題目大意：輸入兩個16進制數，輸出相加的結果
/*
*
*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
*
*
*Source
*校慶杯Warm Up
*
*
*Recommend
*linle
*
*/
#include<iostream>
#include<cmath>
#include<string>
using namespace std;
long long a, b;
int length;
long long htod(string s) {
long long temp(0);
length = s.size();
for (int i = s.size() - 1; i >= 0; --i) {
if ('A' <= s.at(i) && s.at(i) <= 'Z') {
temp += (s.at(i) - 'A' + 10)*pow(16, length - i - 1);
}
else if (s.at(i) == '-') {
temp = -temp;
}
else if (s.at(i) == '+') {
continue;
}
else {
temp += (s.at(i) - '0')*pow(16, length - i - 1);
}
}
return temp;
}
int main() {
string A, B;
while (cin >> A >> B) {
a = htod(A);
b = htod(B);
if (a + b >= 0)
printf("%llX\n", a + b);
else
printf("-%llX\n",-(a + b));
}
system("pause");
return 0;
}