2057 A + B Again

A + B Again Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15391 Accepted Submission(s): 6713 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 思路：当时写的时候，想的是把16进制转化为10进制，相加后再转化为16进制，感觉比较麻烦 就查了网上的博客，才发现16进制也可以加减，就是不能输出正负好，就设置了一个字符c来 控制正负，注意printf("%x\n",sum);和printf（“%X\n",sum);在这道题上是有区别的；刚开始就把 “X"写成"x",一直提交说答案错误！！！正确代码如下 #include<stdio.h> int main() { __int64 a,b,sum; char c; while(scanf("%I64X %I64X",&a,&b)!=EOF) { sum=a+b; if(sum<0) { sum=-sum; c='-'; } else c=0; if(c) putchar(c); printf("%I64X\n",sum); } return 0; }