A + B Again

Problem Description

There must be many A + B problems in our HDOJ , now a new one is coming.
Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too.
Easy ? AC it !

 

Input

The input contains several test cases, please process to the end of the file.
Each case consists of two hexadecimal integers A and B in a line seperated by a blank.
The length of A and B is less than 15.

 

Output

For each test case,print the sum of A and B in hexadecimal in one line.

 

Sample Input

+A -A

+1A 12

1A -9

-1A -12

1A -AA

 

Sample Output

0

2C

11

-2C

-90

 

题目求的是十六进制的加法。刚开始想的是把十六进制转化为十进制，进行加法运算后，再转化为十六进制。

所以这题可以直接用十六进制输入，然后进行十六进制的运算（其实不管是什么进制，在计算机中都是以二进制来计算的，只是按输入输出的格式不同，而强制转化为其它的进制），就像十进制的加法一样。

这里要注意的是输入小于15位，结果超过了二进制中的32位而小于64位。所以这里用__int64的类型。输入输入出格式就是（%I64x,%I64X）。由于%I64X，不能输出负数，所以负数的输出要做处理。

 

#include <stdio.h>

int main(){
    __int64 a;
    __int64 b;
    __int64 c;
    
    while(scanf("%I64X%I64X",&a,&b)!=EOF){
        c=a+b;
        
        if(c<0){
            c=-c;
            printf("-");
        }
        printf("%I64X\n",c);
    }
        
    return 0;
}