1001 A+B Format

1001 A+B Format （20 分）

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

 

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题型分类：字符串处理

题目大意：题目要求计算两个数的和，每三位要输出一个“逗号”。

解题思路：采用string来处理，同时对负号进行特殊处理。result变量中存入最终的输出答案，其中result变量赋值时，高位存入sum变量的低位，每存入3位数字后，存入一个“，”。输出时倒序输出即可。

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#include 

using namespace std;

int main() {
	int a, b;
	scanf("%d %d", &a, &b);
	int sum = a + b;
	if(sum < 0){ //负数则输出负号，并且将和转变为正数（除去符号更方便处理） 
		printf("-");
		sum = -sum;
	}
	string result, tempStr = to_string(sum);
	int cnt = 0;
	for(int i = tempStr.length() - 1; i >= 0; i--, cnt++){
		if(cnt != 0 && cnt % 3 == 0){ //每隔3位加上一个"," 
			result += ',';
		}
		result += tempStr[i];
	}
	for(int i = result.length() - 1; i >= 0; i--){
		printf("%c", result[i]);
	}
	return 0;
}