Question

In the simulation, player 2 will always play according to the same strategy. The number of coins player 2 spends is based on what round it is, as described below. (a) You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on. The method returns 1, 2, or 3 based on the following rules. If round is divisible by 3, then return 3. If round is not divisible by 3 but is divisible by 2, then return 2. If round is not divisible by 3 and is not divisible by 2, then return 1. Complete method getPlayer2Move below by assigning the correct value to result to be returned.

Answer 1

The simulation, player 2 will always play according to the same strategy.

Method getPlayer2Move below is completed by assigning the correct value to result to be returned.

Explanation:

• You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.

#include <bits/stdc++.h>  

using namespace std;

bool getplayer2move(int x, int y, int n)  

{

   int dp[n + 1];  

   dp[0] = false;  

   dp[1] = true;  

   for (int i = 2; i <= n; 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

           dp[i] = false;  

   }  

   return dp[n];  

}  

int main()  

{  

   int x = 3, y = 4, n = 5;  

   if (findWinner(x, y, n))  

       cout << 'A';  

   else

       cout << 'B';  

   return 0;  

}