A + B Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16794 Accepted Submission(s): 7229
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
题目大意：给你两个带符号的16进制数A和B，输出A+B
思路：A和B比较大，用64位整数来存储，输入的时候用X%来控制输入、输出，
但是X%输入、输出的是无符号16进制数，应该再加个判断。
附带：
格式字符 格式字符意义
d 以十进制形式输出带符号整数(正数不输出符号)
o 以八进制形式输出无符号整数(不输出前缀O)
x 以十六进制形式输出无符号整数(不输出前缀OX)
u 以十进制形式输出无符号整数
f 以小数形式输出单、双精度实数
e 以指数形式输出单、双精度实数
g 以%f%e中较短的输出宽度输出单、双精度实数
c 输出单个字符
s 输出字符串
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int main()
{
__int64 a,b;
while(~scanf("%I64X%I64X",&a,&b))
{
if(a + b >= 0)
printf("%I64X\n",a+b);
else
printf("-%I64X\n",-(a+b));
}
return 0;
}
