通过选择数字，根据和的绝对差来预测游戏的获胜者

给定一个由N 个数字组成的数组。两个玩家X和Y玩一个游戏，其中每一步都有一个玩家选择一个数字。一个号码只能选择一次。选择完所有数字后，如果X和Y收集的数字之和的绝对差能被4整除，则玩家X获胜，否则Y获胜。
注意：玩家 X 开始游戏，并在每一步都优化选择数字。
例子：

Input: a[] = {4, 8, 12, 16} 
Output: X 
X chooses 4 
Y chooses 12 
X chooses 8 
Y chooses 16 
|(4 + 8) – (12 + 16)| = |12 – 28| = 16 which is divisible by 4. 
Hence, X wins
Input: a[] = {7, 9, 1} 
Output: Y 
 

方法：可以按照以下步骤解决问题：

• 将count0 、 count1 、 count2和count3初始化为0 。
• 如果a[i] % 4 == 0 , a[i] % 4 == 1 , a[i] % 4 == 2或a[i] % 4 则迭代数组中的每个数字并相应地增加上述计数器== 3 。
• 如果count0 、 count1 、 count2和count3都是偶数，则X获胜，否则Y获胜。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to decide the winner
int decideWinner(int a[], int n)
{
    int count0 = 0;
    int count1 = 0;
    int count2 = 0;
    int count3 = 0;
 
    // Iterate for all numbers in the array
    for (int i = 0; i < n; i++) {
 
        // Condition to count
 
        // If mod gives 0
        if (a[i] % 4 == 0)
            count0++;
 
        // If mod gives 1
        else if (a[i] % 4 == 1)
            count1++;
 
        // If mod gives 2
        else if (a[i] % 4 == 2)
            count2++;
 
        // If mod gives 3
        else if (a[i] % 4 == 3)
            count3++;
    }
 
    // Check the winning condition for X
    if (count0 % 2 == 0
        && count1 % 2 == 0
        && count2 % 2 == 0
        && count3 == 0)
        return 1;
    else
        return 2;
}
 
// Driver code
int main()
{
 
    int a[] = { 4, 8, 5, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    if (decideWinner(a, n) == 1)
        cout << "X wins";
    else
        cout << "Y wins";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to decide the winner
static int decideWinner(int []a, int n)
{
    int count0 = 0;
    int count1 = 0;
    int count2 = 0;
    int count3 = 0;
 
    // Iterate for all numbers in the array
    for (int i = 0; i < n; i++)
    {
 
        // Condition to count
 
        // If mod gives 0
        if (a[i] % 4 == 0)
            count0++;
 
        // If mod gives 1
        else if (a[i] % 4 == 1)
            count1++;
 
        // If mod gives 2
        else if (a[i] % 4 == 2)
            count2++;
 
        // If mod gives 3
        else if (a[i] % 4 == 3)
            count3++;
    }
 
    // Check the winning condition for X
    if (count0 % 2 == 0 && count1 % 2 == 0 &&
        count2 % 2 == 0 && count3 == 0)
        return 1;
    else
        return 2;
}
 
// Driver code
public static void main(String args[])
{
    int []a = { 4, 8, 5, 9 };
    int n = a.length;
    if (decideWinner(a, n) == 1)
        System.out.print("X wins");
    else
        System.out.print("Y wins");
}
}
 
// This code is contributed by Akanksha Rai

Python3

# Python3 implementation of the approach
 
# Function to decide the winner
def decideWinner(a, n):
    count0 = 0
    count1 = 0
    count2 = 0
    count3 = 0
 
    # Iterate for all numbers in the array
    for i in range(n):
 
        # Condition to count
 
        # If mod gives 0
        if (a[i] % 4 == 0):
            count0 += 1
 
        # If mod gives 1
        elif (a[i] % 4 == 1):
            count1 += 1
 
        # If mod gives 2
        elif (a[i] % 4 == 2):
            count2 += 1
 
        # If mod gives 3
        elif (a[i] % 4 == 3):
            count3 += 1
     
    # Check the winning condition for X
    if (count0 % 2 == 0 and count1 % 2 == 0 and
        count2 % 2 == 0 and count3 == 0):
        return 1
    else:
        return 2
 
# Driver code
a = [4, 8, 5, 9]
n = len(a)
if (decideWinner(a, n) == 1):
    print("X wins")
else:
    print("Y wins")
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
class GFG
{
     
// Function to decide the winner
static int decideWinner(int []a, int n)
{
    int count0 = 0;
    int count1 = 0;
    int count2 = 0;
    int count3 = 0;
 
    // Iterate for all numbers in the array
    for (int i = 0; i < n; i++)
    {
 
        // Condition to count
 
        // If mod gives 0
        if (a[i] % 4 == 0)
            count0++;
 
        // If mod gives 1
        else if (a[i] % 4 == 1)
            count1++;
 
        // If mod gives 2
        else if (a[i] % 4 == 2)
            count2++;
 
        // If mod gives 3
        else if (a[i] % 4 == 3)
            count3++;
    }
 
    // Check the winning condition for X
    if (count0 % 2 == 0 && count1 % 2 == 0 &&
        count2 % 2 == 0 && count3 == 0)
        return 1;
    else
        return 2;
}
 
// Driver code
public static void Main()
{
    int []a = { 4, 8, 5, 9 };
    int n = a.Length;
    if (decideWinner(a, n) == 1)
        Console.Write("X wins");
    else
        Console.Write("Y wins");
}
}
 
// This code is contributed by Akanksha Rai

Javascript

// javascript implementation of the approach   
// Function to decide the winner
    function decideWinner(a , n) {
        var count0 = 0;
        var count1 = 0;
        var count2 = 0;
        var count3 = 0;
 
        // Iterate for all numbers in the array
        for (i = 0; i < n; i++) {
 
            // Condition to count
 
            // If mod gives 0
            if (a[i] % 4 == 0)
                count0++;
 
            // If mod gives 1
            else if (a[i] % 4 == 1)
                count1++;
 
            // If mod gives 2
            else if (a[i] % 4 == 2)
                count2++;
 
            // If mod gives 3
            else if (a[i] % 4 == 3)
                count3++;
        }
 
        // Check the winning condition for X
        if (count0 % 2 == 0 && count1 % 2 == 0 && count2 % 2 == 0 && count3 == 0)
            return 1;
        else
            return 2;
    }
 
    // Driver code
     
        var a = [ 4, 8, 5, 9 ];
        var n = a.length;
        if (decideWinner(a, n) == 1)
            document.write("X wins");
        else
            document.write("Y wins");
 
// This code contributed by Rajput-Ji

输出：

X wins

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