【PAT-Advanced】1001 A+B format

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Problem

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

Solution

解题思路很简单，因为输入的数字绝对值最大是$10^6$，因此和最大是七位数。根据和的位数向输出中增加”,”。这里需要注意的是c++中substr和substring函数的区别（一开始弄错了导致debug了很久）

s.substr(pos, n)：返回一个string，包含s中从pos开始的n个字符的拷贝（pos的默认值是0，n的默认值是s.size() - pos，即不加参数会默认拷贝整个s）
s.substring(start, end)：将返回一个包含从 start 到最后（不包含 end ）的子字符串的字符串

更加详细的区别和用法可以参见这篇博客

但是必须要说一下substring其实不是c++的标准函数，不过在java里有。

所以用c++的话还是用substr（java写多了下手就是substring qaq）

代码如下：

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#include <iostream>
#include <string>
using namespace std;

int main(int argc, const char * argv[]) {
    int a, b, c;
    cin >> a >> b;
    c = a + b;
    string output = to_string(c);
    int length = output.length();
    if(abs(c) >= 1000 && abs(c) < 1000000){
        output = output.substr(0, length-3) + "," + output.substr(length-3);
    }else if(abs(c) >= 1000000){
        output = output.substr(0, length-6) + "," + output.substr(length-6, 3) + "," + output.substr(length-3);
    }
    cout << output << endl;
    return 0;
}