杭电 2057 A + B Again【十六进制】【三目运算符】

A + B Again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23035    Accepted Submission(s): 9990


[align=left]Problem Description[/align] 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 ! 
[align=left]Input[/align] 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. 
[align=left]Output[/align] For each test case,print the sum of A and B in hexadecimal in one line.
 
[align=left]Sample Input[/align]

+A -A
+1A 12
1A -9
-1A -12
1A -AA 
[align=left]Sample Output[/align]

0
2C
11
-2C
-90
【不好意思复制粘贴了。。。】
#include<stdio.h>int main(){ __int64 a,b; // 由于定义的数字最多有16位长度，所以定义为int64 while(scanf("%I64X%I64X",&a,&b)!=EOF) printf(a+b>=0?"%I64X\n":"-%I64X\n",a+b>0?a+b:-a-b); // 完全是考察输出格式的为题，三目运算符的绝佳运用。。 return 0;}