Avionics 1002 A + B problem II//algorithm

A + B Problem II

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 429408 Accepted Submission (s): 83462

Problem Descriptioni has a very simple problem for you. Given integers A and B, your job is to calculate the Sum of A + B.

Inputthe 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.

Outputfor Each 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

1 : 1 2 3  2:112233445566778899 998877665544332211  1111111111111111110

Add large numbers and note the format

1#include <iostream>2#include <cstring>3#include <string>4 using namespacestd;5 stringBigadd (stringNUM1,stringnum2) {6 7     stringRes;8     if(num1.size () = =0) {9res =num2;Ten         returnRes; One     } A     if(num2.size () = =0) { -res =NUM1; -         returnRes; the     } -  -res =""; -     intN1 = Num1.size ()-1, N2 = Num2.size ()-1; +     intcarry =0; -      while(N1 >=0|| N2 >=0) { +  A         intA = N1 >=0? num1[n1--]-'0':0; at         intb = N2 >=0? num2[n2--]-'0':0; -  -         intT = carry + A +b; -Carry = t/Ten; -t = t%Ten; -res = to_string (t) +Res; in     } -     //Judging if there's a carry. to      while(carry) { +         intT = carry/Ten; -Carry%=Ten; theres = to_string (carry) +Res; *carry =T; $     }Panax Notoginseng     returnRes; - } the  + intMain () { A     intsum=0, Count; theCin>>count; +     if(count>=1&&count<= -) { -         stringa[ A][2]; $          for(intI=0; i<count; i++) { $cin>>a[i][0]>>a[i][1]; -         } -          for(intI=0; i<count; i++) { thecout<<" Case"<<i+1<<":"<<Endl; -cout<<a[i][0]<<" + "<<a[i][1]<<" = "<<bigadd (a[i][0],a[i][1]) <<Endl;Wuyi             if(i+1!=count) { thecout<<Endl; -             } Wu         } -     } About     return 0; $}

Avionics 1002 A + B problem II//algorithm