Question

C++ Implement Hot Potato game

________________________________________________

A group of people, numbered 1 to N, are sitting in a circle. Starting at person 1, a hot potato is passed. After X passes, the person holding the hot potato is eliminated, the circle closes ranks, and the game continues with the person who was sitting after the eliminated person picking up the hot potato. The last remaining person wins.

For example:

if X = 0 and N = 5, players are eliminated in order, and player 5 wins

If X = 1 and N = 5, the order of elimination is 2, 4, 1, 5.   

Write a program for general values of X and N.

Ask a user for number of people and number of passes

To speed up input, debugging, store names of the people in a file (don’t need a file but it does help)

Output number AND the name of a person being eliminated

Output number and the name of the winner

Allow user to play the game as many times as user wants

Answer 1

#include<iostream> //for cin and cout
#include<vector> //for vector
#include<iterator> //for erase method
using namespace std;
//main method
int main()
{
int x,n; //to store the passes and number of players.
char again; //to store the choice of user to play again

do{
cout<<"Enter number of passes: ";
cin>>x;
cout<<"Enter number of players: ";
cin>>n;

x = (n>0)?x%n:0; //decrements the passes
vector<int> players; //creates a vector which represents players.
//populates the players indices
for(int i=1;i<=n;i++){
players.push_back(i);
}
//iterator helps in erasing(deleting) value from vector
vector<int>::iterator it = players.begin();
int pos = 0;

//loop continues until size is greater than 1
while(players.size()>1){
for(int i=1;i<=x;i++){ //loops passes times
++pos;
if(pos==players.size())//if pos is equal to size then
pos = 0; //assigns 0.
}
it = players.begin()+pos;
players.erase(it); //removes the value from vector
if(pos==players.size())
pos = 0;
}
//prints the winner
cout<<"\nPlayer "<<players[0]<<" Won"<<endl;
//prompts the user for playing again
cout<<"\nDo you want to play again(y/n)? ";
cin>>again;
cout<<endl;
}while(again=='y'); //loop continues if user enters y

return 0;
}

//output screenshot

//any query, post in the comment section