[精品]整理出ACM所有题目及内容答案

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1、.1000A+BProblemProblemDescriptionCalculateA+B.InputEachlinewillcontaintwointegersAandB.Processtoendoffile. OutputForeachcase,outputA+Binoneline. SampleInput11 SampleOutput2 AuthorHDOJ代码：#includeintmain(){inta,b;while(scanf("%d%d",&a,&b)!=EOF)printf("%d",a+b);}1001SumProblemProblemDescript

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

3、ampleInput1100 SampleOutput15050 AuthorDOOMIII解答：#includemain(){intn,i,sum;sum=0;while((scanf("%d",&n)!=-1)){sum=0;for(i=0;i<=n;i++)sum+=i;printf("%d",sum);}}.1002A+BProblemIIProblemDescriptionIhaveaverysimpleproblemforyou.GiventwointegersAandB,yourjobistocalculatetheSumofA+B. InputThef

4、irstlineoftheinputcontainsanintegerT(1<=T<=20)whichmeansthenumberoftestcases.ThenTlinesfollow,eachlineconsistsoftwopositiveintegers,AandB.Noticethattheintegersareverylarge,thatmeansyoushouldnotprocessthembyusing32-bitinteger.Youmayassumethelengthofeachintegerwillnotexceed1000. OutputForeachtestcase,

5、youshouldoutputtwolines.Thefirstlineis"Case#:",#meansthenumberofthetestcase.Thesecondlineistheanequation"A+B=Sum",SummeanstheresultofA+B.Notetherearesomespacesinttheequation.Outputablanklinebetweentwotestcases. SampleInput212112233445566778899998877665544332211 SampleOutputCase1:1+2=3Case2:112233445

6、566778899+998877665544332211=1111111111111111110 AuthorIgnatius.L 代码：#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]={0},b[1001]={0},c[1001]={0};scanf("%s",str1);len_str1=strlen(str1);for(

7、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_str