【杭电oj】2057 - A + B Again（16进制输入输出）

A + B Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22142 Accepted Submission(s): 9569

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

Author

linle

Source

校庆杯Warm Up

这里有个挺有意思的知识点：十六进制直接输入并计算然后输出，不用字符串去做。

因为这里最多为15位，所以用 %I64X或%I64x（x的大小写表示输入与输出十六进制字母的大小写）。

另外还要注意：16进制如果是负数是没法直接输出的，单独处理一下就好了。

代码如下：

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