A + B Problem II---hdu1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260647    Accepted Submission(s): 50397

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

 

Sample Input

2

1 2

112233445566778899 998877665544332211

 

Sample Output

Case 1:

1 + 2 = 3

 

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

 

 

 #include<stdio.h>
 #include<string.h>
 #define as 1000
 int main()
 {
     int as1[as+],as2[as+],cot=,j,t;
     char shuru1[as+],shuru2[as+];
     int n,i;
     scanf("%d",&n);
     getchar();//吸收回车符
     t=n;
     while(n--)
     {
         cot++;
         scanf("%s",shuru1);
         scanf("%s",shuru2);
         memset(as1,,sizeof(as1));
         memset(as2,,sizeof(as2));
         for(i=,j=strlen(shuru1)-;j>=;j--,i++)
         {
             as1[i]=shuru1[j]-'';
 
         }
         for(i=,j=strlen(shuru2)-;j>=;j--,i++)
         {
             as2[i]=shuru2[j]-'';
         }
         for(i=;i<as+;i++)
         {
             as1[i]+=as2[i];
             if(as1[i]>=)//判断是否满十进一
             {
                 as1[i+]++;
                 as1[i]-=;
             }
 
         }
         for(i=as+;(i>=)&&(as1[i]==);i--);//去掉结果前面多余的0
         printf("Case %d:\n",cot);
         printf("%s + %s = ",shuru1,shuru2);
          if(i>=)
          {
              for(;i>=;i--)
              printf("%d",as1[i]);
          }
          else
          printf("");
          printf("\n");
          if(cot!=t)
          printf("\n");//最后不要多换行
     }
     return ;
 }