A + B Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 15391 Accepted Submission(s): 6713

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

思路：当时写的时候，想的是把16进制转化为10进制，相加后再转化为16进制，感觉比较麻烦

就查了网上的博客，才发现16进制也可以加减，就是不能输出正负好，就设置了一个字符c来

控制正负，注意printf("%x\n",sum);和printf（“%X\n",sum);在这道题上是有区别的；刚开始就把

“X"写成"x",一直提交说答案错误！！！正确代码如下

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