题目：
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
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.
Output
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
参考代码：（C++）
#include<stdio.h>
using namespace std;
int main(void){
int a,b;
while(scanf("%d%d",&a,&b)!=EOF){
int c = a+b;
if(c<0){
c=-c;
printf("-");
}
if(c>=1000000)
printf("%d,%03d,%03d ",c/1000000,(c/1000)%1000,c%1000);
else if(c>=1000)
printf("%d,%03d ",c/1000,c%1000);
else
printf("%d ",c);
}
return 0;
}
分析：
对于负数，可以先判断a+b的结果是否为负，若结果为负数，则先输出负号『-』
由题目a，b范围，可知a+b的结果的绝对值小于10000000，所以对于大于1000000的部分c/1000000即可。