7th后杭电ACM总结.docx
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7th后杭电ACM总结.docx
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7th后杭电ACM总结
第一次ACM习题总结
1013DigitalRoots1
1048TheHardestProblemEver3
1089A+BforInput-OutputPractice(I)6
1090A+BforInput-OutputPractice(II)7
1091A+BforInput-OutputPractice(III)8
1092A+BforInput-OutputPractice(IV)9
1093A+BforInput-OutputPractice(V)11
1094A+BforInput-OutputPractice(VI)13
1095A+BforInput-OutputPractice(VII)15
1096A+BforInput-OutputPractice(VIII)16
1170BalloonComes!
18
1013DigitalRoots
TimeLimit:
2000/1000MS(Java/Others) MemoryLimit:
65536/32768K(Java/Others)
TotalSubmission(s):
13885 AcceptedSubmission(s):
3870
ProblemDescription
Thedigitalrootofapositiveintegerisfoundbysummingthedigitsoftheinteger.Iftheresultingvalueisasingledigitthenthatdigitisthedigitalroot.Iftheresultingvaluecontainstwoormoredigits,thosedigitsaresummedandtheprocessisrepeated.Thisiscontinuedaslongasnecessarytoobtainasingledigit.
Forexample,considerthepositiveinteger24.Addingthe2andthe4yieldsavalueof6.Since6isasingledigit,6isthedigitalrootof24.Nowconsiderthepositiveinteger39.Addingthe3andthe9yields12.Since12isnotasingledigit,theprocessmustberepeated.Addingthe1andthe2yeilds3,asingledigitandalsothedigitalrootof39.
Input
Theinputfilewillcontainalistofpositiveintegers,oneperline.Theendoftheinputwillbeindicatedbyanintegervalueofzero.
Output
Foreachintegerintheinput,outputitsdigitalrootonaseparatelineoftheoutput.
SampleInput
24
39
0
SampleOutput
6
3
思路:
1.定义char字符串
2.将各个字符相加,然后再循环相加至小于10
3.输出结果
代码呈现:
#include
#include
intmain()
{
charch[100000];
inti,k;
while(gets(ch)&&strcmp(ch,"0")!
=0)
{
intsum=0;
for(i=0;ch[i]!
='\0';i++)
{
k=ch[i]-'0';
sum+=k;
}
if(sum<10)
printf("%d\n",sum);
else
{
if(sum%9==0)//当sum可以被9整除的时候输出应该是9
printf("9\n");
else//其他sum对9求余所得即可
printf("%d\n",sum%9);
}
}
return0;
}
代码效果:
1048TheHardestProblemEver
TimeLimit:
2000/1000MS(Java/Others) MemoryLimit:
65536/32768K(Java/Others)
TotalSubmission(s):
4787 AcceptedSubmission(s):
2199
ProblemDescription
JuliusCaesarlivedinatimeofdangerandintrigue.ThehardestsituationCaesareverfacedwaskeepinghimselfalive.Inorderforhimtosurvive,hedecidedtocreateoneofthefirstciphers.Thiscipherwassoincrediblysound,thatnoonecouldfigureitoutwithoutknowinghowitworked.
YouareasubcaptainofCaesar'sarmy.ItisyourjobtodecipherthemessagessentbyCaesarandprovidetoyourgeneral.Thecodeissimple.Foreachletterinaplaintextmessage,youshiftitfiveplacestotherighttocreatethesecuremessage(i.e.,iftheletteris'A',theciphertextwouldbe'F').SinceyouarecreatingplaintextoutofCaesar'smessages,youwilldotheopposite:
Ciphertext
ABCDEFGHIJKLMNOPQRSTUVWXYZ
Plaintext
VWXYZABCDEFGHIJKLMNOPQRSTU
Onlylettersareshiftedinthiscipher.Anynon-alphabeticalcharactershouldremainthesame,andallalphabeticalcharacterswillbeuppercase.
Input
Inputtothisproblemwillconsistofa(non-empty)seriesofupto100datasets.Eachdatasetwillbeformattedaccordingtothefollowingdescription,andtherewillbenoblanklinesseparatingdatasets.Allcharacterswillbeuppercase.
Asingledatasethas3components:
Startline-Asingleline,"START"
Ciphermessage-Asinglelinecontainingfromonetotwohundredcharacters,inclusive,comprisingasinglemessagefromCaesar
Endline-Asingleline,"END"
Followingthefinaldatasetwillbeasingleline,"ENDOFINPUT".
Output
Foreachdataset,therewillbeexactlyonelineofoutput.ThisistheoriginalmessagebyCaesar.
SampleInput
START
NSBFW,JAJSYXTKNRUTWYFSHJFWJYMJWJXZQYTKYWNANFQHFZXJX
END
START
NBTZQIWFYMJWGJKNWXYNSFQNYYQJNGJWNFSANQQFLJYMFSXJHTSINSWTRJ
END
START
IFSLJWPSTBXKZQQBJQQYMFYHFJXFWNXRTWJIFSLJWTZXYMFSMJ
END
ENDOFINPUT
SampleOutput
INWAR,EVENTSOFIMPORTANCEARETHERESULTOFTRIVIALCAUSES
IWOULDRATHERBEFIRSTINALITTLEIBERIANVILLAGETHANSECONDINROME
DANGERKNOWSFULLWELLTHATCAESARISMOREDANGEROUSTHANHE
思路:
将ABCDEFGHIJKLMNOPQRSTUVWXYZ
转换成VWXYZABCDEFGHIJKLMNOPQRSTU
1.当ASCII码<=69时,ASCII加上21,当ASCII<=90时,减去5
2.gets输入整条语句,循环判断当其ASCII不在6590范围内则不作改动,否则改动
3.当“EOF”时,不再解码
4.直到输入ENDOFINPUT程序结束
注意:
while(gets(a))是无限输入
还要注意a(non-empty)seriesofupto100datasets这句话
代码呈现:
#include
#include
intmain()
{
chara[11],b[3],ch[201];
inti=0,j;
while(gets(a)&&i!
=100)
{
i++;
if(strcmp(a,"ENDOFINPUT")==0)
break;
else
{
gets(ch);
for(j=0;ch[j]!
='\0';j++)
{
if(ch[j]>=65&&ch[j]<=90)
{
if(ch[j]<70)
ch[j]=ch[j]+21;
else
ch[j]=ch[j]-5;
}
}
gets(b);
if(strcmp(b,"END")==0)
puts(ch);
}
}
return0;
}
代码效果:
1089A+BforInput-OutputPractice(I)
TimeLimit:
2000/1000MS(Java/Others) MemoryLimit:
65536/32768K(Java/Others)
TotalSubmission(s):
20555 AcceptedSubmission(s):
12034
ProblemDescription
YourtaskistoCalculatea+b.
Tooeasy?
!
Ofcourse!
Ispeciallydesignedtheproblemforacmbeginners.
Youmusthavefoundthatsomeproblemshavethesametitleswiththisone,yes,alltheseproblemsweredesignedforthesameaim.
Input
Theinputwillconsistofaseriesofpairsofintegersaandb,separatedbyaspace,onepairofintegersperline.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.
SampleInput
15
1020
SampleOutput
6
30
思路:
略
代码呈现:
#include
intmain()
{
inta,b;
while(scanf("%d%d",&a,&b)!
=EOF)
printf("%d\n",a+b);
return0;
}
代码效果:
1090A+BforInput-OutputPractice(II)
TimeLimit:
2000/1000MS(Java/Others) MemoryLimit:
65536/32768K(Java/Others)
TotalSubmission(s):
15676 AcceptedSubmission(s):
10723
ProblemDescription
YourtaskistoCalculatea+b.
Input
InputcontainsanintegerNinthefirstline,andthenNlinesfollow.Eachlineconsistsofapairofintegersaandb,separatedbyaspace,onepairofintegersperline.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.
SampleInput
2
15
1020
SampleOutput
6
30
思路:
略
代码呈现:
#include
intmain()
{
inta,b,n;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&a,&b);
printf("%d\n",a+b);
}
return0;
}
代码效果:
1091A+BforInput-OutputPractice(III)
TimeLimit:
2000/1000MS(Java/Others) MemoryLimit:
65536/32768K(Java/Others)
TotalSubmission(s):
19984 AcceptedSubmission(s):
9892
ProblemDescription
YourtaskistoCalculatea+b.
Input
Inputcontainsmultipletestcases.Eachtestcasecontainsapairofintegersaandb,onepairofintegersperline.Atestcasecontaining00terminatestheinputandthistestcaseisnottobeprocessed.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.
SampleInput
15
1020
00
SampleOutput
6
30
思路:
略
代码呈现:
#include
intmain()
{
inta,b;
while(scanf("%d%d",&a,&b)!
=EOF&&!
(a==0&&b==0))
printf("%d\n",a+b);
return0;
}
代码效果:
1092A+BforInput-OutputPractice(IV)
TimeLimit:
2000/1000MS(Java/Others) MemoryLimit:
65536/32768K(Java/Others)
TotalSubmission(s):
17012 AcceptedSubmission(s):
9064
ProblemDescription
YourtaskistoCalculatethesumofsomeintegers.
Input
Inputcontainsmultipletestcases.EachtestcasecontainsaintegerN,andthenNintegersfollowinthesameline.Atestcasestartingwith0terminatestheinputandthistestcaseisnottobeprocessed.
Output
Foreachgroupofinputintegersyoushouldoutputtheirsuminoneline,andwithonelineofoutputforeachlineininput.
SampleInput
41234
512345
0
SampleOutput
10
15
思路:
1.循环至0结束
2.定义int数组
3.循环相加输出结果
代码呈现:
#include
intmain()
{
intarr[1000];
intn,i,sum;
while(scanf("%d",&n)!
=EOF&&n!
=0)
{
sum=0;
for(i=0;i { scanf("%d",&arr[i]); sum=sum+arr[i]; } printf("%d\n",sum); } return0; } 代码效果: 1093A+BforInput-OutputPractice(V) TimeLimit: 2000/1000MS(Java/Others) MemoryLimit: 65536/32768K(Java/Others) TotalSubmission(s): 12467 AcceptedSubmission(s): 8460 ProblemDescription Yourtaskistocalculatethesumofsomeintegers. Input InputcontainsanintegerNinthefirstline,andthenNlinesfollow.EachlinestartswithaintegerM,andthenMintegersfollowinthesameline. Output Foreachgroupofinputintegersyoushouldoutputtheirsuminoneline,andwithonelineofoutputforeachlineininput. SampleInput 2 41234 512345 SampleOutput 10 15 思路: 1.循环n次结束 2.定义int数组 3.循环相加输出结果 代码呈现: #include intmain() { intarr[1000]; intn,i,j,sum; scanf("%d",&n); while(n--) { scanf("%d",&j); sum=0; for(i=0;i { scanf("%d",&arr[i]); sum=sum+arr[i]; } printf("%d\n",sum); } return0; } 代码效果: 1094A+BforInput-OutputPractice(VI) TimeLimit: 2000/1000MS(Java/Others) MemoryLimit: 65536/32768K(Java/Others) TotalSubmission(s): 11662 AcceptedSubmission(s): 7911 ProblemDescription Yourtaskistocalculatethesumofsomeintegers. Input Inputcontainsmultipletestcases,andonecaseoneline.EachcasestartswithanintegerN,
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