hdu2057

来源：互联网发布：linux搭建web邮件系统编辑：程序博客网时间：2024/06/08 00:21

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 121A -9-1A -121A -AA

Sample Output

02C11-2C-90

/*注意输出的例如-2C大写字母c,如果用%I64x输出的则是小写c*/

#include <iostream>#include <cstdio>using namespace std;int main(){    __int64 a, b, sum;  //__int64最大能表示16位16进制    while(scanf("%I64X%I64X", &a, &b) != EOF){        sum = a + b;        if(sum < 0)            printf("-%I64X\n", -sum);    //16进制表示负数的时候是补码所以要注意        else            printf("%I64X\n", sum);    }    return 0;}

0 0