Problem:
Score : 200 points

Problem Statement

In the beginning, Takahashi has A cookies, and Aoki has B cookies.
They will perform the following operation alternately, starting from Takahashi:

If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person.

Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.

Constraints

1 \leq A,B \leq 10^9

1 \leq K \leq 100

A,B and K are integers.

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the number of cookies Takahashi has, and the number of cookies Aoki has, in this order, after performing K operations in total.

Sample Input 1

5 4 2

Sample Output 1

5 3

The process will go as follows:

In the beginning, Takahashi and Aoki have 5 and 4 cookies, respectively.

Takahashi eats one cookie and gives two cookies to Aoki. They now have 2 and 6 cookies, respectively.

Aoki gives three cookies to Takahashi. They now have 5 and 3 cookies, respectively.

Sample Input 2

3 3 3

Sample Output 2

1 3

Sample Input 3

314159265 358979323 84

Sample Output 3

448759046 224379523

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <utility>
#include <climits>
#include <bitset>

using namespace std;
#define li long long int
#define FOR(i, a, b) for (int i = a ; i < b ; ++i)
#define REP(i, n) for (int i = 0 ; i < n ; ++i)

li gcd(li a,li b) {
    return a % b == 0 ? b : gcd(b, a % b);
}

int main(void) {
    li A, B, K;
    cin >> A >> B >> K;

    FOR(i, 0, K) {
        if (i % 2 == 0) {
            // 偶数回目なので高橋くん
            if (A % 2 == 1) {
                A = A - 1;
            }
            A = A / 2;
            B = B + A;
        } else {
            // 奇数回目なので青木くん
            if (B % 2 == 1) {
                B = B - 1;
            }            
            B = B / 2;
            A = A + B;

        }
    }

    cout << A << " " << B;
     
    return 0;
}