Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 39021 Accepted Submission(s): 15794
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
过年玩的有点嗨......emmmm....... 一开始打算用字符串储存之后一个个分析结果老是WA。。。。。。逛了一下讨论区看到了dalao们写的ac码 如下
#include <stdio.h>
int main(void)
{
long long i,a,b,sum;
while(scanf("%I64X %I64X",&a,&b) != EOF)
{
sum = a+b;
if(sum >= 0)
{
printf("%I64X\n",sum);
}
else
{
sum = -sum;
printf("-%I64X\n",sum);
}
}
}
发现了%x这个符号，在论坛上找出一串相关符号
（以下转自网站）
转换说明符
%a(%A) 浮点数、十六进制数字和p-(P-)记数法(C99)
%c 字符
%d 有符号十进制整数
%f 浮点数(包括float和doulbe)
%e(%E) 浮点数指数输出[e-(E-)记数法]
%g(%G) 浮点数不显无意义的零"0"
%i 有符号十进制整数(与%d相同)
%u 无符号十进制整数
%o 八进制整数 e.g. 0123
%x(%X) 十六进制整数0f(0F) e.g. 0x1234
%p 指针
%s 字符串
%% "%"
2`标志
左对齐："-" e.g. "%-20s"
右对齐："+" e.g. "%+20s"
空格：若符号为正，则显示空格，负则显示"-" e.g. "% 6.2f"
#：对c,s,d,u类无影响；对o类，在输出时加前缀o；对x类，在输出时加前缀0x；
对e,g,f 类当结果有小数时才给出小数点。
3．格式字符串（格式）
〔标志〕〔输出最少宽度〕〔．精度〕〔长度〕类型
"％-md" ：左对齐，若m比实际少时，按实际输出。
"%m.ns"：输出m位，取字符串(左起)n位，左补空格，当n>m or m省略时m=n
e.g. "%7.2s" 输入CHINA
输出" CH"
"%m.nf"：输出浮点数，m为宽度，n为小数点右边数位
e.g. "%3.1f" 输入3852.99
输出3853.0