題目：

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

Sample Output:

-999,991

題目分析：
簡單的a+b問題，兩個注意點：
(1)在考慮逗號的時候，是每遇到3個數字加一個逗號，但最後一個逗號不一定存在，如600000，如果不把最後的逗號去掉，就會成為，600，000
(2)考慮a+b=0的情況，如果是0，字串為空，這時要處理一下。
 

程式碼如下：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int main(){
 6     int a,b;
 7     cin>>a>>b;
 8     int c = a+b;
 9     int cc = c;
10     c = abs(c);
11     string s = "";
12     int count = 0;
13     while(c){
14
 
         count++;
15         s += ('0'+c%10);
16         c /= 10;
17         if(count%3==0&&c){   //防止最後一位多出逗號，如600000，出現“,600,000”的情況 
18             s += ',';
19         }
20     }
21     if(cc<0){
22         s += '-';
23     }
24     reverse(s.begin(),s.end());
25     if(s == ""){            //注意判斷和為0的情況，否則輸出的s為空 
26         cout<<0<<endl;
27     } else{
28         cout<<s<<endl;
29     }
30     return 0;
31 }