HDU 1002 ????

HDU 1002 大数相加，题目的链接如下：
http://acm.hdu.edu.cn/showproblem.php?pid=1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 315379 Accepted Submission(s): 61186

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 <iostream>#include <cstring>#include <string>using namespace std ;/*int OJ(int m,int n){    int r;    r=m%n;    while(r!=0){        m=n;        n=r;        r=m%n;    }    return n;}*/string add(string a, string b){    if(a == "0" && b=="0") return "0";    if(a=="0") return b;    if(b=="0") return a;    string max , min;    max = a , min = b ;    if(a.length() < b.length()){        max = b ;        min = a ;    }    int c = max.length()-1, d = min.length()-1;    for(int i=d ; i>=0 ; i--){        max[c--] += min[i] - '0';    }    for(int i=max.length()-1 ; i>0 ; i--){        if(max[i] > '9'){            max[i] -= 10 ;            max[i-1]++;        }    }    if(max[0] > '9')    {        max[0] -= 10;        max = '1' + max ;    }    return max ;}string str1 , str2 , str3 ;using namespace std ;int main(){    int n , t = 1;    cin >> n ;    for(int i=0 ; i<n ; i++)    {        cin >> str1 >> str2 ;        cout << "Case " << t << ":" << endl ;        str3 = add(str1, str2);        cout << str1 << " + " << str2 <<  " = " << str3 << endl;        if(t<n) cout << endl ;        t++;    }    return 0 ;}