第二次寒假作业的编程题

(我觉得我大一就只学了怎么做题目呀,还没碰过几次电脑呢,啊啊啊,我实在不知道怎么发截图啊,更不知道怎么发链接,上机课都没教过这个呀)

1001. A+B Format (20)
      时间限制
      400 ms
      内存限制
      65536 kB
      代码长度限制
      16000 B
      判题程序
      Standard
      作者
      CHEN, Yue
      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

解题思路:
看不懂题目,但是幸好有输入输出的例子,再问一下同学大概知道题目问的是a+b,且要三位三位输出,且-1000000 <= a, b <= 1000000,貌似很简单,所以我就写了

include<stdio.h>

int main()
{
int a,b,i,d,e,f;
scanf("%d%d",&a,&b);
d=a+b;
f=d%1000;
d/=1000;
e=d%1000;
d/=1000;
if(d!=0)
printf("%d,%03d,%03d",d,e,f);
else if(e!=0)
printf("%d,%03d",e,f);
else printf("%d",f);
return 0;
}
其中f代表后三位的数,e代表千位到十万位的数,d代表百万位到最高位的数
然后自己试了一下输入例子中的-1000000 9
结果却跑出了-999,-991,多了一个负号!额,于是我就改成:

include<stdio.h>

include<math.h>

int main()
{
int a,b,i,d,e,f;
scanf("%d%d",&a,&b);
d=a+b;
f=d%1000;
d/=1000;
e=d%1000;
d/=1000;
if(d!=0)
printf("%d,%03d,%03d",d,abs(e),abs(f));
else if(e!=0)
printf("%d,%03d",e,abs(f));
else printf("%d",f);
return 0;
}
结果提交上去就全对了