杭电 HDU ACM 1720 A+B Coming



A+B Coming

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6464    Accepted Submission(s): 4248

Problem Description

Many classmates said to me that A+B is must needs.
If you can’t AC this problem, you would invite me for night meal. ^_^

 

Input

Input may contain multiple test cases. Each case contains A and B in one line.
A, B are hexadecimal number.
Input terminates by EOF.

 

Output

Output A+B in decimal number in one line.

 

Sample Input

1 9 A B a b

 

Sample Output

10 21 21

 

这个题无语……表示以前见过 但是忘了！！

知识一：

将十六进制转换为十进特例：

AB转换为dec；

0xAB
= 0x0A * 16 + 0x0B
= 10 * 16 + 11
= 171

c 输入十六进制和八进制方法：

scanf（“%x %x”，&a，&b）；

scanf（“%o %o”，&a，&b）；

c++ 方法：

cin>>hex>>a>>b;

cin>>oct>>a>>b;

输出与之类似 ，

败了 看来要请吃饭了！

AC：

#include<iostream>
using namespace std;
int main()
{
	int a,b;
	while(cin>>hex>>a>>b)
	  cout<<dec<<a+b<<endl;
return 0;
}