PAT-1001（甲级）

博客皆个人学习过程中整理，如有问题，欢迎大家指正。

本文链接： https://blog.csdn.net/qq_42017331/article/details/102079840

题目描述：

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

问题分析：

题目很简单，求a+b的和，然后按指定格式输出。根据样例分析，从后往前每3个字符前需要添加一个’,’。根据规律可以得出思路，将和转换为字符串，逆序每数三个字符，即添加一个’,’。不妨先使用vector存储结果。这里需要注意的是，每三个字符添加一个’,’，若a+b结果是3/6/9……位，则’,'岂不是会被添加到首位？因此需要特殊判断，避免出现这种情况。

代码示例：

#include<iostream>
#include<sstream>
#include<vector>
using namespace std;

string toString(int a){
 	stringstream s;
	s << a;
 	return s.str();
}

int main(){
 	int a, b, sum, count = 1;
 	cin >> a >> b;
 	sum = a+b;
 	if(sum < 0){//负数情况处理 
  		cout << "-";
  		sum = -sum;
 	}
 	string str = toString(sum);
 	vector<char> chr; 
 	for(int i = str.size()-1; i >= 0; i--, ++count){
  		chr.push_back(str[i]);
  		if(count % 3 == 0 && count < str.size())
   			chr.push_back(',');
 	}
 	for(int i = chr.size()-1; i >= 0; i--){//逆序打印
  		cout << chr[i];
 	}
 	return 0;
}