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):69615AcceptedSubmission(s):12678
ProblemDescription
Ihaveaverysimpleproblemforyou.GiventwointegersAandB,yourjobistocalculatetheSumofA+B.

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

Output
Foreachtestcase,youshouldoutputtwolines.Thefirstlineis"Case#:",#meansthenumberofthetestcase.Thesecondlineistheanequation"A+B=Sum",SummeanstheresultofA+B.Notetherearesomespacesinttheequation.Outputablanklinebetweentwotestcases.

SampleInput
2
12

SampleOutput
Case1:
1+2=3

Case2:

Author
Ignatius.L

#include<stdio.h>
#include<string.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_str1>len_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):25250AcceptedSubmission(s):10776

ProblemDescription
输入一个百分制的成绩t，将其转换成对应的等级，具体转换规则如下：
90~100为A;
80~89为B;
70~79为C;
60~69为D;
0~59为E;

Input
输入数据有多组，每组占一行，由一个整数组成。

Output
对于每组输入数据，输出一行。如果输入数据不在0~100范围内，请输出一行：“Scoreiserror!”。

Sam