替换数组元素的游戏

有两个玩家A和B对玩数字游戏感兴趣。玩家在每一步中都选择两个不同的数字，例如a1和a2 ，然后将所有a2替换为a1或将a1替换为a2 。如果他们中的任何一个无法选择两个号码，并且他们无法选择一个数组中的两个不同的号码，则他们停止玩游戏。第一个玩家总是先移动，然后再移动。任务是找到哪个玩家获胜。

例子：

Input :  arr[] = { 1, 3, 3, 2,, 2, 1 }
Output : Player 2 wins
Explanation:
First plays always looses irrespective
of the numbers chosen by him. For example,
say first player picks ( 1 & 3) 
replace all 3 by 1  
Now array Become { 1, 1, 1, 2, 2, 1 }
Then second player picks ( 1  2 )
either he replace 1 by 2 or 2 by 1 
Array Become { 1, 1, 1, 1, 1, 1 }
Now first player is not able to choose.

Input  : arr[] = { 1, 2, 1, 2 }
Output : Player 1 wins

从上面的示例中，我们可以观察到，如果不同元素的数量为偶数，则第一个玩家总是获胜。其他第二名获胜。

让我们再举一个例子：

int arr[] =  1, 2, 3, 4, 5, 6 

此处，不同元素的数量为偶数(n)。如果玩家1选择任意两个数字，说(4，1)，那么我们剩下n-1个不同的元素。因此，玩家以n-1个独特元素排在第二位。这个过程一直进行到不同元素变为1为止。这里n = 6

Player   :  P1    p2    P1   p2    P1     P2    
distinct : [n, n-1, n-2, n-3, n-4, n-5 ]  
 
"At this point no distinct element left, 
so p2 is unable to pick two Dis element."

下面实现上述想法：

C++

// CPP program for Game of Replacement
#include 
using namespace std;
  
// Function return which player win the game
int playGame(int arr[], int n)
{
    // Create hash that will stores
    // all distinct element
    unordered_set hash;
  
    // Traverse an array element
    for (int i = 0; i < n; i++)
        hash.insert(arr[i]);
  
    return (hash.size() % 2 == 0 ? 1 : 2);
}
  
// Driver Function
int main()
{
    int arr[] = { 1, 1, 2, 2, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << "Player " << playGame(arr, n) << " Wins" << endl;
    return 0;
}

Java

// Java program for Game of Replacement 
import java.util.HashSet;
public class GameOfReplacingArrayElements 
{
    // Function return which player win the game 
    public static int playGame(int arr[])
    {
        // Create hash that will stores 
        // all distinct element 
        HashSet set=new HashSet<>();
  
        // Traverse an array element 
        for(int i:arr)
            set.add(i);
        return (set.size()%2==0)?1:2;
    }
  
    public static void main(String args[]) {
        int arr[] = { 1, 1, 2, 2, 2, 2 }; 
        System.out.print("Player "+playGame(arr)+" wins");
    }
}
//This code is contributed by Gaurav Tiwari

Python3

# Python program for Game of Replacement 
  
# Function return which player win the game 
def playGame(arr, n):
  
    # Create hash that will stores 
    # all distinct element 
    s = set()
  
    # Traverse an array element 
    for i in range(n):
        s.add(arr[i])
    return 1 if len(s) % 2 == 0 else 2
  
# Driver code
arr = [1, 1, 2, 2, 2, 2]
n = len(arr)
print("Player",playGame(arr, n),"Wins")
  
# This code is contributed by Shrikant13

C#

// C# program for Game of Replacement 
using System;
using System.Collections.Generic;
  
public class GameOfReplacingArrayElements 
{
    // Function return which player win the game 
    public static int playGame(int []arr)
    {
        // Create hash that will stores 
        // all distinct element 
        HashSet set = new HashSet();
  
        // Traverse an array element 
        foreach(int i in arr)
            set.Add(i);
        return (set.Count % 2 == 0) ? 1 : 2;
    }
  
    // Driver code
    public static void Main(String []args)
    {
        int []arr = { 1, 1, 2, 2, 2, 2 }; 
        Console.Write("Player " + playGame(arr) + " wins");
    }
}
  
// This code has been contributed by 29AjayKumar

输出：

Player 1 Wins

时间复杂度： O(n)