首页 > 代码库 > HDU 1720 A+B Coming
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. ^_^
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.
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 <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;
}HDU 1720 A+B Coming
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。