问题描述
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
问题分析
Problem Analyse
在HDOJ上已经有很多道A+B的题目了，现在有来了道新的。
给你两个16进制数，你的任务就是计算它们的和，然后以16进制输出。简单吗？把它AC掉！
C编程基础知识
Algorithm Analyse
本题的数据量不大，用64位整数不会溢出。
用%I64X就可以对数据读入输出进行操作。但问题在于直接用%I64X，是不能输出负数的，所以我们碰到负数要自己转成正数，在前面加‘-’。
算法实现
判断是否为负数，可以直接判断是否<0，也可以求二进制最高位上是不是为1：(a+b)&((__int64)1<<63)
参考源码
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| #include<stdio.h> | |
| int main(void) | |
| { | |
| __int64 a, b; | |
| while(scanf("%I64x%I64X", &a, &b) != EOF) | |
| printf(a+b<0?"-%I64X\n":"%I64X\n", a+b<0?-a-b:a+b); | |
| return 0; | |
| } |