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??106≤a,b≤106. 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
题目分析：
对于代码中的判断条件" (i + 1) % 3 == len % 3 && i != len - 1 "可以有以下推导过程：
以输入-1000000 9为例，两数之和为-999991，转换成string后如下表所示：
| 标号索引 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 字符串 | — | 9 | 9 | 9 | 9 | 9 | 1 |
题目要求是每满3个数就加一个"，"，首先共有7个字符，7%3=1，所以不满3个一组的数只有1个。
按照上表的情况应该是在标号3和4之间加入"，"，可以发现标号的规律是满足（标号+1）%3==1的时候需要在该数的后面加入"，"，但是要排除该数是字符串的最后一个字符的情况。
此外，还要注意当字符串首位有负号的时候要跳过。
这样一来这题就很简单啦~
程序代码如下：
#include <iostream>
using namespace std;
int main() {
int a, b;
cin >> a >> b;//输入两个整数
int num=a+b;
string str = to_string(num);//将两数之和转换为string类型
int len = str.length();
for (int i = 0; i < len; i++) {
cout << str[i];
if (str[i] == '-') continue;//如果结果有负号"-"则跳过
if ((i + 1) % 3 == len % 3 && i != len - 1) cout << ",";//判断何时添加","
}
return 0;
}