A + B Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15358 Accepted Submission(s): 6700
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 !
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.
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
思路：
这道题中午看了，当时打算吧十六进制转化成十进制，然后十进制转化成十六进制，但是发现这样做有困难。。。所以不会了，看了网上的代码，才明白，计算机自己就会执行十六进制的运算，只不过的不能直接输出负数，需要自己转化。。。题上说的A 和B 的长度不超过15，因为A和B 是十六进制数，如果长度不考虑正负号的话，一个数字表示4位二进制数，所以A 、B的位数是60位，所以定义A、B时，要定义为64位。当然如果考虑正负号的话，A、B含有的的数字最多为14个，这时，A、B的位数为56位,当然还要定义成64位的。。然后，把A和B相加，当然还是十六进制的数，赋值给B,判断B是不是负数，如果是负的，需变成正的，输出时，添个负号即可。。。
代码：
#include<stdio.h>
int main()
{
__int64 A,B;
while(scanf("%I64X %I64X",&A,&B)!=EOF)
{
B=A+B;
if(B>=0)
printf("%I64X\n",B);
if(B<0)
{
B=-B;
printf("-%I64X\n",B);
}
}
return 0;
}
//如果不相信，认为是负数可以直接输出来的话，请看：
