ACM第二学期题目报告.docx

ACM第二学期题目报告.docx

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ACM第二学期题目报告

开放性实验ACM部分实验报告

ZOJProblemSet-1057

1057Undercut

TimeLimit:

1Second    MemoryLimit:

32768KB

Undercutisacardgamewheretwoplayerseachhavefivecardsnumberedonethroughfive.Ateachround,eachplayerselectsacard,thensimultaneouslyrevealsit.Ifthecardsareofequalvalue,thereisnoscore.Otherwise,therearetwocases:

thetwocardsareexactlyonepointapart（thisiscalledanundercut）,orthecardsaremorethanonepointapart.Inthelattercase,thepersonrevealingthelargerofthecardsgetsthenumberofpointsonthelargercard.Inthecaseofanundercuttheplayerwiththelowercardgetsthesumofthetwocards.Theexception[例外情况]tothisiswhenthecardsare1and2,inwhichcasetheplayerwiththelowercardgets6points（insteadofonly3points）.Aftereachround,thecardsarereturnedtothehandsandtheyplayanotherround.

Forexample,ifthereare5roundsandplayerAplays（inthisorder）5,3,1,3,5andplayerBplays3,3,3,3,4,thenthescoringforeachroundwouldbe:

Agets5points,nopoints,Bgets3points,nopoints,Bgets9points.ThetotalswouldbeA:

5,B:

12.

Inthisproblemyouwillbegivencardplaysforbothplayersandmustdeterminethefinalscores.

Input

Therewillbemultipleinputinstances.Eachinstancewillbeonegame.Thefirstlineofinputforagamewillbeanintegern<=20.（Avalueofn=0terminatesinput.）Thenexttwolineswilleachcontainnintegersbetween1and5inclusiveindicatingthecardsplayedoneachofnrounds.ThefirstlineareplayerA'scardplaysandthesecondlineareplayerB'scardplays.

Output

Eachinputinstanceshouldgenerateonelineofoutputoftheform:

Ahasapoints.Bhasbpoints.

wherethevalueofaandbareforyoutodetermine.Ablanklineshouldseparateoutputlines.

SampleInput

5

53135

33334

4

2311

1555

0

SampleOutput

Ahas5points.Bhas12points.

Ahas0points.Bhas21points.

题意：

A、B玩牌，若牌的点数相同则双方均不得分，若相差一分，则牌点数较小的得分（两张点数之和），若大于一分，则点数较大者得到两张牌中点数较大的分值，一个例外情况，当牌点数为1、2时，较低者得6分，其中牌点数在1~5之间

思路：

由以上题目可得ifelse的思路，重点在于空行的位置的输出，应该在第二次n的值输入后打出空行

代码：

#include

intmain（）

{

inta[21],b[21],i,sa,sb,n,j=1;

scanf（"%d",&n）;

while

（1）

{

sa=0;

sb=0;

for（i=0;i

scanf（"%d",&a[i]）;

for（i=0;i

scanf（"%d",&b[i]）;

for（i=0;i

{

if（a[i]==1&&b[i]==2）

sa+=6;

elseif（a[i]==2&&b[i]==1）

sb+=6;

elseif（a[i]-b[i]==1||a[i]-b[i]==-1）

{

if（a[i]

sa+=a[i]+b[i];

else

sb+=a[i]+b[i];

}

elseif（a[i]-b[i]>1||a[i]-b[i]<-1）

{

if（a[i]>b[i]）

sa=sa+a[i];

else

sb+=b[i];

}

}

printf（"Ahas%dpoints.Bhas%dpoints.\n",sa,sb）;

scanf（"%d",&n）;

if（n）

printf（"\n"）;

else

break;

}

return0;

}

代码效果：

ZOJProblemSet-1078

1078PalindromeNumbers

TimeLimit:

1Second    MemoryLimit:

32768KB

StatementoftheProblem

Wesaythatanumberisapalindromifitisthesanewhenreadfromlefttorightorfromrighttoleft.Forexample,thenumber75457isapalindrom.

Ofcourse,thepropertydependsonthebasisinwhichisnumberisrepresented.Thenumber17isnotapalindrominbase10,butitsrepresentationinbase2（10001）isapalindrom.

Theobjectiveofthisproblemistoverifyifasetofgivennumbersarepalindromsinanybasisfrom2to16.

InputFormat

Severalintegernumberscomprisetheinput.Eachnumber0

OutputFormat

YourprogrammustprintthemessageNumberiispalindrominbasiswhereIisthegivennumber,followedbythebasiswheretherepresentationofthenumberisapalindrom.Ifthenumberisnotapalindrominanybasisbetween2and16,yourprogrammustprintthemessageNumberiisnotpalindrom.

SampleInput

17

19

0

SampleOutput

Number17ispalindrominbasis2416

Number19isnotapalindrom

题意：

判断所给数字是否为回文数，判断依据：

若该数字的n（2-16）进制的表示对称，则输出相关的进制

思路：

写出一个十进制转化为n进制的函数，然后判断是否对称，若是则存进一个数组，循环结束后打出

代码：

代码效果：

ZOJProblemSet-1195

1195BlowingFuses

TimeLimit:

1Second    MemoryLimit:

32768KB

Maybeyouarefamiliarwiththefollowingsituation.Youhavepluggedinalotofelectricaldevices,suchastoasters,refrigerators,microwaveovens,computers,stereos,etc,andhavethemallrunning.ButatthemomentwhenyouturnontheTV,thefuseblows,sincethepowerdrawnfrom[从中得到、从……提取]allthemachinesisgreaterthanthecapacity[容量、能力]ofthefuse.Ofcoursethisisagreatsafetyfeature,avoidingthathousesburndowntoooftenduetofiresignitedbyoverheatingwires.Butitisalsoannoyingtowalkdowntothebasement（orsomeotherinconvenientplace）toreplacetofuseorswitchitbackon.

Whatonewouldliketohaveisaprogramthatchecksbeforeturningonanelectricaldevicewhetherthecombinedpowerdrawnbyallrunningdevicesexceedsthefusescapacity（anditblows）,orwhetheritissafetoturniton.

Input

Theinputconsistsofseveraltestcases.Eachtestcasedescribesasetofelectricaldevicesandgivesasequenceofturnon/offoperationsforthesedevices.

Thefirstlineofeachtestcasecontainsthreeintegersn,mandc,wherenisthenumberofdevices（n<=20）,mthenumberofoperationsperformedon[上演]thesedevicesandcisthecapacityofthefuse（inAmperes）.Thefollowingnlinescontainonepositiveintegercieach,theconsumption（inAmperes）ofthei-thdevice.

Thisisfollowedbymlinesalsocontainingoneintegereach,between1andninclusive.Theydescribeasequenceofturnon/turnoffoperationsperformedonthedevices.Foreverynumber,thestateofthatparticulardeviceistoggled,i.e.ifitiscurrentlyrunning,itisturnedoff,andifitiscurrentlyturnedoff,itwillbeswitchedon.Atthebeginningalldevicesareturnedoff.

Theinputwillbeterminatedbyatestcasestartingwithn=m=c=0.Thistestcaseshouldnotbeprocessed.

Output

Foreachtestcase,firstoutputthenumberofthetestcase.Thenoutputwhetherthefusewasblownduringtheoperationsequence.Thefusewillbeblownifthesumofthepowerconsumptionsciofturnedondevicesatsomepointexceedsthecapacityofthefusec.

Ifthefuseisnotblown,outputthemaximalpowerconsumptionbyturnedondevicesthatoccurredduringthesequence.

Outputablanklineaftereachtestcase.

SampleInput

2210

5

7

1

2

3610

2

5

7

2

1

2

3

1

3

000

SampleOutput

Sequence1

Fusewasblown.

Sequence2

Fusewasnotblown.

Maximalpowerconsumptionwas9amperes.

题意：

计算正在使用的电器的电流量是否超过了最大电容量，若是输出“Fusewasblown.”，若不是，则输出“Fusewasnotblown.Maximalpowerconsumptionwas9amperes.”，第一行为电器数、每台电器所需电流量、对相应电器的操作（开关）

思路：

输入操作数的值，然后进行相应的加减，最后判断是否超出最大值，相应输出

代码：

#include

#include

intmain（）

{

intm,n,c,i,j=1,dev[21],oper[21],nc,maxc,t;

while（scanf（"%d%d%d",&n,&m,&c）!

=EOF&&m!

=0&&n!

=0&&c!

=0）

{

intflag=0;

memset（oper,0,sizeof（oper））;

nc=maxc=0;

for（i=1;i<=n;i++）

scanf（"%d",&dev[i]）;

for（i=1;i<=m;i++）

{

scanf（"%d",&t）;

if（flag）

continue;

if（oper[t]==0）

{

oper[t]=1;

nc+=dev[t];

if（nc>c）

flag=1;

elseif（nc>maxc）

maxc=nc;

}

else

{

oper[t]=0;

nc-=dev[t];

}

}

printf（"Sequence%d\n",j++）;

if（flag）

printf（"Fusewasblown.\n\n"）;

else

printf（"Fusewasnotblown.\nMaximalpowerconsumptionwas%damperes.\n\n",maxc）;

}

return0;

}

代码效果：

ZOJProblemSet-1272

1272NumericallySpeaking

TimeLimit:

1Second    MemoryLimit:

32768KB

Adeveloperofcrosswordpuzzles[填字游戏]（andothersimilarwordgames）hasdecidedtodevelopamappingbetweeneverypossiblewordwithfromonetotwentycharactersanduniqueintegers.Themappingisverysimple,withtheorderingbeingdonefirstbythelengthoftheword,andthenalphabetically[照字母顺序排列地].Partofthelistisshownbelow.

a         1

  b         2

   ...

   z         26

   aa        27

   ab        28

   ...

   snowfall  157,118,051,752

   ...

Yourjobinthisproblemistodevelopaprogramwhichcantranslate,bidirectionally[变形的],betweentheuniquewordnumbersandthecorrespondingwords.

Input

Inputtotheprogramisalistofwordsandnumbers,oneperlinestartingincolumnone,followedbyalinecontainingasingleasteriskincolumn[列]one.Anumberwillconsistonlyofdecimaldigits[小数位数]（0through9）followedimmediatelybytheendofline（thatis,therewillbenocommasininputnumbers）.Awordwillconsistofbetweenoneandtwentylowercasealphabeticcharacters[字母符号]（athroughz）.

Output

Theoutputistocontainasinglelineforeachwordornumberintheinputdata.Thislineistocontainthewordstartingincolumnone,followedbyanappropriatenumberofblanks,andthecorrespondingwordnumberstartingincolumn23.Wordnumbersthathavemorethanthreedigitsmustbeseparatedbycommas[逗号]atthousands,millions,andsoforth.

SampleInput

29697684282993

transcendental

28011622636823854456520

computationally

zzzzzzzzzzzzzzzzzzzz

*

SampleOutput

elementary           29,697,684,282,993

transcendental      51,346,529,199,396,181,750

prestidigitation     28,011,622,636,823,854,456,520

computationally      232,049,592,627,851,629,097

zzzzzzzzzzzzzzzzzzzz 20,725,274,851,017,785,518,433,805,270

题意：

对长度在23位之内的字母或字符进行10进制与26进制之间的转换

思路：

不会

代码：

代码效果：

ZOJProblemSet-1489

14892^xmodn=1

TimeLimit:

1Second    MemoryLimit:

32768KB

Giveanumbern,findtheminimumxthatsatisfies2^xmodn=1.

Input

Onepositiveintegeroneachline,thevalueofn.

Output

Iftheminimumxexists,printalinewith2^xmodn=1.

Print2^?

modn=1