有两个玩家 A 和 B 对玩数字游戏感兴趣。在每次移动中，玩家选择两个不同的数字，假设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数字，则他们停止玩游戏，并且无法在数组中选择两个不同数字的玩家输掉游戏。第一个玩家总是先移动，然后再移动。任务是找出哪个玩家获胜。
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从上面的例子中，我们可以观察到，如果不同元素的数量是偶数，那么第一个玩家总是获胜。否则第二个玩家获胜。
让我们再举一个例子：
int arr[] = 1, 2, 3, 4, 5, 6 这里不同元素的数量是偶数（n）。如果玩家 1 选择任意两个数字，比如说 (4, 1)，那么我们留下了 n-1 个不同的元素。所以玩家第二个离开了 n-1 个不同的元素。这个过程一直持续到不同元素变为 1。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 Shrikant13C#
// 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 Javascript
输出：
Player 1 Wins时间复杂度： O(n)
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