两人游戏，其中玩家可以删除所有出现的数字

两名球员，PLAYER1和player2，都打在一个给定的数字序列S一个游戏，PLAYER1开始第一和他们都发挥最佳状态。任务是找出玩家 1 是赢还是输。如果他赢了，打印“是”，否则打印“否”。
游戏规则如下：

• 玩家的交替回合。
• 在每一回合中，当前玩家必须选择当前序列 S 中的一个或多个元素，使得所有所选元素的值相同，并将这些元素从 S 中删除。
• 当一个玩家不能选择任何东西时(序列 S 已经是空的)，这个玩家就输了。

例子：

Input: S = {2, 1, 2, 2} 
Output: Yes 
Explanation: 
The first player can choose the subset of elements with indices {0, 3} {0, 2} or {2, 3} (in the sequence) to ensure his victory in the future.

Input: S = {3, 2, 2, 3, 3, 5} 
Output:No 
Explanation: 
The first player can not win in this sequence. 
If player1 chooses 2 and removes all the 2 from the sequence, then player2 will choose two 3 and remove only two 3 as both players play optimally. 
Then player1 removes either 3 or 5 and player2 will remove the final element. So player1 will always lose. 

处理方法：上述问题的处理方法可以假设为：如果序列中的元素总数为偶数，并且出现多次的元素数也为偶数，则第一个玩家不能获胜。如果出现多次的元素个数为奇数且重复元素的出现次数大于等于4且为2的倍数，则第一个玩家不能获胜。否则， player1可能是赢家。

下面是上述方法的实现：

C++

// C++ implementation for Two player
// game in which a player can remove
// all occurrences of a number
 
#include 
using namespace std;
 
// Function that print whether
// player1 can wins or loses
void game(int v[], int n)
{
    unordered_map m;
 
    // storing the number of occurrence
    // of elements in unordered map
    for (int i = 0; i < n; i++) {
 
        if (m.find(v[i]) == m.end())
            m[v[i]] = 1;
 
        else
            m[v[i]]++;
    }
 
    int count = 0;
 
    // variable to check if the
// occurrence of repeated
// elements is >= 4 and
    // multiple of 2 or not
    int check = 0;
 
    // count elements which
// occur more than once
    for (auto i: m) {
        if (i.second > 1) {
 
            if (i.second >= 4
&& i.second % 2 == 0)
                check++;
 
            count++;
        }
    }
 
    if (check % 2 != 0)
        bool flag = false;
 
    if (check % 2 != 0)
        cout << "Yes" << endl;
 
    else if (n % 2 == 0
&& count % 2 == 0)
        cout << "No" << endl;
 
    else
        cout << "Yes" << endl;
}
 
// Driver code
int main()
{
 
    int arr[] = { 3, 2, 2, 3, 3, 5 };
 
    int size = sizeof(arr)
/ sizeof(arr[0]);
 
    game(arr, size);
 
    return 0;
}

Java

// Java implementation for Two player
// game in which a player can remove
// all occurrences of a number
import java.util.*;
class GFG{
     
// Function that print whether
// player1 can wins or loses
public static void game(int[] v,
                        int n)
{
  HashMap m = new HashMap<>();
 
  // Storing the number of occurrence
  // of elements in unordered map
  for (int i = 0; i < n; i++)
  {
    if (!m.containsKey(v[i]))
      m.put(v[i], 1);
    else
      m.replace(v[i], m.get(v[i]) + 1);
  }
 
  int count = 0;
 
  // variable to check if the
  // occurrence of repeated
  // elements is >= 4 and
  // multiple of 2 or not
  int check = 0;
 
  // count elements which
  // occur more than once
  for (Map.Entry i : m.entrySet())
  {
    if(i.getValue() > 1)
    {
      if(i.getValue() >= 4 &&
         i.getValue() % 2 == 0)
        check++;
 
      count++;
    }
  }
   
  boolean flag;
   
  if (check % 2 != 0)
    flag = false;
  if (check % 2 != 0)
    System.out.println("Yes");
  else if (n % 2 == 0  &&
           count % 2 == 0)
    System.out.println("No");
  else
    System.out.println("Yes");
}
 
public static void main(String[] args)
{
  int[] arr = {3, 2, 2, 3, 3, 5};
  int size = arr.length;
  game(arr, size);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation for two player
# game in which a player can remove
# all occurrences of a number
from collections import defaultdict
 
# Function that print whether
# player1 can wins or loses
def game(v, n):
 
    m = defaultdict(int)
     
    # Storing the number of occurrence
    # of elements in unordered map
    for i in range(n):
        if (v[i] not in m):
            m[v[i]] = 1
             
        else:
            m[v[i]] += 1
             
    count = 0
 
    # Variable to check if the
    # occurrence of repeated
    # elements is >= 4 and
    # multiple of 2 or not
    check = 0
 
    # Count elements which
    # occur more than once
    for i in m.values():
        if (i > 1):
            if (i >= 4 and i % 2 == 0):
                check += 1
            count += 1
 
    if (check % 2 != 0):
        flag = False
 
    if (check % 2 != 0):
        print("Yes")
 
    elif (n % 2 == 0 and count % 2 == 0):
        print("No")
         
    else:
        print("Yes")
 
# Driver code
if __name__ == "__main__":
     
    arr = [ 3, 2, 2, 3, 3, 5 ]
    size = len(arr)
     
    game(arr, size)
 
# This code is contributed by chitranayal

C#

// C# implementation for Two player
// game in which a player can remove
// all occurrences of a number
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function that print whether
// player1 can wins or loses
public static void game(int[] v,
                        int n)
{
    Dictionary m = new Dictionary();
     
    // Storing the number of occurrence
    // of elements in unordered map
    for(int i = 0; i < n; i++)
    {
        if (!m.ContainsKey(v[i]))
            m.Add(v[i], 1);
        else
            m[v[i]]= m[v[i]] + 1;
    }
     
    int count = 0;
     
    // Variable to check if the
    // occurrence of repeated
    // elements is >= 4 and
    // multiple of 2 or not
    int check = 0;
     
    // Count elements which
    // occur more than once
    foreach(KeyValuePair i in m)
    {
        if (i.Value > 1)
        {
            if (i.Value >= 4 &&
                i.Value % 2 == 0)
                check++;
             
            count++;
        }
    }
     
    bool flag;
     
    if (check % 2 != 0)
        flag = false;
    if (check % 2 != 0)
        Console.WriteLine("Yes");
    else if (n % 2 == 0 &&
         count % 2 == 0)
        Console.WriteLine("No");
    else
        Console.WriteLine("Yes");
}
 
// Driver code
public static void Main(String[] args)
{
    int[] arr = { 3, 2, 2, 3, 3, 5 };
    int size = arr.Length;
     
    game(arr, size);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

// Java implementation for Two player
// game in which a player can remove
// all occurrences of a number
 
// Function that print whether
// player1 can wins or loses
function game(v,n)
{
  let m = new Map();
  
  // Storing the number of occurrence
  // of elements in unordered map
  for (let i = 0; i < n; i++)
  {
    if (!m.has(v[i]))
      m.set(v[i], 1);
    else
      m.set(v[i], m.get(v[i]) + 1);
  }
  
  let count = 0;
  
  // variable to check if the
  // occurrence of repeated
  // elements is >= 4 and
  // multiple of 2 or not
  let check = 0;
  
  // count elements which
  // occur more than once
  for (let [key, value] of m.entries())
  {
    if(value> 1)
    {
      if(value >= 4 &&
         value % 2 == 0)
      check++;
  
      count++;
    }
  }
    
  let flag;
    
  if (check % 2 != 0)
    flag = false;
  if (check % 2 != 0)
    document.write("Yes<br>");
  else if (n % 2 == 0  &&
           count % 2 == 0)
    document.write("No<br>");
  else
    document.write("Yes<br>");
}
 
let arr=[3, 2, 2, 3, 3, 5];
let size = arr.length;
game(arr, size);
 
 
// This code is contributed by avanitrachhadiya2155

输出：

No

时间复杂度：上述方法的复杂度为O(N)。

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