A + B Again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52139    Accepted Submission(s): 21390

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

题目大意：给你两个带符号的16进制数A和B，输出A+B

思路：A和B比较大，用64位整数来存储，输入的时候用X%来控制输入、输出，

但是X%输入、输出的是无符号16进制数，应该再加个判断。

附带：

格式字符  格式字符意义 

d   以十进制形式输出带符号整数(正数不输出符号) 

o 以八进制形式输出无符号整数(不输出前缀O) 

x   以十六进制形式输出无符号整数(不输出前缀OX) 

u   以十进制形式输出无符号整数 

f   以小数形式输出单、双精度实数 

e   以指数形式输出单、双精度实数 

g   以%f%e中较短的输出宽度输出单、双精度实数 

c   输出单个字符 

s   输出字符串

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
 
int main()
{
    __int64 a,b;
    while(~scanf("%I64X%I64X",&a,&b))
    {
        if(a + b >= 0)
            printf("%I64X\n",a+b);
        else
            printf("-%I64X\n",-(a+b));
    }
    return 0;
}