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 <iomanip>
#include <math.h>
#include <complex>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <functional>
#include <assert.h>
using ll=long long;
#define int ll
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define SZ(x) ((int)x.size())
#define fi first
#define se second
const int INF = 1e15;
using namespace std;

template<class T> using vc=vector<T>;
template<class T> using vvc=vc<vc<T>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class T>
inline bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<class T>
inline bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout<<fixed<<setprecision(20);

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