2057 A + B Again-优快云博客

本文介绍了一种处理十六进制数加法的方法，通过直接在十六进制下进行加法运算，避免了将十六进制转换为十进制的复杂过程。文章提供了完整的C语言代码实现，用于读取两个十六进制整数，计算它们的和，并以十六进制形式输出结果。

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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<stdio.h>
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;
}

转载于:https://www.cnblogs.com/NYNU-ACM/p/4236789.html