A+B超过long long 大数相加问题

转自： http://blog.csdn.net/summerfanl/article/details/52074316

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  

像大数这一类题目一般都是用 string 来做，这个题首要把两个字符串用 reverse 倒序过来运算（由于数字运算是从个位，十位 .....）。最后再倒序回来。

注意：1.这里的运算并不是字符间直接运算，而是转化成 int 型在加减运算。

2.转化成 int 型的两个数运算后可能出现超过 10 ，一定要记得进位。

[cpp]view plain copy     
 #include<iostream>     
 #include<string>     
 using namespace std;     
 char add(char temps,char tempstr,int &tempaddc)     
 {     
     int temp;      
     temp = (temps - '0') + (tempstr - '0') + tempaddc;  //实现进位。     
     tempaddc = temp / 10;  // 算出进位数     
     return temp % 10 + '0';     
 }     
 int main()     
 {     
     string str,s;     
     int tempaddc;     
     char c;     
     int n,lenstr,lens;     
     while(cin>>n)     
     {     
         while(n--)     
         {     
             cin>>str>>s;     
             tempaddc = 0;     
             c = '0';     
             if(s.length() > str.length())swap(s,str);     
             lenstr = str.length();     
             lens = s.length();     
             lens--;     
             lenstr--;     
             while (lenstr >= 0)     
             {     
                 if(lens < 0) str[lenstr] = add(c,str[lenstr],tempaddc);    //字符串长度较长部分的计算     
                 else     
                 {     
                     str[lenstr] = add(s[lens],str[lenstr],tempaddc);     
                     lens--;     
                 }     
                 lenstr--;     
             }     
             cout<<str<<endl;     
         }     
     }     
     return 0;     
 }    