PAT(Advanced)1001 A+B Format (20 分)

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

读题：给出两个有符号整数，将它们按照三位一个逗号的方式输出，有负号打负号。整数的范围是-1000000到1000000。

解题思路：因为逗号的判断是按位判断的，如果顺序遍历每一位会很麻烦，用字符串的函数to_string()将数字转换为字符串，已知字符串的大小，就可以从后往前数，不用把逗号存到字符串里，直接打印就好了。

AC代码

#include<iostream>
#include<string>
using namespace std;
int main(){
    int a,b,c;
    string result;
    cin >> a >> b;
    c = a+b;
    result = to_string(c);
    for( int i = 0 ; i < result.size() ; i ++ ){
        cout << result[i];
        if( abs(c) >= 1e6 && i == result.size() - 7 ){
            cout << ',';
    }
        if( abs(c) >= 1e3 && i == result.size() - 4 ){
        cout << ',';
    }
    }
    return 0;
}