A + B Problem II 大数加法

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：

 #include<iostream>
 #include<string>
 using namespace std;
 string ans;
 void bintadd(string a,string b)
 {
     int alen = a.length() - ,blen = b.length() - ;//alen,blen分别存放a,b字符串的最大下标
     int up = ,i = ,j,te;//up存放进位值，te存放余数

     while(alen >=  || blen >= )
     {
         if(alen >=  && blen >= )//两个字符串都没有计算完
             te = a[alen--] + b[blen--] + up -'' - '';//倒着计算字符串里的数，记得要加上进位值up
         else
             if(alen >=  && blen < )//如果b字符串已经计算完
                 te = a[alen--] + up -'';
             else
                 if(alen <  && blen >= )//如果a字符串已经计算完
                     te = b[blen--] + up - '';
         if(te > )
         {
             up = te / ;//得到进位值
             te = te % ;//得到余数
         }
         else
             up = ;
         ans.push_back(te + '');//将结果放进结果字符串
     }
     if(up)
         ans.push_back(up + '');//如果出现两个字符串一样长，而还有进位值
 }

 int main()
 {
     string a,b;
     int t,ca = ,i;
     cin >> t;
     for(ca = ;ca <= t;ca++)
     {
         cin >> a >> b;
         bintadd(a,b);
         if(ca != )
             cout << endl;
         cout << "Case" << " " << ca << ":" << endl;
         cout << a << " + " << b << " = ";
         for(i = ans.length() - ;i >= ;i--)//反向输出结果字符串
             cout << ans[i];
         cout << endl;
         ans.erase();
     }
 }

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。。。

参考地址：http://www.haogongju.net/art/1227474