杭电acm部分答案

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1、ProblemDescriptionCalculate A+B. InputEachlinewillcontaintwointegers A and B.Processtoendoffile. OutputForeachcase,output A+B inoneline. SampleInput11 SampleOutput2#includevoidmain(){inta,b;while(scanf("%d%d",&a,&b)!=EOF){printf("%d",a+b);}}ProblemDescriptionHe

2、y,welcometoHDOJ(HangzhouDianziUniversityOnlineJudge).Inthisproblem,yourtaskistocalculateSUM(n)=1+2+3+...+n.InputTheinputwillconsistofaseriesofintegersn,oneintegerperline.OutputForeachcase,outputSUM(n)inoneline,followedbyablankline.Youmayassumetheresultwillbeintherangeof32

3、-bitsignedinteger.SampleInput1100SampleOutput15050#includevoidmain(){intn,sum,i;while(scanf("%d",&n)!=EOF){sum=0;for(i=0;i<=n;i++)sum+=i;printf("%d",sum);}}ProblemDescriptionIhaveaverysimpleproblemforyou.GiventwointegersAandB,yourjobistocalculatetheSumofA+B. 

4、InputThefirstlineoftheinputcontainsanintegerT(1<=T<=20)whichmeansthenumberoftestcases.ThenTlinesfollow,eachlineconsistsoftwopositiveintegers,AandB.Noticethattheintegersareverylarge,thatmeansyoushouldnotprocessthembyusing32-bitinteger.Youmayassumethelengthofeachintegerwill

5、notexceed1000. OutputForeachtestcase,youshouldoutputtwolines.Thefirstlineis"Case#:",#meansthenumberofthetestcase.Thesecondlineistheanequation"A+B=Sum",SummeanstheresultofA+B.Notetherearesomespacesinttheequation.Outputablanklinebetweentwotestcases. SampleInput2121122334455

6、66778899998877665544332211 SampleOutputCase1:1+2=3Case2:112233445566778899+998877665544332211=1111111111111111110#include#includeintmain(){charstr1[1001],str2[1001];intt,i,len_str1,len_str2,len_max,num=1,k;scanf("%d",&t);getchar();while(t--){inta[1001]=

7、{0},b[1001]={0},c[1001]={0};scanf("%s",str1);len_str1=strlen(str1);for(i=0;i<=len_str1-1;++i)a[i]=str1[len_str1-1-i]-'0';scanf("%s",str2);len_str2=strlen(str2);for(i=0;i<=len_str2-1;++i)b[i]=str2[len_str2-1-i]-'0';if(len_str1>len_str2)len_max=len_str1;elselen_max=len_str2

8、;k=0;for(i=0;i<=len_max-1;++i){c[i]=(a[i]+b[i]+k)%10;k=(a[i]+b[i]+k)/10;}if(k!=0)c[len_max]=1;pr