【2057 A + B Again ？】

A + B Again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12201    Accepted Submission(s): 5331

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

 

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
		_int64 a,b,sum;
		while(scanf("%I64X %I64X",&a,&b)!=EOF){
			 sum=a+b;
			if(sum<0){
				sum=-sum;
				printf("-");
			}
			printf("%I64X\n",sum);
		}		
}

这个十六进制可以直接加减的。所以无需转换的。

在VC++中  longlong 用_int64 表示。输入格式为

输出符 %I64x

[signed]        long long   [int]        64        -2^63 ~ 2^63-1             %I64d

unsigned      long long   [int]        64        0 ~ 2^64-1                    %I64u、%I64o、%I64x

 

所以直接输出的a+b 是两个正数的和。

至于是无符号，输入时间把符号位至1，按正数计算。至于sum=a+b。出现负数，是因为十六进制本身可以自动加减，只是在输出时间把符号位按正数处理了。

如果理解有误请指出。