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=,t;
     for(int i=;i<s1.size();i++)
     {
         if(s1[i]>=''&&s1[i]<='')
             t=s1[i]-'';
         else if(s1[i]>='A'&&s1[i]<='Z')
             t=s1[i]-'A'+;
         else if(s1[i]>='a'&&s1[i]<='z')
             t=s1[i]-'a'+;
         num=num*+t;
     }
     return num;
 }
 int main()
 {
     int sum;
     string a,b;
     while(cin>>a>>b)
     {
         sum=change(a)+change(b);
         cout<<sum<<endl;
     }
     return ;
 }

16进制转10进制，水题。