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<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>
#include<sstream>
using namespace std;
typedef  long long int lint;
typedef pair<int, int> P;
typedef pair<lint, lint> LLP;
typedef pair<char, char>CP;
#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n; i >= 0; i--)
#define sort(v) sort((v).begin(), (v).end())
#define reverse(v) reverse((v).begin(), (v).end())
#define upper(v,hoge) upper_bound(v.begin(),v.end(),hoge)
#define lower(v,hoge) lower_bound(v.begin(),v.end(),hoge)
#define llower(v,hoge) *lower_bound(v.begin(), v.end(), hoge)
#define lupper(v,hoge) *upper_bound(v.begin(), v.end(), hoge)

const lint MOD = pow(10, 9) + 7;

lint gcd(lint a, lint b) {
	//aとbの最大公約数を返す
	if (a < b)swap(a, b);
	if (a % b == 0)return b;
	return gcd(a % b, b);
}

lint lcm(lint a, lint b) {
	//aとbの最小公倍数を返す
	return a / gcd(a, b) * b;
}

lint sgm(lint n) {
	//0からの総和を返す
	return n * (n + 1) / 2;
}

lint fact(lint n) {
	//n!を返す
	if (n == 0) return 1;
	else return n * fact(n - 1);
}

lint power(lint x, lint n, lint m) {
	//(x^n)%mを返す
	lint res = 1;
	x %= m;
	while (n > 0) {
		if (n % 2 == 1) {
			res = res * x % m;
		}
		x = x * x % m;
		n /= 2;
	}
	return res;
}

/*
vector<char>al(26);
al = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
vector<char>AL(26);
AL = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
vector<char>NUM(10);
	NUM = { '0','1','2' ,'3' ,'4' ,'5' ,'6' ,'7' ,'8' ,'9' };
*/


int main() {
	lint A, B, K;
	cin >> A >> B >> K;
	rep(i, K) {
		if (i % 2 == 0) {
			if (A % 2 != 0) {
				A--;
			}
			A -= (A / 2);
			B += A;
		}
		else if (i % 2 != 0) {
			if (B % 2 != 0) {
				B--;
			}
			B -= (B / 2);
			A += B;
		}
		//cout << A << ' ' << B << endl;
	}
	cout << A << ' ' << B << endl;
}