【Problem Description】:
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 -10 6<a,b<10 6​. 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

【解题思路】
利用stringstream将两数之和转化为字符串，反转字符串，新建一字符串并在符合条件的位置增加逗号，反转该字符串。
【C++代码】

#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
#include <string>
using namespace std;
int main() {
 int a,b,c;
 cin>>a>>b;
 c=a+b;
 if(c<0) {
  cout<<"-";
  c=abs(c);
 }
 stringstream ss;
 ss<<c;
 string str=ss.str();
 reverse(str.begin(),str.end());
 string ans="";
 for(int i=0; i<str.size(); i++) {
  if((i%3==0)&&(i!=0)) {
   ans+=",";
   ans+=str[i];
  } else ans+=str[i];
 }
 reverse(ans.begin(),ans.end());
 cout<<ans<<endl;
 return 0;
}

【Java代码】

import java.util.Scanner;
public class Main {   
 public static void main(String[] args) {        
  Scanner scan = new Scanner(System.in);        
  int a = scan.nextInt();        
  int b = scan.nextInt();        
  scan.close();        
  System.out.printf("%,d",a+b);//由于是标准格式可以直接输出    
 }
}