PTA 1001

题目连接：https://pintia.cn/problem-sets/994805342720868352/problems/994805528788582400

Quesetion Describe：

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 −. 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

题目大意：

             求A+B的值，结果按照三位数字一个逗号隔开的格式输出【例：1，000】

思路：

　　　　这题是典型的A+B，范围也不大，两者相加最大也不过是int范围，主要是三位一个逗号的格式，需要单独考虑一下最后结果少于四位不要输出逗号了，防止误判。详情可见Code注释。

 AC代码：

 1 #include <iostream>
 2 
 3 int main()
 4 {
 5     int a,b;
 6     int ch[50];
 7     std::cin >> a >> b;
 8     int num =0 ;
 9     int c = a+b;           //直接相加
10     if(c / 1000 !=0)       //判断是否大于等于1000，否则易出现1+1=002
11     {
12         if(c < 0)              //负数将其负号去掉，方便操作
13         {
14             std::cout << '-';
15             c=0-c;
16         }
17         while(c!=0)          //将c按照3位3位分开存入数组
18         {
19             ch[++num] = c %1000;
20             c/=1000;
21         }
22         std::cout << ch[num--] ;    //先输出首位，防止出现001，200的情况
23         for(int i = num;i>=1;i--)
24         {
25             std::cout<< ',';
26             std::cout.fill('0');           //补位0
27             std::cout.width(3);            //限制3位
28             std::cout << ch[i]  ;
29         }
30     }
31     else
32     {
33         std::cout << c << std::endl;       //小于1000直接输出
34     }
35 
36 }

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