杭电 OJ 1720 A+B Coming

A+B Coming

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9659    Accepted Submission(s): 6309

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

///////1

#include<stdio.h>

int main()

{

    int a,b;

    while(~scanf("%x %x",&a,&b))

    {

        printf("%d\n",a+b);

    }   

     

    return 0;

}

///////2

#include <iostream>

using namespace std;

int main()

{

    int a,b;

    while( cin>>hex>>a>>b)

    {

        cout << a+b<<endl;

    }

    return 0;

}

////拓展

cin>>oct>>i; //输入为八进制数
 4 cin>>hex>>j; //输入为十六进制数
 5 cin>>k; //输入仍为十六进制数
 6 cin>>dec>>l; //输入为十进制数
 7 cout<<”hex:”<<”i=”<<hex<<i<<endl;
 8 cout<<”dec:”<<”j=”<<dec<<j<<′\t′<<”k=”<<k<<endl;
 9 cout<<”oct:”<<”l=”<<oct<<l;
10 cout<<dec<<endl; //恢复十进制输出状态

//本题就是一个十六进制数的加法，上面附带了一些可能用到的输入。