【2057 A + B Again ？】A + B Again

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我是靠谱客的博主拉长电源，最近开发中收集的这篇文章主要介绍【2057 A + B Again ？】A + B Again，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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

#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("%I64Xn",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