A和B在玩游戏。一开始有n个硬币。给定另外两个数字x和y。玩家在每一步中都可以选择x或y或l个硬币。总是开始游戏。选择最后一个硬币的玩家将赢得比赛。对于给定的n值，如果两者都处于最佳状态，则求A是否会赢得比赛。
例子：
Input : n = 5, x = 3, y = 4
Output : A
There are 5 coins, every player can pick 1 or
3 or 4 coins on his/her turn.
A can win by picking 3 coins in first chance.
Now 2 coins will be left so B will pick one
coin and now A can win by picking the last coin.
Input : 2 3 4
Output : B让我们以x = 3，y = 4的n个示例值为例。
n = 0 A不能选择任何硬币，因此他会损失
n = 1 A可以选择1个硬币并赢得比赛
n = 2 A只能选择1个硬币。现在B将选择1个硬币并赢得比赛
n = 3 4 A将赢得3或4个硬币，从而赢得比赛
n = 5、6 A将选择3或4个硬币。现在，B将必须从2个硬币中进行选择，这样A才能获胜。
我们可以观察到，只有当A输掉硬币n-1，nx和ny时，A才能赢取n个硬币。
C++
// C++ program to find winner of game
// if player can pick 1, x, y coins
#include
using namespace std;
// To find winner of game
bool findWinner(int x, int y, int n)
{
// To store results
int dp[n + 1];
// Initial values
dp[0] = false;
dp[1] = true;
// Computing other values.
for (int i = 2; i <= n; i++) {
// If A losses any of i-1 or i-x
// or i-y game then he will
// definitely win game i
if (i - 1 >= 0 and !dp[i - 1])
dp[i] = true;
else if (i - x >= 0 and !dp[i - x])
dp[i] = true;
else if (i - y >= 0 and !dp[i - y])
dp[i] = true;
// Else A loses game.
else
dp[i] = false;
}
// If dp[n] is true then A will
// game otherwise he losses
return dp[n];
}
// Driver program to test findWinner();
int main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
cout << 'A';
else
cout << 'B';
return 0;
} Java
// Java program to find winner of game
// if player can pick 1, x, y coins
import java.util.Arrays;
public class GFG {
// To find winner of game
static boolean findWinner(int x, int y, int n)
{
// To store results
boolean[] dp = new boolean[n + 1];
Arrays.fill(dp, false);
// Initial values
dp[0] = false;
dp[1] = true;
// Computing other values.
for (int i = 2; i <= n; i++) {
// If A losses any of i-1 or i-x
// or i-y game then he will
// definitely win game i
if (i - 1 >= 0 && dp[i - 1] == false)
dp[i] = true;
else if (i - x >= 0 && dp[i - x] == false)
dp[i] = true;
else if (i - y >= 0 && dp[i - y] == false)
dp[i] = true;
// Else A loses game.
else
dp[i] = false;
}
// If dp[n] is true then A will
// game otherwise he losses
return dp[n];
}
// Driver program to test findWinner();
public static void main(String args[])
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n) == true)
System.out.println('A');
else
System.out.println('B');
}
}
// This code is contributed by Sumit GhoshPython3
# Python3 program to find winner of game
# if player can pick 1, x, y coins
# To find winner of game
def findWinner(x, y, n):
# To store results
dp = [0 for i in range(n + 1)]
# Initial values
dp[0] = False
dp[1] = True
# Computing other values.
for i in range(2, n + 1):
# If A losses any of i-1 or i-x
# or i-y game then he will
# definitely win game i
if (i - 1 >= 0 and not dp[i - 1]):
dp[i] = True
elif (i - x >= 0 and not dp[i - x]):
dp[i] = True
elif (i - y >= 0 and not dp[i - y]):
dp[i] = True
# Else A loses game.
else:
dp[i] = False
# If dp[n] is true then A will
# game otherwise he losses
return dp[n]
# Driver Code
x = 3; y = 4; n = 5
if (findWinner(x, y, n)):
print('A')
else:
print('B')
# This code is contributed by Azkia AnamC#
// C# program to find winner of game
// if player can pick 1, x, y coins
using System;
public class GFG {
// To find winner of game
static bool findWinner(int x, int y, int n)
{
// To store results
bool[] dp = new bool[n + 1];
for(int i = 0; i < n+1; i++)
dp[i] =false;
// Initial values
dp[0] = false;
dp[1] = true;
// Computing other values.
for (int i = 2; i <= n; i++)
{
// If A losses any of i-1 or i-x
// or i-y game then he will
// definitely win game i
if (i - 1 >= 0 && dp[i - 1] == false)
dp[i] = true;
else if (i - x >= 0 && dp[i - x] == false)
dp[i] = true;
else if (i - y >= 0 && dp[i - y] == false)
dp[i] = true;
// Else A loses game.
else
dp[i] = false;
}
// If dp[n] is true then A will
// game otherwise he losses
return dp[n];
}
// Driver program to test findWinner();
public static void Main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n) == true)
Console.WriteLine('A');
else
Console.WriteLine('B');
}
}
// This code is contributed by vt_m.PHP
= 0 and !$dp[$i - 1])
$dp[$i] = true;
else if ($i - $x >= 0 and !$dp[$i - $x])
$dp[$i] = true;
else if ($i - $y >= 0 and !$dp[$i - $y])
$dp[$i] = true;
// Else A loses game.
else
$dp[$i] = false;
}
// If dp[n] is true then A will
// game otherwise he losses
return $dp[$n];
}
// Driver program to test findWinner();
$x = 3; $y = 4; $n = 5;
if (findWinner($x, $y, $n))
echo 'A';
else
echo 'B';
// This code is contributed by anuj_67.
?>Javascript
输出：
A