通过选择数字
根据和的绝对差预测比赛的胜负
给定一组 N 个数字。两个玩家 X 和 Y 玩一个游戏，每走一步都有一个玩家选择一个号码。一个号码只能选择一次。在所有号码都被选中后，如果 X 和 Y 收集的号码总和之间的绝对差被 4 整除，则玩家 X 获胜，否则 Y 获胜。 注意:玩家 X 开始游戏，每一步都是对数字进行优化选择。 例:
输入: a[] = {4，8，12，16} 输出: X X 选择 4 Y 选择 12 X 选择 8 Y 选择 16 |(4+8)–(12+16)| = | 12–28 | = 16 可被 4 整除。 于是，X 胜 输入: a[] = {7，9，1} 输出: Y
接近:可以按照以下步骤解决问题:
- 初始化计数 0 、计数 1 、计数 2 和计数 3 至 0 。
- 如果 a[i] % 4 == 0 ， a[i] % 4 == 1 ， a[i] % 4 == 2 或 a[i] % 4 == 3 ，则迭代数组中的每个数字并相应增加上述计数器。
- 如果 count0 、 count1 、 count2 和 count3 均为偶数，则 X 获胜，否则 Y 将获胜。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to decide the winner
function decideWinner($a, $n)
{
$count0 = 0;
$count1 = 0;
$count2 = 0;
$count3 = 0;
// Iterate for all numbers in the array
for ($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
$a = array( 4, 8, 5, 9 );
$n = count($a);
if (decideWinner($a, $n) == 1)
echo "X wins";
else
echo "Y wins";
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// 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
</script>
Output:
X wins
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