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

1. #include<iostream>  
2. #include<string>  
3. using namespace std;  
4. char add(char temps,char tempstr,int &tempaddc)  
5. {  
6.     int temp;   
7.     temp = (temps - '0') + (tempstr - '0') + tempaddc;  //实现进位。  
8.     tempaddc = temp / 10;  // 算出进位数  
9.     return temp % 10 + '0';  
10. }  
11. int main()  
12. {  
13.     string str,s;  
14.     int tempaddc;  
15.     char c;  
16.     int n,lenstr,lens;  
17.     while(cin>>n)  
18.     {  
19.         while(n--)  
20.         {  
21.             cin>>str>>s;  
22.             tempaddc = 0;  
23.             c = '0';  
24.             if(s.length() > str.length())swap(s,str);  
25.             lenstr = str.length();  
26.             lens = s.length();  
27.             lens--;  
28.             lenstr--;  
29.             while (lenstr >= 0)  
30.             {  
31.                 if(lens < 0) str[lenstr] = add(c,str[lenstr],tempaddc);    //字符串长度较长部分的计算  
32.                 else  
33.                 {  
34.                     str[lenstr] = add(s[lens],str[lenstr],tempaddc);  
35.                     lens--;  
36.                 }  
37.                 lenstr--;  
38.             }  
39.             cout<<str<<endl;  
40.         }  
41.     }  
42.     return 0;  
43. }