,then

t=bandterminate.Otherwise,a=b,b=r,andturntostep

two.Soifwefigureoutthelargestcommondivisort,wecan

figureouttheleastcommonmultipleofatimesbovert.//

sourcecodeisasfollows:

#include<iostream.H>

Int,int,

Voidmain()

TheInput2Numbers:

Inta,b,t;

Cin>a>.

T=gray(a,b);

Maxcommondivisor.

<<<<<<<<<<<<<<<<<<<><<<<<<<<<<<

<<<<<<>>>

}

Intm,intn//tossandturntomaximizethecommondivisor.

(

Intml,r;

If(m<n){ml=m;M=n;N=ml;}

R=m%n;

While(r!=0){m=n;N=r;R=m%n;}

Return(n);

//programrunresults:

Input2Numbers:

23445

Maxcommondivisor:9

Mincommonmultiple:1170

Pressanykeytocontinue

1-14ABCDEFGHeightpeoplestandinarow,thenumbershown

infigure1.10startsat1.Whofirstreportedto19431005?

Answer:#include<stdio.H>

Voidmain()

(

Longanum=19431005;

Constlongnum=8;

Theswitch(numanum%)

Case0:printf("H\n");Break;

Case1:printf("A\n");Break;

Case2:printf("B\n");Break;

Case3:printf("C\n");Break;

Case4:printf("D'n");Break;

Case5:printf("E\n");Break;

Case6:printf("F\n");Break;

Case7:printf("G\n");Break;

Default:printf("wrong!\n");Break;

1-15ofancientChinesemathematicsmagnumopus"ninechapter

arithmetic“volume138entitled〃atotaloffiveWells,the

hometwoBing(pumpingwithcoldwater),suchas(in)aBing

b;bthreeBingisinsufficient,suchasacBing;fourBing

cisinsufficient,suchasreadingaBing;butylfiveBingis

insufficient,suchashepatitisaBing;esixBingis

insufficient,suchasarmorBing,and.

Askthewell,thedirector'sgeometry.Answer:sevenzhangs2

feetoneinchholedepth,aBingtwozhangssixfeetfiveinches

long,bBingtenninefeetoneinchlong,cBingshakuhachia

post,thefourincheslong,butylBingtenthreefeet9inches

long,eBingsevenfeet,sixincheslong.""

Answer:a,b,c,d),fivestringx,y,z,s,t,welldepth,u,

thenlisttheequations:(2x+y=u|3y+z=u+s={4

zu|5s+t=u(t+6x=usolutionequationstox,y,z,

s,t,respectively,equalto265/721mmmm,191/721,191/721,

191/721,76/721m,andwelldeeply1m.

1-16,howmanywaysaretheretochangeadollartoanickel

for5,2,and1

Answer:541kinds#include<iomanip>

#include<iostream>

UsingnamespaceSTD.

Themain(){

Intn=0;

For(int1=0;I<=20;I++)

For(intj=0;j<=50;j++){

Intisequalto100minusItimes5minusjtimes2.

Ifk,>=0.

Cout<<setw++n(4)<<<<":1<<"setw(3)<<<<k

",2"<<setw(3)<<<<j",5"<<setw(3)<<I<<"zhang

"<<endl;

If(!(n%200))getchar();

)

}

Onein17scientistshadamathproblem:therewereacow.It

hasasmallcowatthebeginningofeveryyear;Everysmal1cow

willhaveasmallcowatthebeginningofeveryyearsincethe

fourthyear.Inthislaw,ifnocattledie,howmanycowsare

thereinthe20thy