PAT Advanced Level 1001

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本文介绍了一个编程问题：计算两个整数a和b的和，并将结果以标准格式输出，即每三位数字用逗号分隔。文章提供了一个C++实现示例，包括输入处理、格式化输出等功能。

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1001. A+B Format (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

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

提交代码

这是鄙人的第一个Advanced Level级别的代码C++

#include<iostream>
#include<string>
#include<sstream>
using namespace std;
void format(int data);
template<class in, class out>
out convert(const in& a);
int main()
{
    int a,b;
    while(cin>>a>>b)
    {
        format(a + b);
    }
    return 0;
}
void format(int data)
{
    if (data < 0)
    {
        cout<<"-";
        data = data * (-1);
    }
    string str;
    string colloma(",");
    str = convert<int, string>(data);
    int flag = str.length() % 3;
    for(int i = str.length() - 3; i >= flag; i-=3)
    {
        str.insert(i, colloma);
    }
    if (str[0] == ',')
    {
        str = str.substr(1);
    }
    cout<<str<<endl;
}

template<class in, class out>
out convert(const in& a)
{
    stringstream temp;
    temp<<a;
    out b;
    temp>>b;
    return b;
}