华东师范大学计算机机试真题Word文档格式.docx
《华东师范大学计算机机试真题Word文档格式.docx》由会员分享，可在线阅读，更多相关《华东师范大学计算机机试真题Word文档格式.docx（73页珍藏版）》请在冰豆网上搜索。
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
Writeaprogramwhichreplaceallthelower-caselettersofagiventextwiththecorrespondingcaptitalletters.
Atextincludinglower-caseletters,periods,andspace.
OutputTheconvertedtext.
welcometoeastchinanormaluniversity.
WELCOMETOEASTCHINANORMALUNIVERSITY.
string.h>
charstr[1000];
intl;
while（gets（str））
l=strlen（str）;
inti;
for（i=0;
i<
l;
i++）
if（str[i]>
='
a'
&
str[i]<
z'
）
printf（"
%c"
str[i]-32）;
else
str[i]）;
\n"
）;
素数对
PrimesPair
Wearrangethenumbersbetween1andN（1<
=N<
=10000）inincreasingorderanddecreasingorderlikethis:
123456789...N
N...987654321
Twonumbersfacedeachotherformapair.YourtaskistocomputethenumberofpairsPsuchthatbothnumbersinthepairsareprime.
Thefirstlineofinputgivesthenumberofcases,C（1≤C≤100）.Ctestcasesfollow.
EachtestcaseconsistsofanintegerNinoneline.
Foreachtestcase,outputP.
4
1
7
51
6
boolprime[10005];
voidinit（）
intj;
prime[0]=prime[1]=false;
//不是素数
prime[2]=true;
//是素数
for（i=3;
=10005;
i+=2）
prime[i]=true;
prime[i+1]=false;
//不是素数除0和2之外的偶数都不是素数
if（prime[i]==true）//是素数
j=i+i;
while（j<
=10005）
prime[j]=false;
j+=i;
intc;
init（）;
//初始化
c）!
while（c--）
scanf（"
n）;
intsum=0;
for（i=2;
=n/2;
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;
sum）;
求最大公约数和最小公倍数
GCDandLCM
Writeaprogramwhichcomputesthegreatestcommondivisor（GCD）andtheleastcommonmultiple（LCM）ofgivenaandb（0<
a,b≤44000）.
Eachtestcasecontainstwointergeraandbseparatedbyasinglespaceinaline.
Foreachtestcase,printGCDandLCMseparatedbyasinglespaceinaline.
86
50003000
224
100015000
intgetgcd（inta,intb）
intgcd;
intt1,t2;
t1=a;
t2=b;
gcd=t1%t2;
while（gcd!
=0）
t1=t2;
t2=gcd;
returnt2;
%d%d\n"
getgcd（a,b）,a*b/（getgcd（a,b）））;
排序后求位置处的数
Sortit…
Thereisadatabase,partychenwantyoutosortthedatabase’sdataintheorderfromtheleastuptothegreatestelement,thendothequery:
"
Whichelementisi-thbyitsvalue?
"
-withibeinganaturalnumberinarangefrom1toN.
Itshouldbeabletoprocessquicklyquerieslikethis.
Thestandardinputoftheproblemconsistsoftwoparts.Atfirst,adatabaseiswritten,andthenthere'
sasequenceofqueries.Theformatofdatabaseisverysimple:
inthefirstlinethere'
sanumberN（1<
=N<
=100000）,inthenextNlinestherearenumbersofthedatabaseoneineachlineinanarbitraryorder.Asequenceofqueriesiswrittensimplyaswell:
inthefirstlineofthesequenceanumberofqueriesK（1<
=K<
=100）iswritten,andinthenextKlinestherearequeriesoneineachline.Thequery"
iscodedbythenumberi.
TheoutputshouldconsistofKlines.Ineachlinethereshouldbeananswertothecorrespondingquery.Theanswertothequery"
i"
isanelementfromthedatabase,whichisi-thbyitsvalue（intheorderfromtheleastuptothegreatestelement）.
5
121
123
5
algorithm>
usingnamespacestd;
intnum[100010];
intpos[105];
intk;
for（i=1;
=n;
num[i]）;
k）;
=k;
pos[i]）;
sort（num+1,num+1+n）;
num[pos[i]]）;
*路由器连接
HubConnectionplan
Partychenisworkingassystemadministratorandisplanningtoestablishanewnetworkinhiscompany.TherewillbeNhubsinthecompany,theycanbeconnectedtoeachotherusingcables.Sinceeachworkerofthecompanymusthaveaccesstothewholenetwork,eachhubmustbeaccessiblebycablesfromanyotherhub（withpossiblysomeintermediatehubs）.
Sincecablesofdifferenttypesareavailableandshorteronesarecheaper,itisnecessarytomakesuchaplanofhubconnection,thatthecostisminimal.partychenwillprovideyouallnecessaryinformationaboutpossiblehubconnections.Youaretohelppartychentofindthewaytoconnecthubssothatallaboveconditionsaresatisfied.
Thefirstlineoftheinputcontainstwointegernumbers:
N-thenumberofhubsinthenetwork（2<
=1000）andM-thenumberofpossiblehubconnections（1<
=M<
=15000）.Allhubsarenumberedfrom1toN.ThefollowingMlinescontaininformationaboutpossibleconnections-thenumbersoftwohubs,whichcanbeconnectedandthecablecostrequiredtoconnectthem.costisapositiveintegernumberthatdoesnotexceed106.Therewillalwaysbeatleastonewaytoconnectallhubs.
Outputtheminimizecostofyourhubconnectionplan.
46
121
131
142
231
341
241
structEdge{
intcost;
}E[15010];
intTree[1010];
intfindRoot（intx）
if（Tree[x]==-1）
returnx;
inttmp=findRoot（Tree[x]）;
Tree[x]=tmp;
returntmp;
boolCmp（Edgea,Edgeb）
returna.cost<
b.cost;
intm;
m）;
=m;
%d%d%d"
E[i].a,&
E[i].b,&
E[i].cost）;
sort（E+1,E+1+m,Cmp）;
//排序
Tree[i]=-1;
intans=0;
inta=findRoot（E[i].a）;
intb=findRoot（E[i].b）;
if（a!
=b）
Tree[a]=b;
ans+=E[i].cost;
ans）;
*编译原理
PrinciplesofCompiler
AfterlearntthePrinciplesofCompiler,partychenthoughtthathecansolveasimpleexpressionproblem.Sohegiveyoustringsoflessthan100characterswhichstrictlyadheretothefollowinggrammar（giveninEBNF）:
A:
='
（'
B'
）'
|'
x'
.
B:
=AC.
C:
={'
+'
A}.
Canyousolvethemtoo?
ThenextNlineswilleachcontainastringasdescribedabove.
Foreachtestcase,iftheexpressionisadapttotheEBNFaboveoutput“Good”,elseoutput“Bad”.
（x）
（x+（x+x））
（）（x）
Good
Bad
#include<
cstdio>
cstring>
cstdlib>
vector>
cmath>
iostream>
functional>
string>
map>
cctype>
charex[110];
intindex;
boolA（）;
boolB（）;
boolC（）;
boolA（）
if（ex[index]=='
index++;
while（ex[index]=='
'
）index++;
returntrue;
if（B（）&
ex[index]=='
returnfalse;
boolB（）
returnA（）&
C（）;
boolC（）
//returnA（）;
if（!
A（））
intN;
N）;
getchar（）;
while（N--）
gets（ex）;
index=0;
%s\n"
A（）&
\0'
?
Good"
:
Bad"
return0;
*分开连接
SeparateConnections
Partychenareanalyzingacommunicationsnetworkwithatmost18nodes.Characterinamatrixi,j（i,jboth0-based,asmatrix[i][j]）denoteswhethernodesiandjcancommunicate（'
Y'
foryes,'
N'
forno）.Assuminganodecannotcommunicatewithtwonodesatonce,returnthemaximumnumberofnodesthatcancommunicatesimultaneously.Ifnodeiiscommunicatingwithnodejthennodejiscommunicatingwithnodei.
Ineachtestcase,thefirstlineisthenumberofnodesM（1≤M≤18）,thenthereareagridbyM*Mdescribledthematrix.
Foreachtestcase,outputthemaximumnumberofnodesthatcancommunicatesimultaneously
NYYYY
YNNNN
YNNNY
YNYYN
Hint
Thefirsttestcase:
Allcommunicationsmustoccurwithnode0.Sincenode0canonlycommunicatewith1nodeatatime,theoutputvalueis2.
Thesecondtestcase:
Inthissetup,wecanletnode0communicatewithnode1,andnode3communicatewithnode4.
#i
升级会员 