通过在范围 [1, K] 中选择一个总和为 N 的数字来找到将获胜的玩家

给定两个整数K和N ，并且还假设Alice和Bob正在玩游戏。在单步棋中，玩家可以在[1, K]范围内选择一个数字，其数字使总和等于N的玩家获胜。如果Alice赢得游戏，则打印Alice ，否则打印Bob ，如果Alice和Bob都交替且最优地玩游戏并且Alice开始游戏。

例子：

Input: K = 7, N = 8
Output: Bob
Explanation: There is no way for Alice to win the game. When Alice picks any number from 1 to 7 (inclusive both), the opponent wins the game in the next turn by making the total 8 . Suppose you choose number 5, the opponent chooses 3 and wins the game, making the total 5+3=8 .

Input: K = 7, N = 50
Output: Alice

方法：这个问题可以用博弈论的概念来解决。观察获胜状态。这里要抓住的诀窍是，无论Bob选择什么， Alice总是可以重复(K+1) 的循环。假设在某个时刻，如果Bob选择a ， Alice可以选择[(K+1)-a ]将选择保持在1 到 K 的范围内，从而形成一个(K+1) 的循环。现在，由于Alice有第一次机会， Alice应该选择Remainder left on Dividing N by (K+1) at the Starting 以便之后，她可以继续重复(K+1) 的循环并达到总数为N 。
现在，观察结果是，如果N%(K+1) 为 0 ，这是唯一不可能为 Alice 赢得比赛的情况。
请按照以下步骤解决问题。

• 检查N%(K+1) 是否为 0，打印Bob 。
• 在任何其他情况下，打印Alice 。

下面是上述方法的实现。

C++14

// C++ program for above approach
#include 
using namespace std;
 
// Function to predict the winner
void predictTheWinner(int K, int N)
{
 
    if (N % (K + 1) == 0)
        cout << "Bob";
    else
        cout << "Alice";
}
 
// Driver Code
int main()
{
 
    // Given Input
    int K = 7, N = 50;
 
    // Function call
    predictTheWinner(K, N);
 
    return 0;
}

Java

// Java program for the above approach
 
import java.io.*;
 
class GFG {
   
// Function to predict the winner
static void predictTheWinner(int K, int N)
{
 
    if (N % (K + 1) == 0)
         System.out.println( "Bob");
    else
         System.out.println("Alice");
}
 
// Driver Code
    public static void main (String[] args) {
        // Given Input
    int K = 7, N = 50;
 
    // Function call
    predictTheWinner(K, N);
    }
}
// This code is contributed by Potta Lokesh

Python3

# Python 3 program for above approach
 
# Function to predict the winner
def predictTheWinner(K, N):
    if (N % (K + 1) == 0):
        print("Bob")
    else:
        print("Alice")
 
# Driver Code
if __name__ == '__main__':
    # Given Input
    K = 7
    N = 50
 
    # Function call
    predictTheWinner(K, N)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
 
using System;
 
public class GFG {
   
    // Function to predict the winner
    static void predictTheWinner(int K, int N)
    {
     
        if (N % (K + 1) == 0)
             Console.WriteLine( "Bob");
        else
             Console.WriteLine("Alice");
    }
 
    // Driver Code
    public static void Main (string[] args) {
        // Given Input
        int K = 7, N = 50;
 
        // Function call
        predictTheWinner(K, N);
    }
}
 
// This code is contributed by AnkThon

Javascript

 
// javascript program for above approach
 
// Function to predict the winner
function predictTheWinner(K, N)
{
 
    if (N % (K + 1) == 0)
        document.write("Bob");
    else
        document.write("Alice");
}
 
// Driver Code
    // Given Input
    var K = 7, N = 50;
 
    // Function call
    predictTheWinner(K, N);
 
// This code is contributed by ipg2016107.

输出：

Alice

时间复杂度： O(1)
辅助空间： O(1)