1001 A+B Format (20分)（一个字短！）-优快云博客

本文链接：https://blog.youkuaiyun.com/Ychaoding/article/details/106431920

本文详细解析了1001A+B问题，这是一个经典的编程练习题，旨在掌握整数求和及格式化输出技能。文章通过具体示例介绍了如何读取两个整数，计算它们的和，并按特定格式输出结果，包括处理负数和添加千位分隔符。

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1001 A+B Format (20分)

题目

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

题意

输入两个绝对值不大于10⁶的整数，求和，并且按照每三位之前添加一个逗号的格式输出。

分析

求和，输出符号并取绝对值，转化为字符串；取模确定第一个不是三位的输出序列的长度是多少，输出，然后按照格式输出剩下的各位上的数

代码

#include<iostream>
#include<string>
using namespace std;
int main() {	
	int a, b;
 	scanf("%d %d", &a, &b);
 	a += b;
 	if (a < 0) {
  	printf("-");
  	a = -a;
 	}
 	string s = to_string(a);
 	int i = s.size() % 3;
 	i = i == 0 ? 3 : i;
 	printf("%s", s.substr(0, i).c_str());
 	while (i != s.size()) {
 		printf(",%s", s.substr(i, 3).c_str());
  		i += 3;
 	}
 	return 0;
}