寻找移除奇数或替换偶数数组元素游戏的赢家
给定一个由 N 个整数组成的数组 arr[] 。两个玩家玩家 1 和玩家 2 轮流玩，其中一个玩家可以选择以下两个动作之一:
- 将偶数数组元素转换为任何其他整数。
- 移除奇数数组元素。
不能移动的玩家输掉游戏。任务是打印游戏的获胜者。打印 -1 如果游戏可能永远继续。
示例:
输入: arr[] = {3，1，9，7} 输出:玩家 2 说明:由于所有阵元都是奇数，所以不可能转换。 回合 1: 玩家 1 删除 3。 回合 2: 玩家 2 删除 1。 转 3: 玩家 1 删除 9。 回合 4: 玩家 2 删除 7。现在，玩家 1 已经没有招式了。因此玩家 2 赢得游戏。
输入: arr[]={4，8} 输出: -1
方法:按照以下步骤解决问题:
- 遍历数组。
- 统计数组中出现的偶数和奇数元素的数量。
- 如果偶数元素的数量为零，执行以下操作:
- 如果奇数计数为偶数，则玩家 2 将赢得游戏。
- 否则玩家 1 将赢得游戏。
- 如果奇数计数为奇数，并且数组中只有一个偶数元素，则玩家 1 将赢得游戏。
- 否则每次都会有平局。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to evaluate the
// winner of the game
void findWinner(int arr[], int N)
{
// Stores count of odd
// array elements
int odd = 0;
// Stores count of even
// array elements
int even = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// If array element is odd
if (arr[i] % 2 == 1) {
odd++;
}
// Otherwise
else {
even++;
}
}
// If count of even is zero
if (even == 0) {
// If count of odd is even
if (odd % 2 == 0) {
cout << "Player 2" << endl;
}
// If count of odd is odd
else if (odd % 2 == 1) {
cout << "Player 1" << endl;
}
}
// If count of odd is odd and
// count of even is one
else if (even == 1 && odd % 2 == 1) {
cout << "Player 1" << endl;
}
// Otherwise
else {
cout << -1 << endl;
}
}
// Driver Code
int main()
{
int arr[] = { 3, 1, 9, 7 };
int N = sizeof(arr)
/ sizeof(arr[0]);
findWinner(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to evaluate the
// winner of the game
static void findWinner(int arr[], int N)
{
// Stores count of odd
// array elements
int odd = 0;
// Stores count of even
// array elements
int even = 0;
// Traverse the array
for(int i = 0; i < N; i++)
{
// If array element is odd
if (arr[i] % 2 == 1)
{
odd++;
}
// Otherwise
else
{
even++;
}
}
// If count of even is zero
if (even == 0)
{
// If count of odd is even
if (odd % 2 == 0)
{
System.out.println("Player 2");
}
// If count of odd is odd
else if (odd % 2 == 1)
{
System.out.println("Player 1");
}
}
// If count of odd is odd and
// count of even is one
else if (even == 1 && odd % 2 == 1)
{
System.out.println("Player 1");
}
// Otherwise
else
{
System.out.println(-1);
}
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 3, 1, 9, 7 };
int N = arr.length;
findWinner(arr, N);
}
}
// This code is contributed by ipg2016107
Python 3
# Python3 program for the above approach
# Function to evaluate the
# winner of the game
def findWinner(arr, N):
# Stores count of odd
# array elements
odd = 0
# Stores count of even
# array elements
even = 0
# Traverse the array
for i in range(N):
# If array element is odd
if (arr[i] % 2 == 1):
odd += 1
# Otherwise
else:
even += 1
# If count of even is zero
if (even == 0):
# If count of odd is even
if (odd % 2 == 0):
print("Player 2")
# If count of odd is odd
elif (odd % 2 == 1):
print("Player 1")
# If count of odd is odd and
# count of even is one
elif (even == 1 and odd % 2 == 1):
print("Player 1")
# Otherwise
else:
print(-1)
# Driver code
if __name__ == '__main__':
arr = [ 3, 1, 9, 7 ]
N = len(arr)
findWinner(arr, N)
# This code is contributed by Shivam Singh
C
// C# program for the above approach
using System;
class GFG{
// Function to evaluate the
// winner of the game
static void findWinner(int []arr, int N)
{
// Stores count of odd
// array elements
int odd = 0;
// Stores count of even
// array elements
int even = 0;
// Traverse the array
for(int i = 0; i < N; i++)
{
// If array element is odd
if (arr[i] % 2 == 1)
{
odd++;
}
// Otherwise
else
{
even++;
}
}
// If count of even is zero
if (even == 0)
{
// If count of odd is even
if (odd % 2 == 0)
{
Console.WriteLine("Player 2");
}
// If count of odd is odd
else if (odd % 2 == 1)
{
Console.WriteLine("Player 1");
}
}
// If count of odd is odd and
// count of even is one
else if (even == 1 && odd % 2 == 1)
{
Console.WriteLine("Player 1");
}
// Otherwise
else
{
Console.WriteLine(-1);
}
}
// Driver Code
public static void Main()
{
int []arr = { 3, 1, 9, 7 };
int N = arr.Length;
findWinner(arr, N);
}
}
// This code is contributed by bgangwar59
java 描述语言
<script>
// JavaScript program to implement
// the above approach
// Function to evaluate the
// winner of the game
function findWinner(arr, N)
{
// Stores count of odd
// array elements
let odd = 0;
// Stores count of even
// array elements
let even = 0;
// Traverse the array
for(let i = 0; i < N; i++)
{
// If array element is odd
if (arr[i] % 2 == 1)
{
odd++;
}
// Otherwise
else
{
even++;
}
}
// If count of even is zero
if (even == 0)
{
// If count of odd is even
if (odd % 2 == 0)
{
document.write("Player 2");
}
// If count of odd is odd
else if (odd % 2 == 1)
{
document.write("Player 1");
}
}
// If count of odd is odd and
// count of even is one
else if (even == 1 && odd % 2 == 1)
{
document.write("Player 1");
}
// Otherwise
else
{
document.write(-1);
}
}
// Driver Code
let arr = [ 3, 1, 9, 7 ];
let N = arr.length;
findWinner(arr, N);
</script>
Output:
Player 2
时间复杂度:O(N) T5辅助空间** : O(1)
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