7th后浙大ACM总结.docx
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7th后浙大ACM总结
第一次ACM总结（7thACM）
ZOJProblemSet–1001A+BProblem1
ZOJProblemSet–1037Gridland2
ZOJProblemSet–1045HangOver4
ZOJProblemSet–1048FinancialManagement6
ZOJProblemSet–1049IThinkINeedaHouseboat8
ZOJProblemSet–1067ColorMeLess11
ZOJProblemSet–1073RoundandRoundWeGo14
ZOJProblemSet–1113uCalculatee17
ZOJProblemSet–1115DigitalRoots19
ZOJProblemSet–1151WordReversal21
ZOJProblemSet–1216Deck23
ZOJProblemSet-1240IBMMinusOne25
ZOJProblemSet-1241GeometryMadeSimple27
ZOJProblemSet-1251BoxofBricks30
ZOJProblemSet-1292IntegerInquiry32
ZOJProblemSet-1331PerfectCubes34
ZOJProblemSet-1337Pi36
ZOJProblemSet-1350TheDrunkJailer39
ZOJProblemSet-1382ASimpleTask41
ZOJProblemSet-1712SkewBinary43
ZOJProblemSet-1730CrazyTeaParty45
ZOJProblemSet-1760Doubles47
ZOJProblemSet-1763ASimpleQuestionofChemistry49
ZOJProblemSet-1797LeastCommonMultiple52
ZOJProblemSet-1871Steps54
ZOJProblemSet-1879JollyJumpers//又是没有代码56
ZOJProblemSet-1915AboveAverage58
ZOJProblemSet-2001AddingReversedNumbers60
ZOJProblemSet-2201NoBrainer63
ZOJProblemSet-2388BeattheSpread!
65
ZOJProblemSet-2947Abbreviation67
ZOJProblemSet-2965AccuratelySay"CocaCola"!
70
ZOJProblemSet-2969EasyTask72
ZOJProblemSet-2970Faster,Higher,Stronger74
ZOJProblemSet-2987Misspelling77
ZOJProblemSet-2988Conversions79
ZOJProblemSet–3313Clock//没有代码……82
ZOJProblemSet–3322WhoisOlder?
83
ZOJProblemSet-3323SomaliPirates85
ZOJProblemSet-3328WuXing87
ZOJProblemSet-3333GuessthePrice90
附录：
ACM常见单词表92
ZOJProblemSet–1001A+BProblem
TimeLimit:
1Second MemoryLimit:
32768KB
Calculate[计算]a+b
Input
Theinputwillconsistofaseriesofpairsofintegersaandb,separatedbyaspace,onepairofintegersperline.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.
SampleInput
15
SampleOutput
6
思路：
略
代码呈现：
#include
intmain（）
{
inta,b;
while（scanf（"%d%d",&a,&b）!
=EOF）//无限输入直到遇见shift+F6停止
printf（"%d\n",a+b）;
return0;
}
代码效果：
ZOJProblemSet–1037Gridland
TimeLimit:
1Second MemoryLimit:
32768KB
Background
Foryears,computerscientistshavebeentryingtofindefficientsolutionstodifferentcomputing[计算]problems.Forsomeofthemefficientalgorithms[算法]arealreadyavailable,thesearethe"easy"problemslikesorting[整理；排序；分类拣选],evaluatingapolynomial[多项式]orfindingtheshortestpathinagraph.Forthe"hard"onesonlyexponential-timealgorithms[指数时间算法]areknown.Thetraveling-salesmanproblembelongstothislattergroup.GivenasetofNtownsandroadsbetweenthesetowns,theproblemistocomputetheshortestpathallowingasalesmantovisiteachofthetownsonceandonlyonceandreturntothestartingpoint.
Problem
ThepresidentofGridlandhashiredyoutodesignaprogramthatcalculatesthelengthoftheshortesttraveling-salesmantourforthetownsinthecountry.InGridland,thereisonetownateachofthepointsofarectangulargrid[直角坐标网].RoadsrunfromeverytowninthedirectionsNorth,Northwest,West,Southwest,South,Southeast,East,andNortheast,providedthatthereisaneighboringtowninthatdirection.ThedistancebetweenneighboringtownsindirectionsNorth-SouthorEast-Westis1unit.ThelengthoftheroadsismeasuredbytheEuclideandistance[欧几里得距离].Forexample,Figure7shows2*3-Gridland,i.e.,arectangulargridofdimensions[规模，大小]2by3.In2*3-Gridland,theshortesttourhaslength6.
Figure7:
Atraveling-salesmantourin2*3-Gridland.
Input
Thefirstlinecontainsthenumberofscenarios[剧情（情况）].
Foreachscenario,thegriddimensionsmandnwillbegivenastwointegernumbersinasingleline,separatedbyasingleblank,satisfying1Output
Theoutputforeachscenariobeginswithalinecontaining"Scenario#i:
",whereiisthenumberofthescenariostartingat1.Inthenextline,printthelengthoftheshortesttraveling-salesmantourroundedtotwodecimaldigits.Theoutputforeveryscenarioendswithablankline.
SampleInput
2
22
23
SampleOutput
Scenario#1:
4.00
Scenario#2:
6.00
思路：
如果城镇数是偶数，即分布的行和列至少有一个是偶数的时候，我们可以一个个走，正好最后一次回到起始城镇，即m*n;如果是奇数，就是说分布的行和列都是奇数，那么我们倒数第二次到的城镇和起点城镇之间相距斜线，即m*n-1+根号2。
代码呈现：
#include
#include
intmain（void）
{
intk,i=1;
floatm,n;
scanf（"%d",&k）;//循环执行的次数
while（k--）
{
scanf（"%f%f",&m,&n）;
if（（int）m%2!
=0&&（int）n%2!
=0）
printf（"Scenario#%d:
\n%.2f\n\n",i,m*n-1+sqrt
（2））;
else
printf（"Scenario#%d:
\n%.2f\n\n",i,n*m）;
}
return0;
}
代码效果：
ZOJProblemSet–1045HangOver
TimeLimit:
1Second MemoryLimit:
32768KB
Howfarcanyoumakeastackofcardsoverhangatable?
Ifyouhaveonecard,youcancreateamaximumoverhangofhalfacardlength.（We'reassuming[假设]thatthecardsmustbeperpendiculartothetable.）Withtwocardsyoucanmakethetopcardoverhangthebottomonebyhalfacardlength,andthebottomoneoverhangthetablebyathirdofacardlength,foratotalmaximumoverhangof1/2+1/3=5/6cardlengths.Ingeneralyoucanmakencardsoverhangby1/2+1/3+1/4+...+1/（n+1）cardlengths,wherethetopcardoverhangsthesecondby1/2,thesecondoverhangsthethirdby1/3,thethirdoverhangsthefourthby1/4,etc.,andthebottomcardoverhangsthetableby1/（n+1）.Thisisillustrated[有插图的]inthefigurebelow.
Theinputconsistsofoneormoretestcases,followedbyalinecontainingthenumber0.00thatsignalstheendoftheinput.Eachtestcaseisasinglelinecontainingapositivefloating-pointnumbercwhosevalueisatleast0.01andatmost5.20;cwillcontainexactlythreedigits.
Foreachtestcase,outputtheminimumnumberofcardsnecessarytoachieveanoverhangofatleastccardlengths.Usetheexactoutputformatshownintheexamples.
Exampleinput:
1.00
3.71
0.04
5.19
0.00
Exampleoutput:
3card（s）
61card（s）
1card（s）
273card（s）
思路：
题意：
输入的小数在0.01-5.20之间，输出满足1/2+1/3+……+1/（n+1）稍大于小数的n的值
1.无限输入小数
2.判断该小数是否在0.01-5.20之间
3.对1/2+1/3+……+1/（n+1）进行累加，同时判断该式子是否大于小数，是则输出n。
不是则继续循环直至是
代码呈现：
#include
intmain（）
{
floata;
doubles;
intn;
while（scanf（"%f",&a）!
=EOF&&a!
=0.00）
{
s=0.00;
n=1;
if（a>=0.01&&a<=5.20）
{
for（;a>s;n++）
s+=1.0/（n+1）;
printf（"%dcard（s）\n",n-1）;
}
}
return0;
}
代码效果：
ZOJProblemSet–1048FinancialManagement
TimeLimit:
1Second MemoryLimit:
32768KB
Larrygraduatedthisyearandfinallyhasajob.He'smakingalotofmoney,butsomehowneverseemstohaveenough.Larryhasdecidedthatheneedstograbholdof[控制、抓住]hisfinancialportfolioandsolvehisfinancingproblems.Thefirststepistofigureoutwhat'sbeengoingonwithhismoney.Larryhashisbankaccountstatementsandwantstoseehowmuchmoneyhehas.HelpLarrybywritingaprogramtotakehisclosingbalancefromeachofthepasttwelvemonthsandcalculatehisaverageaccountbalance[账户余额].
InputFormat:
Theinputwillbetwelvelines.Eachlinewillcontaintheclosingbalance[终期余额]ofhisbankaccount[银行存款]foraparticularmonth.Eachnumberwillbepositiveanddisplayedtothepenny.Nodollarsignwillbeincluded.
OutputFormat:
Theoutputwillbeasinglenumber,theaverage（mean）oftheclosingbalancesforthetwelvemonths.Itwillberoundedtothenearestpenny,precededimmediatelybyadollarsign,andfollowedbytheend-of-line.Therewillbenootherspacesorcharactersintheoutput.
SampleInput:
100.00
489.12
12454.12
1234.10
823.05
109.20
5.27
1542.25
839.18
83.99
1295.01
1.75
SampleOutput:
$1581.42
思路：
求12个浮点型实数的和，然后求平均输出
注：
输出$时的格式是printf（"$%.2f",s）;
代码呈现：
#include
voidmain（）
{
inti;
floata,s;
s=0.0;
for（i=0;i<12;i++）
{
scanf（"%f",&a）;
s=s+a;
}
s=s/12;
printf（"$%.2f\n",s）;
}
代码效果：
ZOJProblemSet–1049IThinkINeedaHouseboat
TimeLimit:
1Second MemoryLimit:
32768KB
FredMapperisconsideringpurchasing[购买,获得]somelandinLouisianatobuildhishouseon.Intheprocessofinvestigating[调查；审查]theland,helearnedthatthestateofLouisianaisactuallyshrinking[退缩]by50squaremileseachyear,duetoerosion[侵蚀，腐蚀]causedbytheMississippiRiver.SinceFredishopingtoliveinthishousetherestofhislife,heneedstoknowifhislandisgoingtobelosttoerosion.
Afterdoingmoreresearch,Fredhaslearnedthatthelandthatisbeinglostformsasemicircle[半圆，半圆形].Thissemicircleispartofacirclecenteredat（0,0）,withthelinethatbisects[二等分]thecirclebeingtheXaxis[X轴].LocationsbelowtheXaxisareinthewater.Thesemicirclehasanareaof0atthebeginningofyear1.（SemicircleillustratedintheFigure.）
InputFormat:
Thefirstlineofinputwillbeapositiveintegerindicatinghowmanydatasetswillbeincluded（N）.
EachofthenextNlineswillcontaintheXandYCartesiancoordinates[笛卡儿坐标]ofthelandFredisconsidering.Thesewillbefloatingpointnumbersmeasuredinmiles.TheYcoordinatewillbenon-negative.（0,0）willnotbegiven.
OutputFormat:
Foreachdataset,asinglelineofoutputshouldappear.Thislineshouldtaketheformof:
��PropertyN:
ThispropertywillbeginerodinginyearZ.��
WhereNisthedataset（countingfrom1）,andZisthefirstyear（startfrom1）thispropertywillbewithinthesemicircleATTHEENDOFYEARZ.Zmustbeaninteger.
Afterthelastdataset,thisshouldprintout��ENDOFOUTPUT.�
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