hdu 2057 A + B Again

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

解：这个题可以用字符串转成10进制数，相加后再输出16进制数。有一种更简单的实现方法，直接用%X输入输出，但问题是16进制数没有符号位，所以正负要分开。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <deque>
#include <cmath>
#include <map>

using namespace std;
typedef long long ll;

const double inf=1e20;
const int maxn=1e5+10;
const int mod=1e9+7;

int main(){
    __int64 a,b,c;
    while(scanf("%I64X %I64X ",&a,&b)!=EOF){
        c=a+b;
        if(c>=0)printf("%I64X\n",c);
        else printf("-%I64X\n",-c);
    }
    return 0;
}

 

来源：https://www.cnblogs.com/wz-archer/p/12453921.html