题目描述：
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
代码如下：
1 #include<iostream> 2 #include<string> 3 using namespace std; 4 string ans; 5 void bintadd(string a,string b) 6 { 7 int alen = a.length() - 1,blen = b.length() - 1;//alen,blen分别存放a,b字符串的最大下标 8 int up = 0,i = 0,j,te;//up存放进位值，te存放余数 9 10 while(alen >= 0 || blen >= 0) 11 { 12 if(alen >= 0 && blen >= 0)//两个字符串都没有计算完 13 te = a[alen--] + b[blen--] + up -'0' - '0';//倒着计算字符串里的数，记得要加上进位值up 14 else 15 if(alen >= 0 && blen < 0)//如果b字符串已经计算完 16 te = a[alen--] + up -'0'; 17 else 18 if(alen < 0 && blen >= 0)//如果a字符串已经计算完 19 te = b[blen--] + up - '0'; 20 if(te > 9) 21 { 22 up = te / 10;//得到进位值 23 te = te % 10;//得到余数 24 } 25 else 26 up = 0; 27 ans.push_back(te + '0');//将结果放进结果字符串 28 } 29 if(up) 30 ans.push_back(up + '0');//如果出现两个字符串一样长，而还有进位值 31 } 32 33 int main() 34 { 35 string a,b; 36 int t,ca = 0,i; 37 cin >> t; 38 for(ca = 1;ca <= t;ca++) 39 { 40 cin >> a >> b; 41 bintadd(a,b); 42 if(ca != 1) 43 cout << endl; 44 cout << "Case" << " " << ca << ":" << endl; 45 cout << a << " + " << b << " = "; 46 for(i = ans.length() - 1;i >= 0;i--)//反向输出结果字符串 47 cout << ans[i]; 48 cout << endl; 49 ans.erase(); 50 } 51 }
代码分析：
上面的代码是我根据网上代码改正后的，我自己写的代码虽然答案正确，但没有上面的代码那么简洁，所以就用了网上的代码来分析，不过网上的这代码有错误，但竟然AC了。。。我改正后，也AC了。。
大数加法，主要就在于将数字存储为字符串，然后根据小学学的满十进一的原则，得出答案，最后反向输出。。。