题目描述：

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).
翻译：计算a+b然后按照标准格式输出——这代表数字必须被分成3个一组，之间用逗号相连（除非少于4个数字）

INPUT FORMAT：

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.
翻译：每个输入文件包括一组测试数据。每组测试数据包括一对整数a和b， -1000000 <= a, b <= 1000000，数字之间用空格分开。

OUTPUT FORMAT：

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.
翻译：对于每组测试数据，你需要在一行内输出a+b的和，和必须用标准格式表示。

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Sample Input

-1000000 9

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Sample Output

-999,991

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解题思路：

将sum保存为数组输出，注意处理正负号。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define INF 99999999
using namespace std;
int a,b;
int ans[15],ccount=0;
int main(){
  scanf("%d%d",&a,&b);
  int sum=a+b;
  if(sum==0)printf("0\n");
  if(sum<0){
    printf("-");
    sum=-sum;  
  }
  while(sum!=0){
    ans[ccount++]=sum%10;
    sum=sum/10;
  }
  for(int i=ccount-1;i>=0;i--){
    printf("%d",ans[i]);
    if(i!=0&&i%3==0)printf(",");
    if(i==0)printf("\n");  
  } 
  exit(0);
}