HelveticCodingContest2017onlinemirror(teamsallowed,unrated)暴力都不会的蒟蒻

Asit'sthefirstofApril,Heidiissuspectingthatthenewsshereadstodayarefake,andshedoesnotwanttolooksillyinfrontofallthecontestants.Sheknowsthatanewspieceisfakeifitcontainsheidiasasubsequence.HelpHeidiassesswhetherthegivenpieceistrue,butpleasebediscreetaboutit...

Thefirstandonlylineofinputcontainsasinglenonemptystringsoflengthatmost1000composedoflowercaseletters(a-z).

OutputYESifthestringscontainsheidiasasubsequenceandNOotherwise.

abcheaibcdioutputYESinputhiedioutputNONoteAstringscontainsanotherstringpasasubsequenceifitispossibletodeletesomecharactersfromsandobtainp.

看下给你的字符串是否存在“heidi”

#includechars[]="heidi",st[100005];intn,num=0;intmain(){scanf("%s",st+1);for(inti=1;st[i];++i){if(st[i]==s[num])++num;if(num==5)return0*puts("YES");}puts("NO");return0;}M.AprilFools'Problem(easy)timelimitpertest2secondsmemorylimitpertest256megabytesinputstandardinputoutputstandardoutputThemarmotshavepreparedaveryeasyproblemforthisyear'sHC2–thisone.Itinvolvesnumbersn,kandasequenceofnpositiveintegersa1,a2,...,an.Theyalsocameupwithabeautifulandrivetingstoryfortheproblemstatement.Itexplainswhattheinputmeans,whattheprogramshouldoutput,anditalsoreadslikeagoodcriminal.

HoweverI,Heidi,willhavenoneofthat.Asmyjokefortoday,Iamremovingthestoryfromthestatementandreplacingitwiththesetwounhelpfulparagraphs.Nowsolvetheproblem,fools!

Thefirstlineoftheinputcontainstwospace-separatedintegersnandk(1≤k≤n≤2200).Thesecondlinecontainsnspace-separatedintegersa1,...,an(1≤ai≤104).

Outputonenumber.

8511111111output5input1031682451225632128641output7input512010503046output10input66666666output36input11100output100我在做这些水题真的好么？只当练英语了，n个数选k个最小的和，不开心的时候就A水题？#includeinta[2205];usingnamespacestd;intmain(){intn,k;scanf("%d%d",&n,&k);for(inti=0;i

Heidi'sfriendJennyisaskingHeiditodeliveranimportantlettertooneoftheircommonfriends.SinceJennyisIrish,Heidithinksthatthismightbeaprank.Moreprecisely,shesuspectsthatthemessagesheisaskedtodeliverstates:"Sendthefoolfurther!",anduponreadingittherecipientwillaskHeiditodeliverthesamemessagetoyetanotherfriend(thattherecipienthasincommonwithHeidi),andsoon.

Heidibelievesthatherfriendswanttoavoidawkwardsituations,soshewillnotbemadetovisitthesameperson(includingJenny)twice.Shealsoknowshowmuchitcoststotravelbetweenanytwoofherfriendswhoknoweachother.Shewantstoknow:whatisthemaximalamountofmoneyshewillwasteontravelifitreallyisaprank

Heidi'snfriendsarelabeled0throughn-1,andtheirnetworkofconnectionsformsatree.Inotherwords,everytwoofherfriendsa,bknoweachother,possiblyindirectly(thereisasequenceoffriendsstartingfromaandendingonbandsuchthateachtwoconsecutivefriendsinthesequenceknoweachotherdirectly),andthereareexactlyn-1pairsoffriendswhoknoweachotherdirectly.

Jennyisgiventhenumber0.

Thefirstlineoftheinputcontainsthenumberoffriendsn(3≤n≤100).Thenextn-1lineseachcontainthreespace-separatedintegersu,vandc(0≤u,v≤n-1,1≤c≤104),meaningthatuandvarefriends(knoweachotherdirectly)andthecostfortravellingbetweenuandvisc.

Itisguaranteedthatthesocialnetworkoftheinputformsatree.

Outputasingleinteger–themaximumsumofcosts.

4014022233output5input6123021001420373510output105input1110166420881324670421555511870621265722888722669215361063378output5551NoteInthesecondexample,theworst-casescenariogoeslikethis:JennysendsHeiditothefriendlabeledbynumber2(incurringacostof100),thenfriend2sendshertofriend1(costingHeidi3),andfinallyfriend1relayshertofriend4(incurringanadditionalcostof2).

就是简单的DFS一遍，找到从根到叶的最长路，这个自己当然是想不出来的，别人的代码真的很巧妙

#include#include#include#include#includetypedeflonglongll;#definerep(i,a,b)for(inti=a;i<=b;++i)constintN=5005;usingnamespacestd;intn,ter[N],e,nxt[N],lnk[N],w[N];voidadd(intx,inty,intz){ter[++e]=y,nxt[e]=lnk[x],lnk[x]=e,w[e]=z;}intdfs(intp,intlas){intret=0;for(inti=lnk[p];i;i=nxt[i])if(ter[i]!=las)ret=max(ret,dfs(ter[i],p)+w[i]);returnret;}intmain(){scanf("%d",&n);intx,y,z;rep(i,1,n-1){scanf("%d%d%d",&x,&y,&z);++x,++y;add(x,y,z);add(y,x,z);}printf("%d\n",dfs(1,0));return0;}

YoursearchforHeidiisover–youfinallyfoundheratalibrary,dressedupasahuman.Infact,shehasspentsomuchtimetherethatshenowrunstheplace!Herjobistobuybooksandkeepthematthelibrarysothatpeoplecanborrowandreadthem.Therearendifferentbooks,numbered1throughn.

Wewilllookatthelibrary'soperationduringnconsecutivedays.Heidiknowsinadvancethatonthei-thday(1≤i≤n)preciselyonepersonwillcometothelibrary,requesttoborrowthebookai,readitinafewhours,andreturnthebooklateronthesameday.

Heididesperatelywantstopleaseallherguests,soshewillmakesuretoalwayshavethebookaiavailableinthelibraryonthei-thday.Duringthenightbeforethei-thday,shehastheoptionofgoingtothebookstore(whichoperatesatnightstoavoidcompetitionwiththelibrary)andbuyinganybookforthepriceof1CHF.Ofcourse,ifshealreadyhasabookatthelibrary,shedoesnotneedtobuyitagain.Initially,thelibrarycontainsnobooks.

Thereisaproblem,though.Thecapacityofthelibraryisk–thismeansthatatanytime,therecanbeatmostkbooksatthelibrary.IfbuyinganewbookwouldcauseHeiditohavemorethankbooks,shemustfirstgetridofsomebookthatshealreadyhas,inordertomakeroomforthenewbook.Ifshelaterneedsabookthatshegotridof,shewillneedtobuythatbookagain.

Youaregivenkandthesequenceofrequestsforbooksa1,a2,...,an.Whatistheminimumcost(inCHF)ofbuyingnewbookstosatisfyalltherequests

Thefirstlineofinputwillcontaintwointegersnandk(1≤n,k≤80).Thesecondlinewillcontainnintegersa1,a2,...,an(1≤ai≤n)–thesequenceofbookrequests.

Onasinglelineprinttheminimumcostofbuyingbooksatthestoresoastosatisfyallrequests.

4801221output2input411221output3input421231output3NoteInthefirsttestcase,Heidiisabletokeepallbooksforever.Therefore,sheonlyneedstobuythebook1beforethefirstdayandthebook2beforethesecondday.

Inthesecondtestcase,shecanonlykeeponebookatatime.Thereforeshewillneedtobuynewbooksonthefirst,secondandfourthday.

Inthethirdtestcase,beforebuyingbook3onthethirdday,shemustdecidewhichofthebooks1and2sheshouldgetridof.Ofcourse,sheshouldkeepthebook1,whichwillberequestedonthefourthday.

Q巨和我们讲这个是页面置换算法，其中最完美的是OPT，但是好像硬件并没有这样实现

举例如下：

#includeusingnamespacestd;inta[88];intn,k;intmain(){scanf("%d%d",&n,&k);for(inti=0;is;intt=0;for(inti=0;iM){M=j;p=*it;}}s.erase(p);}s.insert(a[i]);t++;}}printf("%d\n",t);}