PAT-甲级-1001 A+B Format C++

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：

题目意思是计算c=a+b，但是要按照规范输出c，从末位开始，每隔三个输出一个逗号。

分析：

水题。关键在于处理第一个逗号出现的位置。c可能是正数、可能是负数，首先判断是否需要输出-，然后统一处理符号位后面的数字。后面的数字关键在于判断输出逗号的间隔interval，如果数字位数>3，interval需要额外处理，即用%的方式来判断第一个逗号出现的位置，后面剩余的数字位数为3的整数倍，可以固定输出逗号；如果数字位数＜3，不需要输出逗号。

代码：

//
// Created by yaodong on 2023/7/5.
//
#include "iostream"
#include "cstring"
int main() {
    int a, b;
    std::cin >> a >> b;
    int sum = a + b;
    std::string sumString = std::to_string(a + b);
//    std::cout << sumString.length() << std::endl;
    int len = sumString.length();
    if (sum < 0) {
        printf("-");
        sumString = sumString.substr(1, len - 1);
        len = sumString.length();
    }
    int interval = len > 3 ? (len % 3 == 0 ? 3 : len % 3) : len;
    int i = 0;
    while (i < len) {
        if (interval-- > 0)
            std::cout << sumString[i++];
        else {
            interval = 2;
            std::cout << "," << sumString[i++];
        }
    }
//    printf("%s", sumString.c_str());
}