[Python/Java](PAT)1001 A+B Format (20)

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

题目大意
    计算输入的A和B的和，按照每三位以','分割的格式输出
    把a+b的和转为字符串s。除了第一位是负号的情况，只要当前位的下标i满足(i + 1) % 3 == len % 3并且i不是最后一位，就在逐位输出的时候在该位输出后的后面加上一个逗号。

分析

使用python实现就直接相加然后格式化输出就可以了。

使用java实现，则先判断相加后是否是负数。是负数先输出负号。然后倒置字符串，每隔三个字符加一个逗号，且如果处于最后的位置则不用添加逗号。最后再翻转过来，输出即可。

python实现

if __name__ == "__main__":
    line = input().split(" ")
    a, b = int(line[0]), int(line[1])
    print('{:,}'.format(a+b))

 java实现

import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] line = br.readLine().split(" ");
		br.close();
		int temp = Integer.valueOf(line[0]) + Integer.valueOf(line[1]);
		if(temp < 0){
			System.out.print("-");
		}
		String s = String.valueOf(Math.abs(temp));
		String a = new StringBuffer(s).reverse().toString();
		String get = "";
		for(int i=0;i<a.length();i++){
			get = get + a.charAt(i);
			if((i+1) %3 == 0 && i != a.length() - 1){
				get = get + ',';
			}
		}
		String out = new StringBuffer(get).reverse().toString();
		System.out.print(out);
	}
}

通过这个例子可以看出，如果不想用C/C++来写PAT的话，python也勉强可以考虑用来解题，但是java真的太慢了

https://www.amoshuang.com/patadvanced