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📜  预测每回合移除 K 张牌的纸牌游戏的获胜者,使得 K 和堆大小的按位与为 0

📅  最后修改于: 2021-09-24 05:03:54             🧑  作者: Mango

有两个玩家AB以及一堆N叠在一起的牌。任务是找到游戏的获胜者,假设两个玩家都按照以下准则进行最佳游戏:

  • 玩家A总是开始游戏,然后玩家轮流进行。
  • 在每一回合中,如果K & n = 0,玩家可以移除K ( 1 ≤ K ≤ N) 张牌,其中n是当前堆的大小。
  • 如果玩家在游戏中的任何时候都无法移动,则该玩家失败,游戏结束。

例子:

方法:该想法基于以下观察:如果N的二进制表示中1的计数在遇到0之前是奇数,则A赢得比赛。如果整个二进制字符串不存在这样的10组合,则B获胜。请按照以下步骤解决问题:

  • 初始化一个变量countOne来存储1的计数。
  • N转换为其二进制表示并将其存储在字符串binString 中
  • 遍历字符串binString并执行以下操作:
    • 如果遇到“ 1 ”,则增加countOne
    • 如果遇到’ 0 ‘,检查countOne是奇数还是偶数,如果countOne是奇数A获胜,并跳出循环,否则将countOne重置为0并继续遍历。
  • 如果整个字符串被遍历而没有中断,则B获胜。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the winner of the
// game if both player plays optimally
void findWinner(int N)
{
     
    // Stores the count of 1s
    int onesBeforeZero = 0;
    int flag = 1;
     
    // Convert N to binary representation
    int binString[32];
 
    int i = 0;
     
    while (N > 0)
    {
         
        // Storing remainder in binary array
        binString[i] = N % 2;
        N = N / 2;
        i++;
    }
 
    int l = sizeof(binString) /
            sizeof(binString[0]);
 
    // Traverse the binary string
    for(int j = 0; j < l; j++)
    {
         
        // If 1 is encountered,
        // increment count of 1s
        if (binString[j] == 1)
        {
            onesBeforeZero += 1;
        }
 
        // If 0 is encountered, check
        // if count of 1s is odd
        else
        {
             
            // If count of 1s is odd,
            // then winner is A
            if (onesBeforeZero & 1)
            {
                cout << "A";
                flag = 0;
                break;
            }
 
            // If count of 1s is even,
            // reset it to 0
            else
                onesBeforeZero = 0;
        }    
    }
     
    // If entire loop is traversed
    // without breaking, then
    // B is the winner
    if (flag == 1)
        cout << "B";
}
 
// Driver Code
int main()
{
    int N = 4;
     
    // Function Call
    findWinner(N);
 
    return 0;
}
 
// This code is contributed by jana_sayantan


C
// C program for the above approach
#include 
 
// Function to find the winner of the
// game if both player plays optimally
void findWinner(unsigned long long N)
{
    // Stores the count of 1s
    int onesBeforeZero = 0;
    int flag = 1, j = 0;
    char binString[32];
     
    // Converting N into a binary string
    for(int i = 31; i >= 0; i--)
    {
        unsigned long long temp = N >> i;
         
        if (temp & 1)
            binString[j] = '1';
        else
            binString[j] = '0';
             
        j += 1;
    }
     
    // Traverse the binary string
    for(int i = 0; i < 32; i++)
    {
        if (binString[i] == '1')
         
            // If 1 is encountered
            // increment ones count
            onesBeforeZero += 1;
        else
        {
             
            // If 0 is encountered check
            // if ones count is odd
            if (onesBeforeZero & 1)
            {
                 
                // If ones count is odd
                // winner is A break
                printf("A");
                flag = 0;
                break;
            }
            else
                // If ones count is even
                // reset it to 0 and continue
                onesBeforeZero = 0;
        }
    }
     
    // If entire loop is traversed
    // without breaking, then
    // B is the winner
    if (flag == 1)
        printf("B");
}   
 
// Driver code
int main()
{
    unsigned long long N = 4;
     
    // Function Call
    findWinner(N);
     
    return 0;
}
 
// This code is contributed by Praneeth Kapila


Java
// Java program for the above approach
class GFG{
     
// Function to find the winner
static void findWinner(long N)
{
    // Stores the count of 1s
    int onesBeforeZero = 0, flag = 1, j = 0;
     
    String[] binString = new String[32];
     
    // Converting N into a binary string
    for(int i = 31; i >= 0; i--)
    {
        long temp = N >> i;
         
        if ((temp & 1) == 1)
            binString[j] = "1";
        else
            binString[j] = "0";
             
        j += 1;
    }
 
    for(int i = 0; i < 32; i++)
    {
        if (binString[i] == "1")
         
            // If 1 is encountered
            // increment ones count
            onesBeforeZero += 1;
        else
        {
             
            // If 0 is encountered check
            //if ones count is odd
            if ((onesBeforeZero & 1) == 1)
            {
                 
                // If ones count is odd winner
                // is A break
                System.out.println("A");
                flag = 0;
                break;
            }
            else
                // If ones count is even
                // reset it to 0 and continue
                onesBeforeZero = 0;
        }
    }
     
    // If entire loop is traversed
    // without breaking, then
    // B is the winner
    if (flag == 1)
        System.out.println("B");
}
 
// Driver code
public static void main(String[] args)
{
    long N = 4;
     
    // Function Call
    findWinner(N);
}
}
 
// This code is contributed by Praneeth Kapila


Python3
# Python3 program for the above approach
 
# Function to find the winner of the
# game if both player plays optimally
def findWinner(N):
 
    # Stores the count of 1s
    onesBeforeZero = 0
    flag = 1
 
    # Convert N to binary representation
    binString = bin(N).replace("0b", "")
 
    l = len(binString)
 
    # Traverse the binary string
    for j in range(l):
 
        # If 1 is encountered,
        # increment count of 1s
        if binString[j] == '1':
            onesBeforeZero += 1
 
        # If 0 is encountered, check
        # if count of 1s is odd
        else:
 
            # If count of 1s is odd,
            # then winner is A
            if onesBeforeZero & 1:
                print("A")
                flag = 0
                break
 
            # If count of 1s is even,
            # reset it to 0
            else:
                onesBeforeZero = 0
 
    # If entire loop is traversed
    # without breaking, then
    # B is the winner
    if flag == 1:
        print("B")
 
 
# Driver Code
N = 4
 
# Function Call
findWinner(N)


C#
// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the winner
static void findWinner(long N)
{
     
    // Stores the count of 1s
    int onesBeforeZero = 0, flag = 1, j = 0;
     
    String[] binString = new String[32];
     
    // Converting N into a binary string
    for(int i = 31; i >= 0; i--)
    {
        long temp = N >> i;
         
        if ((temp & 1) == 1)
            binString[j] = "1";
        else
            binString[j] = "0";
             
        j += 1;
    }
 
    for(int i = 0; i < 32; i++)
    {
        if (binString[i] == "1")
         
            // If 1 is encountered
            // increment ones count
            onesBeforeZero += 1;
        else
        {
             
            // If 0 is encountered check
            //if ones count is odd
            if ((onesBeforeZero & 1) == 1)
            {
                 
                // If ones count is odd winner
                // is A break
                Console.WriteLine("A");
                flag = 0;
                break;
            }
            else
             
                // If ones count is even
                // reset it to 0 and continue
                onesBeforeZero = 0;
        }
    }
     
    // If entire loop is traversed
    // without breaking, then
    // B is the winner
    if (flag == 1)
        Console.WriteLine("B");
}
 
// Driver code
public static void Main(String[] args)
{
    long N = 4;
     
    // Function Call
    findWinner(N);
}
}
 
// This code is contributed by shivanisinghss2110


Javascript


输出:
A

时间复杂度: O(log N)
辅助空间: O(log N)

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