Big integer addition

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 shoshould 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 shoshould 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

The first solution is to store multiple integers. obviously, no variable of the C/C ++ inherent type can save it. the most intuitive idea is to use a string to save it. A string is essentially a character array. then perform operations from the low position to the high position in sequence. The addition result is greater than 9, that is, carry.
Complete code:

/*************************************** * **/* Hangdian acm1002: A + B Problem II/* 2010/10/1/* VC 6.0 by 52 coder/* persever2009@gmail.com ******************** * *********************/# include <iostream> using namespace STD; string sum_calc (string S1, string S2) {int length; // compare S1, S2 length if (s1.length ()> = s2.length () {length = s1.length (); string TMP (s1.length ()-s2.length (), '0'); S2 = TMP + S2;} else {length = s2.len Struct (); string TMP (s2.length ()-s1.length (), '0'); S1 = TMP + S1;} string sum (length, '0 '); // carry flag int carry = 0; For (int K = length-1; k! =-1; -- k) {// the ASCII code corresponding to the character '0' is 48 sum [k] = (S1 [k] + S2 [k] + carry-96) % 10 + 48; carry = (S1 [k] + S2 [k] + carry-96> 9);} // after the highest bit carries, the carry processing if (carry) sum = '1' + sum; return sum;} void main () {int t; // The number of test cases; CIN> T; // use the dynamic array when not knowing the dimension before compiling; string * A = new string [T]; string * B = new string [T]; string * sum = new string [T]; for (INT I = 0; I! = T; ++ I) {CIN> A [I]> B [I]; sum [I] = sum_calc (A [I], B [I]) ;}for (Int J = 0; J! = T; ++ J) {cout <"case" <j + 1 <": "<Endl <A [J] <" + "<B [J] <" = "<sum [J]; if (I! = T-1) cout <Endl ;}

Article from: I love programmers http://www.52coder.net/archives/1055.html copyright all. This site in addition to the source, are the author of the original articles, can be freely cited, but please indicate the source.