思路分析：
1、将a=a+b之和，当a>1000时，用ans[i++]数组存储a%1000的值，a=a/1000;
2、首先将a带符号输出，若i>0，输出“，”，然后逆向输出”ans[i]’数组，类似地，若i>0，输出“，”。
#include"stdio.h"
#include<stdlib.h>
#include <math.h>
int main()
{int a, b, ans[10] = { 0 }, i = 0;scanf("%d %d", &a, &b);a = a + b;while (abs(a / 1000) > 0) {ans[i++] = abs(a) % 1000;a = a / 1000;}printf("%d", a);if (i > 0) printf(",");for (i--; i >= 0; i--) {printf("%03d", ans[i]);if (i > 0) printf(",");}return 0;
}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