Femme Dew You should know the original de code by heart and make the necessary updates for any twist of the problem in c

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Femme Dew You Should Know The Original De Code By Heart And Make The Necessary Updates For Any Twist Of The Problem In C 1 (28.12 KiB) Viewed 36 times

Femme Dew You Should Know The Original De Code By Heart And Make The Necessary Updates For Any Twist Of The Problem In C 2 (41.02 KiB) Viewed 36 times

Femme Dew You Should Know The Original De Code By Heart And Make The Necessary Updates For Any Twist Of The Problem In C 3 (41.02 KiB) Viewed 36 times

Femme Dew You Should Know The Original De Code By Heart And Make The Necessary Updates For Any Twist Of The Problem In C 4 (37.86 KiB) Viewed 36 times

Femme Dew You Should Know The Original De Code By Heart And Make The Necessary Updates For Any Twist Of The Problem In C 5 (36.02 KiB) Viewed 36 times

Femme Dew You should know the original de code by heart and make the necessary updates for any twist of the problem in case there is a question in this in the excam node cost include cww. ini wetent three to ADC represented any of 3 vector in candidate, to, trom, sve ove count the move tuber coute et ODD number of rings to move 1 crccedi Viitate the towers tot 10-push_baku [ Luah baek (1) ta.push back) Initialize toets and tandsdate from teit candidate whiler ))+11 W there are ring to transfer to tone 11 coucove munbe "mova<<": frater tung" accandidate from tower c theo65to towerchar(65)<cendi de stive the data sound sitel from-pop back) et net om te to the street " - which is it 1 get next to tour to get a candidate candidate - t back Tet 0

Fill in the blanks below. You should know the original lecture code by heart and make the necessary updates for any twist of the problem in case there is a question like this in the exam. #include <iostream> #include<vector> using namespace std; int main() { vector int> t[3]; // three towers A,B,C represented as an array of 3 vectors int n, candidate, to, from, move=0; // move counts the move number cout<<"Please enter an ODD number of rings to move: "; cin>>n; cout<<endl; Vintitialize the 3 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); Z initialize towers and candidate

//intitialize the 3 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); // initialize towers and candidate from=0; to=1; candidate=1; while( t[1].size() <n+1) { // there are still rings to transfer to tower B = t[1] cout<<"move number "<<++move<<": Transfer ring <<candidate<< " from tower " « char (from+65)<<" to tower "<<char(to+65)<<endl; //do it: move the disks around t[to]. t[from].pop_back(); 1 //get next "from tower" It's not the most recent "to" So which is it?

//get next "from tower" It's not the most recent "to" So which is it? if ) from else from // get next "to tower" if ( ) to = else to = //get next candidate candidate t[ J.back(); } return 0;