HDU 1720 A+B Coming

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,仅仅要把算得的结果用十进制输出就可以。

只是还是非常恶心的wa了几次。。

。

AC代码：

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
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)<<endl;
	}
	return 0;
}