华东师范大学计算机机试真题.docx
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华东师范大学计算机机试真题
2009机试
计算和的数位
Sumofdigit
Description
Writeaprogramwhichcomputesthedigitnumberofsumoftwointegersaandb.
Input
Thefirstlineofinputgivesthenumberofcases,N（1≤N≤100）.Ntestcasesfollow.
Eachtestcaseconsistsoftwointegersaandbwhichareseparetedbyaspaceinaline.（0<=a,b<=100000000）.
Output
Foreachtestcase,printthenumberofdigitsofa+b.
SampleInput
3
57
199
1000999
SampleOutput
2
3
4
#include
intmain（）
{
intn;
inta,b;
intsum;
while（scanf（"%d",&n）!
=EOF）
{
while（n--）
{
intan=0;
scanf（"%d%d",&a,&b）;
sum=a+b;
while（sum）
{
an++;
sum/=10;
}
printf（"%d\n",an++）;
}
}
return0;
}
大写改小写
Capitalize
Description
Writeaprogramwhichreplaceallthelower-caselettersofagiventextwiththecorrespondingcaptitalletters.
Input
Atextincludinglower-caseletters,periods,andspace.
Output
OutputTheconvertedtext.
SampleInput
welcometoeastchinanormaluniversity.
SampleOutput
WELCOMETOEASTCHINANORMALUNIVERSITY.
#include
#include
charstr[1000];
intmain（）
{
intl;
while（gets（str））
{
l=strlen（str）;
inti;
for（i=0;i{
if（str[i]>='a'&&str[i]<='z'）
printf（"%c",str[i]-32）;
else
printf（"%c",str[i]）;
}
printf（"\n"）;
}
return0;
}
素数对
PrimesPair
Description
Wearrangethenumbersbetween1andN（1<=N<=10000）inincreasingorderanddecreasingorderlikethis:
123456789...N
N...987654321
Twonumbersfacedeachotherformapair.YourtaskistocomputethenumberofpairsPsuchthatbothnumbersinthepairsareprime.
Input
Thefirstlineofinputgivesthenumberofcases,C（1≤C≤100）.Ctestcasesfollow.
EachtestcaseconsistsofanintegerNinoneline.
Output
Foreachtestcase,outputP.
SampleInput
4
1
4
7
51
SampleOutput
0
2
2
6
#include
#include
boolprime[10005];
voidinit（）
{
inti;
intj;
prime[0]=prime[1]=false;//不是素数
prime[2]=true;//是素数
for（i=3;i<=10005;i+=2）
{
prime[i]=true;//是素数
prime[i+1]=false;//不是素数除0和2之外的偶数都不是素数
}
for（i=3;i<=10005;i+=2）
{
if（prime[i]==true）//是素数
{
j=i+i;
while（j<=10005）
{
prime[j]=false;//不是素数
j+=i;
}
}
}
}
intmain（）
{
intc;
intn;
init（）;//初始化
while（scanf（"%d",&c）!
=EOF）
{
while（c--）
{
scanf（"%d",&n）;
intsum=0;
inti;
for（i=2;i<=n/2;i++）
{
if（prime[i]==true&&prime[n+1-i]==true）
sum++;
}
sum*=2;
if（n%2==1）//n为奇数
{
if（prime[n/2+1]==true）
sum+=1;
}
printf（"%d\n",sum）;
}
}
return0;
}
求最大公约数和最小公倍数
GCDandLCM
Description
Writeaprogramwhichcomputesthegreatestcommondivisor（GCD）andtheleastcommonmultiple（LCM）ofgivenaandb（0Input
Thefirstlineofinputgivesthenumberofcases,N（1≤N≤100）.Ntestcasesfollow.
Eachtestcasecontainstwointergeraandbseparatedbyasinglespaceinaline.
Output
Foreachtestcase,printGCDandLCMseparatedbyasinglespaceinaline.
SampleInput
2
86
50003000
SampleOutput
224
100015000
#include
intgetgcd（inta,intb）
{
intgcd;
intt1,t2;
t1=a;
t2=b;
gcd=t1%t2;
while（gcd!
=0）
{
t1=t2;
t2=gcd;
gcd=t1%t2;
}
returnt2;
}
intmain（）
{
intn;
inta,b;
while（scanf（"%d",&n）!
=EOF）
{
while（n--）
{
scanf（"%d%d",&a,&b）;
printf（"%d%d\n",getgcd（a,b）,a*b/（getgcd（a,b）））;
}
}
return0;
}
排序后求位置处的数
Sortit…
Description
Thereisadatabase,partychenwantyoutosortthedatabase’sdataintheorderfromtheleastuptothegreatestelement,thendothequery:
"Whichelementisi-thbyitsvalue?
"-withibeinganaturalnumberinarangefrom1toN.
Itshouldbeabletoprocessquicklyquerieslikethis.
Input
Thestandardinputoftheproblemconsistsoftwoparts.Atfirst,adatabaseiswritten,andthenthere'sasequenceofqueries.Theformatofdatabaseisverysimple:
inthefirstlinethere'sanumberN（1<=N<=100000）,inthenextNlinestherearenumbersofthedatabaseoneineachlineinanarbitraryorder.Asequenceofqueriesiswrittensimplyaswell:
inthefirstlineofthesequenceanumberofqueriesK（1<=K<=100）iswritten,andinthenextKlinestherearequeriesoneineachline.Thequery"Whichelementisi-thbyitsvalue?
"iscodedbythenumberi.
Output
TheoutputshouldconsistofKlines.Ineachlinethereshouldbeananswertothecorrespondingquery.Theanswertothequery"i"isanelementfromthedatabase,whichisi-thbyitsvalue（intheorderfromtheleastuptothegreatestelement）.
SampleInput
5
7
121
123
7
121
3
3
2
5
SampleOutput
121
7
123
#include
#include
usingnamespacestd;
intnum[100010];
intpos[105];
intmain（）
{
intn;
inti;
intk;
while（scanf（"%d",&n）!
=EOF）
{
for（i=1;i<=n;i++）
scanf（"%d",&num[i]）;
scanf（"%d",&k）;
for（i=1;i<=k;i++）
scanf（"%d",&pos[i]）;
sort（num+1,num+1+n）;
for（i=1;i<=k;i++）
printf（"%d\n",num[pos[i]]）;
}
return0;
}
*路由器连接
HubConnectionplan
Description
Partychenisworkingassystemadministratorandisplanningtoestablishanewnetworkinhiscompany.TherewillbeNhubsinthecompany,theycanbeconnectedtoeachotherusingcables.Sinceeachworkerofthecompanymusthaveaccesstothewholenetwork,eachhubmustbeaccessiblebycablesfromanyotherhub（withpossiblysomeintermediatehubs）.
Sincecablesofdifferenttypesareavailableandshorteronesarecheaper,itisnecessarytomakesuchaplanofhubconnection,thatthecostisminimal.partychenwillprovideyouallnecessaryinformationaboutpossiblehubconnections.Youaretohelppartychentofindthewaytoconnecthubssothatallaboveconditionsaresatisfied.
Input
Thefirstlineoftheinputcontainstwointegernumbers:
N-thenumberofhubsinthenetwork（2<=N<=1000）andM-thenumberofpossiblehubconnections（1<=M<=15000）.Allhubsarenumberedfrom1toN.ThefollowingMlinescontaininformationaboutpossibleconnections-thenumbersoftwohubs,whichcanbeconnectedandthecablecostrequiredtoconnectthem.costisapositiveintegernumberthatdoesnotexceed106.Therewillalwaysbeatleastonewaytoconnectallhubs.
Output
Outputtheminimizecostofyourhubconnectionplan.
SampleInput
46
121
131
142
231
341
241
SampleOutput
3
#include
#include
usingnamespacestd;
structEdge{
inta,b;
intcost;
}E[15010];
intTree[1010];
intfindRoot（intx）
{
if（Tree[x]==-1）
returnx;
else
{
inttmp=findRoot（Tree[x]）;
Tree[x]=tmp;
returntmp;
}
}
boolCmp（Edgea,Edgeb）
{
returna.cost}
intmain（）
{
intn;
intm;
inti;
while（scanf（"%d",&n）!
=EOF）
{
scanf（"%d",&m）;
for（i=1;i<=m;i++）
scanf（"%d%d%d",&E[i].a,&E[i].b,&E[i].cost）;
sort（E+1,E+1+m,Cmp）;//排序
for（i=1;i<=n;i++）
Tree[i]=-1;
intans=0;
for（i=1;i<=m;i++）
{
inta=findRoot（E[i].a）;
intb=findRoot（E[i].b）;
if（a!
=b）
{
Tree[a]=b;
ans+=E[i].cost;
}
}
printf（"%d\n",ans）;
}
return0;
}
*编译原理
PrinciplesofCompiler
Description
AfterlearntthePrinciplesofCompiler,partychenthoughtthathecansolveasimpleexpressionproblem.Sohegiveyoustringsoflessthan100characterswhichstrictlyadheretothefollowinggrammar（giveninEBNF）:
A:
='（'B'）'|'x'.
B:
=AC.
C:
={'+'A}.
Canyousolvethemtoo?
Input
Thefirstlineofinputgivesthenumberofcases,N（1≤N≤100）.Ntestcasesfollow.
ThenextNlineswilleachcontainastringasdescribedabove.
Output
Foreachtestcase,iftheexpressionisadapttotheEBNFaboveoutput“Good”,elseoutput“Bad”.
SampleInput
3
（x）
（x+（x+x））
（）（x）
SampleOutput
Good
Good
Bad
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
usingnamespacestd;
charex[110];
intindex;
boolA（）;
boolB（）;
boolC（）;
boolA（）
{
if（ex[index]=='x'）
{
index++;
while（ex[index]==''）index++;
returntrue;
}
if（ex[index]=='（'）
{
index++;
while（ex[index]==''）index++;
if（B（）&&ex[index]=='）'）
{
index++;
while（ex[index]==''）index++;
returntrue;
}
}
returnfalse;
}
boolB（）
{
returnA（）&&C（）;
}
boolC（）
{
while（ex[index]=='+'）
{
index++;
while（ex[index]==''）index++;
//returnA（）;
if（!
A（））
returnfalse;
}
returntrue;
}
intmain（）
{
intN;
scanf（"%d",&N）;
getchar（）;
while（N--）
{
gets（ex）;
index=0;
printf（"%s\n",A（）&&ex[index]=='\0'?
"Good":
"Bad"）;
}
return0;
}
*分开连接
SeparateConnections
Description
Partychenareanalyzingacommunicationsnetworkwithatmost18nodes.Characterinamatrixi,j（i,jboth0-based,asmatrix[i][j]）denoteswhethernodesiandjcancommunicate（'Y'foryes,'N'forno）.Assuminganodecannotcommunicatewithtwonodesatonce,returnthemaximumnumberofnodesthatcancommunicatesimultaneously.Ifnodeiiscommunicatingwithnodejthennodejiscommunicatingwithnodei.
Input
Thefirstlineofinputgivesthenumberofcases,N（1≤N≤100）.Ntestcasesfollow.
Ineachtestcase,thefirstlineisthenumberofnodesM（1≤M≤18）,thenthereareagridbyM*Mdescribledthematrix.
Output
Foreachtestcase,outputthemaximumnumberofnodesthatcancommunicatesimultaneously
SampleInput
2
5
NYYYY
YNNNN
YNNNN
YNNNN
YNNNN
5
NYYYY
YNNNN
YNNNY
YNNNY
YNYYN
SampleOutput
2
4
Hint
Thefirsttestcase:
Allcommunicationsmustoccurwithnode0.Sincenode0canonlycommunicatewith1nodeatatime,theoutputvalueis2.
Thesecondtestcase:
Inthissetup,wecanletnode0communicatewithnode1,andnode3communicatewithnode4.
#include
#i
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