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 很简单，数字转字符串即可

1 #define _CRT_SECURE_NO_WARNINGS 2  3 #include 
     
       4 #include 
      
        5 #include 
       
         6 #include 
        
          7 #include 
         
           8  9 using namespace std;10 11 int main()12 {13     int a, b,sum;14     string str;15     scanf("%d %d", &a, &b);16     sum = a + b;17     stringstream ss;18     ss << sum;19     ss >> str;20 21     ///此处可以通过sum/1000，然后转为字符串22     stack
          
           res;23 int k;24 string s;25 for (k = str.length(); k > 3;k-=3)26 { 27 s.assign(str.begin() + k - 3, str.begin() + k);28 res.push(s);29 res.push(",");30 }31 s.assign(str.begin(), str.begin() + k);32 if (s == "-")33 res.pop();34 res.push(s);35 while (!res.empty())36 {37 cout << res.top();38 res.pop();39 }40 return 0;41 }
          
         
        
       
      
     

更简洁点

1 #include 
     
       2 using namespace std; 3 int main() { 4     int a, b; 5     cin >> a >> b; 6     string s = to_string(a + b); 7     int len = s.length(); 8     for (int i = 0; i < len; i++) { 9         cout << s[i];10         if (s[i] == '-') continue;11         if ((i + 1) % 3 == len % 3 && i != len - 1) cout << ",";12     }13     return 0;14 }