HDU - 2057 A + B Again
来源:互联网 发布:刺客信条叛变优化设置 编辑:程序博客网 时间:2024/06/01 08:42
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 !
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.
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
#include <cstdio>#include <cmath>#include <cstdlib>int main(){ long long int x, y; while(~scanf("%llx %llx", &x, &y)){ if(x+y < 0) printf("-%llX\n", -(x+y)); else printf("%llX\n", x+y); } return 0;}
0 0
- hdu 2057 A+B Again
- hdu 2057 A + B again
- HDU-2057A + B Again
- hdu 2057 A + B Again
- HDU 2057 A + B Again
- hdu-2057-A + B Again
- HDU 2057 A + B Again
- HDU 2057 A + B Again
- hdu 2057 A + B Again
- HDU 2057 A + B Again
- HDU - 2057 A + B Again
- hdu 2057 a + b again
- HDU-2057 A+B Again
- hdu 2057 A + B Again
- HDU 2057 A + B Again
- hdu 2057 A + B Again
- HDU 2057 (A + B Again)
- hdu 2057 A + B Again
- 条件注释判断浏览器<!--[if !IE]><!--[if IE]><!--[if lt IE 6]><!--[if gte IE 6]>
- 常用sublime text3插件
- JAVA获取客户端的MAC地址
- 2015年学习及工作计划(一)
- H.264视频编解码技术总结
- HDU - 2057 A + B Again
- Think In Java Chapter18 IO系统 练习12 读取文件到list,添加行号并写入另一文件
- linux命令
- 第一章 工欲善其事 必先利其器—Android SDK工具(1)
- 作为一个程序员怎么通过android开发赚钱
- SQLserver重新启动
- 黑苹果安装及相关注意事项之二,基础知识
- 首次进入页面刷新1次|JSP获取上一个访问页面URL的方法
- Makefile选项CFLAGS,LDFLAGS,LIBS