The addition operation of large number, Hangzhou electric oj-1002

Original title Address: http://acm.hdu.edu.cn/showproblem.php?pid=1002 "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 Integ Er 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 be exceed 1000. "Output" for each test case, you should output of both 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 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" AC Code "

1#include <stdio.h>2#include <string.h>3 #defineMax (A, b) a > B? A:b4 5 Chara[1024x768];6 Charb[1024x768];7 Charc[1024x768];8 inti;9 Ten voidReverseChar*a) One { A     intAA =strlen (a); -     CharT; -      for(i=0; i<aa/2; i++) the     { -t =A[i]; -A[i] = a[aa-1-i]; -a[aa-1-I.] =T; +     } - } + voidAddChar*a,Char*b,Char*c) A { at         intCC =0, AA = strlen (a), BB =strlen (b); -         intLen =max (AA, BB); -             for(i=0; i<aa; i++) -            { -A[i] = a[i]-'0'; -         } in             for(i=0; i<bb; i++) -         { toB[i] = b[i]-'0'; +         } -          for(i=0; i<len; i++) the         { *C[i] = (A[I]+B[I]+CC)%Ten+'0'; $CC = (A[I]+B[I]+CC)/Ten;Panax Notoginseng         } -            if(cc) c[i++] = cc +'0'; theC[i] =' /'; + } A voidPrintChar*c) the { +      for(i = strlen (c)-1; i>=0; i--) -printf"%c", C[i]);  $printf"\ n"); $ } - Main () - { the     intN, J; -scanf"%d", &n);Wuyi      for(j=1; j<=n; J + +) the     { -Memset (A,0,sizeof(a)); Wumemset (b,0,sizeof(b)); -Memset (c,0,sizeof(c)); Aboutscanf"%s%s", A, b); $printf"Case %d:\n", j); -printf"%s +%s =", A, b); - reverse (a), reverse (b); - Add (A, B, c); A print (c); +           if(J! = N) printf ("\ n"); the     } -      $}

Addition operation of large number, Hangzhou electric oj-1002