A + B Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16794 Accepted Submission(s): 7229
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
题目大意:给你两个带符号的16进制数A和B,输出A+B
思路:A和B比较大,用64位整数来存储,输入的时候用X%来控制输入、输出,
但是X%输入、输出的是无符号16进制数,应该再加个判断。
附带:
格式字符
格式字符意义
d 以十进制形式输出带符号整数(正数不输出符号)
o 以八进制形式输出无符号整数(不输出前缀O)
x 以十六进制形式输出无符号整数(不输出前缀OX)
u 以十进制形式输出无符号整数
f 以小数形式输出单、双精度实数
e 以指数形式输出单、双精度实数
g 以%f%e中较短的输出宽度输出单、双精度实数
c 输出单个字符
s 输出字符串
#include<iostream> #include<algorithm> #include<cstdio> using namespace std; int main() { __int64 a,b; while(~scanf("%I64X%I64X",&a,&b)) { if(a + b >= 0) printf("%I64X\n",a+b); else printf("-%I64X\n",-(a+b)); } return 0; }