Dada una array de N números. Dos jugadores X e Y juegan un juego donde en cada paso un jugador selecciona un número. Un número se puede seleccionar sólo una vez. Una vez seleccionados todos los números, el jugador X gana si la diferencia absoluta entre la suma de los números recogidos por X e Y es divisible por 4 , de lo contrario, gana Y.
Nota: el jugador X comienza el juego y los números se seleccionan de manera óptima en cada paso.
Ejemplos:
Entrada: a[] = {4, 8, 12, 16}
Salida: X
X elige 4
Y elige 12
X elige 8
Y elige 16
|(4 + 8) – (12 + 16)| = |12 – 28| = 16 que es divisible por 4.
Por lo tanto, X gana
Entrada: a[] = {7, 9, 1}
Salida: Y
Enfoque : Se pueden seguir los siguientes pasos para resolver el problema:
- Inicialice count0 , count1 , count2 y count3 a 0 .
- Iterar para cada número en la array y aumentar los contadores anteriores en consecuencia si a[i] % 4 == 0 , a[i] % 4 == 1 , a[i] % 4 == 2 o a[i] % 4 == 3 .
- Si count0 , count1 , count2 y count3 son todos números pares entonces X gana, de lo contrario Y ganará.
A continuación se muestra la implementación del enfoque anterior:
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
// 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
PHP
<?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
?>
Javascript
<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>
Producción:
X wins
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)