题目1003：A+B

题目描述：

给定两个整数A和B，其表示形式是：从个位开始，每三位数用逗号","隔开。

现在请计算A+B的结果，并以正常形式输出。

输入：

输入包含多组数据数据，每组数据占一行，由两个整数A和B组成（-10^9 < A,B < 10^9）。

输出：

请计算A+B的结果，并以正常形式输出，每组数据占一行。

样例输入：

-234,567,890 123,456,7891,234 2,345,678

样例输出：

-1111111012346912

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <cstdlib>

using namespace std;

int main()
{
    //-234,567,890 123,456,789
    int num1,num2,num3;
    long number1,number2,sum;
    int weishu ;
    num1 = 1000;
        num2 = 1000;
        num3 = 1000;
    while(scanf("%d,%d,%d",&num1,&num2,&num3) != EOF)
    {

        weishu = 0;
        if(num3 <= 999)
            weishu++;
        if(num2 <= 999)
            weishu++;
        if(num1 <=999)
            weishu++;

        if(weishu == 3)
            number1 = abs(num1) * 1000000 + num2 * 1000 + num3;
        if(weishu == 2)
            number1 = abs(num1) * 1000 + num2;
        if(weishu == 1)
            number1 = abs(num1);
        if(num1 <0)
            number1 = number1 * -1;

        num1 = 1000;
        num2 = 1000;
        num3 = 1000;
        scanf("%d,%d,%d",&num1,&num2,&num3);
        weishu = 0;
        if(num3 <= 999)
            weishu++;
        if(num2 <= 999)
            weishu++;
        if(num1 <=999)
            weishu++;

        if(weishu == 3)
            number2 = abs(num1) * 1000000 + num2 * 1000 + num3;
        if(weishu == 2)
            number2 = abs(num1) * 1000 + num2;
        if(weishu == 1)
            number2 = abs(num1);
        if(num1 < 0)
            number2 = number2 * -1;

        sum = number1 + number2;
        cout << sum << endl;

        num1 = 1000;
        num2 = 1000;
        num3 = 1000;
        sum=0;
    }

    return 0;
}