# A + B problem II---hdu1002

Source: Internet
Author: User

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 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 both positive integers, A and B. Notice that the integers is very large, that M EANs should not process them by using 32-bit integer. Assume the length of each integer would not exceed 1000.

Output

For the test case, you should output of the lines. The first line was "Case #:", # means the number of the the test case. The second line is the equation "A + b = sum", sum means the result of A + B. Note there is some spaces int the Equati On. Output a blank line between the test cases.

Sample Input

2 1 2) 112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3 Case 2:1.,122,334,455,667,79e,+17 + 998877665544332211 = 1111111111111111110
`1#include <stdio.h>2#include <string.h>3 #defineAs 10004 intMain ()5 {6     intas1[ as+ -],as2[ as+ -],cot=0, j,t;7     Charshuru1[ as+ -],shuru2[ as+ -];8     intn,i;9scanf"%d",&n);Ten GetChar ();//absorbing carriage return Onet=N; A      while(n--) -     { -cot++; thescanf"%s", shuru1); -scanf"%s", SHURU2); -memset (AS1,0,sizeof(AS1)); -memset (AS2,0,sizeof(AS2)); +          for(i=0, J=strlen (SHURU1)-1; j>=0; J--, i++) -         { +as1[i]=shuru1[j]-'0'; A              at         } -          for(i=0, J=strlen (SHURU2)-1; j>=0; J--, i++) -         { -as2[i]=shuru2[j]-'0'; -         } -          for(i=0;i< as+ -; i++) in         { -as1[i]+=As2[i]; to             if(as1[i]>=Ten)//To determine whether the full 10 into a +             { -as1[i+1]++; theas1[i]-=Ten; *             } \$             Panax Notoginseng         } -          for(i= as+ +;(i>=0) && (as1[i]==0); i--)///Remove the extra 0 in front of the result theprintf"Case %d:\n", cot); +printf"%s +%s =", SHURU1,SHURU2); A          if(i>=0) the          { +               for(; i>=0; i--) -printf"%d", As1[i]); \$          } \$          Else -printf"0"); -printf"\ n"); the          if(cot!=t) -printf"\ n")///Finally, do not change more linesWuyi     } the     return 0; -}`

A + B problem II---hdu1002

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