给定一个整数数组arr [] ，两个玩家A和B正在玩一个游戏，其中A可以从数组中删除3的倍数的任意数量的非零元素。同样， B可以删除5的倍数。无法删除任何元素的玩家将输掉比赛。任务是如果A首先开始并且双方都以最佳状态进行比赛，则找到游戏的赢家。
例子：
Input: arr[] = {1, 2, 3, 5, 6}
Output: A
3 and 6 are the elements that A can remove.
5 is the only element that B can remove.
A can remove 3 in his first move then B will have to remove 5. In the next turn, A will remove 6 and B will be left with no more moves to make.
Input: arr[] = {3, 5, 15, 20, 6, 9}
Output: A
方法：存放在movesA元素只被3整除的计数，在movesB元素只有5整除的数和元素双方在movesBoth整除。现在，
- 如果moveBoth = 0，则两个玩家都只能删除按其各自数字可整除的元素，并且只有当moveA> moveBB时， A才能赢得比赛。
- 如果moveBoth> 0，则为了发挥最佳效果， A将删除所有可被3和5整除的元素，这样B便不会从公共元素中删除任何元素，则仅当moveA + 1时A才是赢家>动作B
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the winner of the game
string getWinner(int arr[], int n)
{
int movesA = 0, movesB = 0, movesBoth = 0;
for (int i = 0; i < n; i++) {
// Increment common moves
if (arr[i] % 3 == 0 && arr[i] % 5 == 0)
movesBoth++;
// Increment A's moves
else if (arr[i] % 3 == 0)
movesA++;
// Increment B's moves
else if (arr[i] % 5 == 0)
movesB++;
}
// If there are no common moves
if (movesBoth == 0) {
if (movesA > movesB)
return "A";
return "B";
}
// 1 is added because A can remove all the elements
// that are part of the common moves in a single move
if (movesA + 1 > movesB)
return "A";
return "B";
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << getWinner(arr, n);
return 0;
} Java
// Java implementation of the approach
class GfG
{
// Function to return the winner of the game
static String getWinner(int arr[], int n)
{
int movesA = 0, movesB = 0, movesBoth = 0;
for (int i = 0; i < n; i++)
{
// Increment common moves
if (arr[i] % 3 == 0 && arr[i] % 5 == 0)
movesBoth++;
// Increment A's moves
else if (arr[i] % 3 == 0)
movesA++;
// Increment B's moves
else if (arr[i] % 5 == 0)
movesB++;
}
// If there are no common moves
if (movesBoth == 0)
{
if (movesA > movesB)
return "A";
return "B";
}
// 1 is added because A can remove
// all the elements that are part
// of the common moves in a single move
if (movesA + 1 > movesB)
return "A";
return "B";
}
// Driver code
public static void main(String []args)
{
int arr[] = { 1, 2, 3, 5, 6 };
int n = arr.length;
System.out.println(getWinner(arr, n));
}
}
// This code is contributed by Rituraj JainPython3
# Python3 implementation of the approach
# Function to return the winner of the game
def getWinner(arr, n):
movesA, movesB, movesBoth = 0, 0, 0
for i in range(0, n):
# Increment common moves
if arr[i] % 3 == 0 and arr[i] % 5 == 0:
movesBoth += 1
# Increment A's moves
elif arr[i] % 3 == 0:
movesA += 1
# Increment B's moves
elif arr[i] % 5 == 0:
movesB += 1
# If there are no common moves
if movesBoth == 0:
if movesA > movesB:
return "A"
return "B"
# 1 is added because A can
# remove all the elements
# that are part of the common
# moves in a single move
if movesA + 1 > movesB:
return "A"
return "B"
# Driver code
if __name__ == "__main__":
arr = [1, 2, 3, 5, 6]
n = len(arr)
print(getWinner(arr, n))
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
class GfG
{
// Function to return the winner of the game
static String getWinner(int []arr, int n)
{
int movesA = 0, movesB = 0, movesBoth = 0;
for (int i = 0; i < n; i++)
{
// Increment common moves
if (arr[i] % 3 == 0 && arr[i] % 5 == 0)
movesBoth++;
// Increment A's moves
else if (arr[i] % 3 == 0)
movesA++;
// Increment B's moves
else if (arr[i] % 5 == 0)
movesB++;
}
// If there are no common moves
if (movesBoth == 0)
{
if (movesA > movesB)
return "A";
return "B";
}
// 1 is added because A can remove
// all the elements that are part
// of the common moves in a single move
if (movesA + 1 > movesB)
return "A";
return "B";
}
// Driver code
public static void Main(String []args)
{
int []arr = { 1, 2, 3, 5, 6 };
int n = arr.Length;
Console.WriteLine(getWinner(arr, n));
}
}
// This code is contributed by
// Rajput-JiPHP
$movesB)
return "A";
return "B";
}
// 1 is added because A can remove all the elements
// that are part of the common moves in a single move
if ($movesA + 1 > $movesB)
return "A";
return "B";
}
// Driver code
$arr = array( 1, 2, 3, 5, 6 );
$n = sizeof($arr) / sizeof($arr[0]);
echo getWinner($arr, $n);
// This code is contributed by ajit.
?>输出：
A