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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int INF = 1e9;
const ll LINF = 1e18;
const int mod = 1e9+7;

#define pi pair<int,int>
#define pl pair<lld,lld>

#define dump(x) cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define SANKOU(n,a,b) cout << ((n) ? (#a) : (#b) ) << endl

#define mem0(x) memset(x,0,sizeof(x))
#define fillnum(x,n) fill(begin(x),end(x),n)
#define asort(x) sort(x.begin(),x.end())
#define dsort(x,t) sort(x.begin(),x.end(),greater<t>())
#define vuniq(x) x.erase(unique(x.begin(), x.end()), x.end())

#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)

int main() {
    ll A,B,K;
    cin >> A >> B >> K;
    ll cnt = 0;
    for (int i = 0; i < K; i++) {
        if(i % 2 == 0){
            if(A % 2 == 1){
                A--;
            }
            A /= 2;
            B += A;
        } else {
            if(B % 2 == 1){
                B--;
            }
            B /= 2;
            A += B;
        }
    }
    cout << A << " " << B << endl;
}