16进制转10进制 HDU-1720

A+B Coming

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9461    Accepted Submission(s): 6173

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

 题意：一个最简单的进制转换问题，只要将两个十六进制的数转化为十进制数相加就好，需注意的是大小写均表示十六进制数。

  思路不用介绍太多，直接上代码说明问题。

#include<bits/stdc++.h>
using namespace std;
int change(string s1)
{
    int num=0,t;
    for(int i=0;i<s1.size();i++)
    {
        if(s1[i]>='0'&&s1[i]<='9')
            t=s1[i]-'0';
        else if(s1[i]>='A'&&s1[i]<='Z')
            t=s1[i]-'A'+10;
        else if(s1[i]>='a'&&s1[i]<='z')
            t=s1[i]-'a'+10;
        num=num*16+t;
    }
    return num;
}
int main()
{
    int sum;
    string a,b;
    while(cin>>a>>b)
    {
        sum=change(a)+change(b);
        cout<<sum<<endl;
    }
    return 0;
}

16进制转10进制，水题。