Towers of Hanoi.cpp

#include <iostream>
#include <vector>
using namespace std;

int main() {
   vector<int> t[3];
   int n;
   cout << "Please enter the number of rings to move: ";
   cin >> n;
   cout << endl;
   // The initial value of to depends on whether n is odd or even
   int from = 0, candidate = 1, move = 0;
   
   int to;
   if(n%2==1)to=1;
   else to=2;
  
   // Initialize the towers
   for(int i = n + 1; i >= 1; --i)
      t[0].push_back(i);
   
   t[1].push_back(n+1);
   t[2].push_back(n+1);
   
   while (t[1].size() < n+1) { // while t[1] does not contain all of the rings
      cout << "Move #" << ++move << ": Transfer ring " << candidate << " from tower " << char(from+'A') << " to tower " << char(to+'A') << "\n";
      
      // Move the ring from the "from tower" to the "to tower" (first copy it, then delete it from the "from tower")
      t[to].push_back(t[from].back());
      t[from].pop_back();
      
      // from = the index of the tower with the smallest ring that has not just been moved: (to+1)%3 or (to+2)%3
      if (t[(to+1)%3].back()<t[(to+2)%3].back())
         from = (to+1)%3;
      else
         from =(to+2)%3;
         
      // candidate = the ring on top of the t[from] tower
      candidate = t[from].back();
         
      // to = the index of the closest tower on which the candidate can be placed: (from+1)%3 or (from+2)%3
      // (compare the candidate with the ring on the closer tower; which tower is "closer" depends on whether n is odd or even)
      if (n%2==1){
        if(t[from].back()<t[(to+1)%3].back())
            to = (from+1)%3;
        else
            to = (from+2)%3;
        
    	if(t[from].back()<t[(from+1)%3].back())
        	to = (from+1)%3;
        else
         	to = (from+2)%3; 
        candidate = t[from].back();
      }else{
        if(t[(from+1)%3].back()<t[(to+2)%3].back())
            to = (from+1)%3;
        else
            to = (from+2)%3;
        
		if(t[from].back()<t[(from+2)%3].back())
        	to = (from+2)%3;
        else
         	to = (from+1)%3;
            
        candidate = t[from].back();
      }
   }
    return 0;
}