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
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#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)<