杭电ACM水题题目及代码.docx

1002A+BProblemII

TimeLimit:2000/1000MS(Java/Others)????MemoryLimit:65536/32768K(Java/Others)

TotalSubmission(s):69615????AcceptedSubmission(s):12678

ProblemDescription

Ihaveaverysimpleproblemforyou.GiventwointegersAandB,yourjobistocalculatetheSumofA+B.

Input

ThefirstlineoftheinputcontainsanintegerT(1=T=20)whichmeansthenumberoftestcases.ThenTlinesfollow,eachlineconsistsoftwopositiveintegers,AandB.Noticethattheintegersareverylarge,thatmeansyoushouldnotprocessthembyusing32-bitinteger.Youmayassumethelengthofeachintegerwillnotexceed1000.

Output

Foreachtestcase,youshouldoutputtwolines.ThefirstlineisCase#:,#meansthenumberofthetestcase.ThesecondlineistheanequationA+B=Sum,SummeanstheresultofA+B.Notetherearesomespacesinttheequation.Outputablanklinebetweentwotestcases.

SampleInput

2

12

SampleOutput

Case1:

1+2=3

Case2:

Author

Ignatius.L

#includestdio.h

#includestring.h

intmain(){

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(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_str1len_str2)

len_max=len_str1;

else

len_max=len_str2;

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;

printf(Case%d:\n,num);

num++;

printf(%s+%s=,str1,str2);

if(c[len_max]==1)

printf(1);

for(i=len_max-1;i=0;--i){

printf(%d,c[i]);

}

printf(\n);

if(t=1)

printf(\n);

}

return0;

}

成绩转换

TimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)

TotalSubmission(s):25250Accepted