16进制A+B
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. ^_^
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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.
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Output
Output A+B in decimal number in one line.
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Sample Input
1 9 A B a b |
Sample Output
10 21 21 |
注:输入十六进制,需要转化
#include<iostream> using namespace std; int htod(int x) //转化为十进制 { int temp,sum; temp = 1; sum=0; while(x!=0) { sum+=(temp*(x%16)); x/=16; temp*=16; } return sum; } int main() { int a,b; int m,n; int i = 0; while(cin >> hex >> a >>b) { m = htod(a); n = htod(b); cout << m+n<<endl; i++; } return 0; }