A+B Coming 解题报告

http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=1563

A+B Coming

Time Limit: 1 Sec   Memory Limit: 125 MB
Submissions: 32   Solved: 15
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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

代码：

#include <iostream>
using namespace std;

int a,b;

int main(){
    while(cin>>hex>>a>>b)
        cout<<dec<<a+b<<endl;
    return 0;
}
/**************************************************************
    Problem: 1563
    User: mmoaay
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1352 kb
****************************************************************/

刚开始的时候自己写的16进制转10进制，结果无耻地错了，然后更无耻的是：原来如此简单……

有必要总结一下c++输入输出数的进制问题： 

dec-十进制（默认）

oct-八进制

hex-十六进制

在cin和cout中指明相应的数据形式。

例：

int i,j,k,l;
cout<<"Input i(oct) j(hex) k(hex) l(dec):"<<endl;
cin>>oct>>i;                                            //输入为八进制数
cin>>hex>>j;                                            //输入为十六进制数
cin>>k;                                                 //输入仍为十六进制数
cin>>dec>>l;                                            //输入为十进制数
cout<<"hex:"<<"i="<<hex<<i<<endl;
cout<<"dec:"<<"j="<<dec<<j<<'/t'<<"k="<<k<<endl;
cout<<"oct:"<<"l="<<oct<<l;
cout<<dec<<endl;                                        //恢复十进制输出状态