A+B Coming

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5856 Accepted Submission(s): 3839

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

Author 威士忌
Source HZIEE 2007 Programming Contest

解题思路：水题一枚，就是十六进制的A+B,只要把算得的结果用十进制输出即可。不过还是很恶心的wa了几次。。。

AC代码：

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       using namespace std; int main(){ // freopen("in.txt", "r",stdin); string a, b; int x, y; while(cin>>a>>b){ int len = a.size(); int cnt1 = 0; int k = 0; while(len){ char x = a[len-1]; if(x>='0' && x<='9') cnt1 += (x - '0')*pow(16, k); else if(x>='a' && x<='f') cnt1 += (x - 'a' + 10)*pow(16, k); else cnt1 += (x - 'A' + 10)*pow(16, k); len --; k ++; } len = b.size(); int cnt2 = 0; k = 0; while(len){ char x = b[len-1]; if(x>='0' && x<='9') cnt2 += (x - '0')*pow(16, k); else if(x>='a' && x<='f') cnt2 += (x - 'a' + 10)*pow(16, k); else cnt2 += (x - 'A' + 10)*pow(16, k); len --; k ++; } cout<<(cnt1 + cnt2)<

HDU 1720 A+B Coming

A+B Coming