杭电OJ2057-A + B Again

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A + B Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22799 Accepted Submission(s): 9890

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

AC代码：（用c语言交）

#include<stdio.h>void main(){__int64 a, b, sum;while(scanf("%I64X%I64X", &a, &b) != EOF){sum = a + b;if(sum >= 0)printf("%I64X\n", sum);elseprintf("-%I64X\n", -sum);}}

同样的代码，用C++交就wrong了，我也不知道为什么。

看来hex和%I64X还是有区别的

应该把两个都搞清楚，用在合适的地方

其次，这道题能这么直接秒A也是在我意料之外的

我把它想复杂了我以为还要把符号拆开算

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