Notes on the 2009 Exam

Transcription

1 Notes on the 2009 Exam A hastily compiled review of this year s exam. Apologies if there are errors. The stuff inside borders is all that you needed to write. The rest is commentary. Please also read notes on previous years exams especially the 2005 set of notes. Question 1 (a) Multiplication table 6 marks The key thing here is to note that all the work is done by nested for loops. The only calculating is i*j once i and j are set up by the for loops. Then just remember to start a new line after each row. int main() { const int NUMROWS = 5; const int NUMCOLS = 10; int i, j; /* Now print rows 1 to NUMROWS */ for (j=1;j<=numrows;j++) { /* print table row */ for (i=1;i<=numcols;i++) { cout << setw(5) << i*j; //end that row cout << "\n"; (b) Perfect numbers up to marks Nested for loops again. The outer one handles each integer 1 to 500. The inner one works through integers looking for divisors of the current n. Note that you were asked to print out perfect numbers. Imperfect numbers should have been ignored so either you cout << n; or you don t. Nothing else. int main() { int n,i,sum; for (n=1; n<500; n++) { //test if n is perfect

2 sum=1; // sum of divisors so far for (i=2;i<n;i++) { if (n%i==0) //if i is a divisor of n sum = sum + i; // add it on //end of looking for divisors of n if (n == sum) cout << i << " "; //end of testing numbers cout << endl; (c) guessing game 7 marks Marks as follows 3 for the loop control; 1 for the if statement; one for proper declarations and call to random; 1 for counting guesses. Many of you had no loop!! This solution is a copy of guess.cpp (see while loops topic) except for the call to (undefined) random function instead of more complex formula. int main() { int guess, rand; bool done; rand = random(100); done = false; while (!done) { cout << "Guess (0..99):"; cin >> guess; cout << endl; if (guess == rand) { cout << "Correct.\n"; done = true; else if (guess < rand) cout << "Too low.\n"; else cout << "Too high.\n"; //end while

3 Question 2 These functions were provided for Practical 12 (2008/09). Note that these functions are constructed so that they don t need to know about details of leap years and month lengths. Leave all that detail to the functions. (a) 6 marks; very few marks if you didn t use loops for both the years and the months. Basically the algorithm is for each year that has passed, add on the number of days in that year; for each month of the final year that has passed, add on the number of days in that month, then add on the days of the current month. int daynumber (int day, int month, int year) { int y, m, daynumber; daynumber = 0; for (y = 1900; y < year; y++) daynumber = daynumber + year_length (y); for (m = 1; m < month; m++) daynumber = daynumber + month_length(m, year); daynumber = daynumber + day; return daynumber; It uses a year_length function as follows: int year_length (int y) { if (leap_year (y)) return 366; else return 365; (b) 8 marks; this should have been called dymthyr, not datetodaynumber. 2 marks reserved for using a void function and call-by-ref parameters (the &s). In this case we are working back from a daynumber. We need to keep going, moving through years (or months), until there are not enough days left to get us to the next year (or month). This requires a while loop: While there are enough days to get us to the end of this year, subtract this year s days, and move on to next year.

4 Note that it is imperative that you know what year the date is in before you start on to calculating the month, so that the call to month_length works correctly for Febraury in leap years.. void dymthyr (int numdays, int& day, int& month, int& year){ year = 1900; while (numdays > year_length (year)) { numdays = numdays - year_length (year); year++; //now we have the right year. What month?? month = 1; while (numdays > month_length (month, year)) { numdays = numdays - month_length (month, year); month++; //numdays now is the date within month of year day = numdays; (c) 6 marks; this tested that you could call the above functions correctly. The daynumber function returns an int, and the value needs to be stored somewhere using an assignment (i.e. number = ). The other function dymthyr uses call-by-reference parameters, and doesn t return anything. In the following, the values of day, month, and year entered by the user are overwritten by the call to dymthyr. So the cout prints the new values. int main () { int number, value, day, month, year; cout << "Enter day month year"; cin >> day >> month >> year; number = daynumber(day, month, year); number = number+90; //Add 90 days dymthyr(number, day, month, year); cout << "90 days later is " << day << "/" << month << "/" << year << endl; Note that you could achieve the three key lines above in one line as follows: Question 3 dymthyr(daynumber(day, month, year)+90, day, month, year);

5 This was practical 7 in 2008/2009. Having seen it before it shouldn t have been too hard. There are simpler formulations but this is the one I provided. It s not easy to do it correctly with a for loop. Many of you kept going all the way till there were no apartments rented, but this was wasteful. Marking was approximately 6/7 marks for reading in and setting up all the variables, 4 marks for the while loop, 7 marks for the (4) updates inside the while loop, 2/3 marks for the output, which had to be perfect not out by one round of the loop. int main () { int tot_units; double all_rented_rent; double incr;//increment which results in one less rental double maint_per_unit; cin >> tot_units >> all_rented_rent >> incr >> maint_per_unit; int n = tot_units; double rent = all_rented_rent; double profit_for_n_units, profit_for_n_minus_1_units; profit_for_n_units = (rent - maint_per_unit)*n; profit_for_n_minus_1_units = (rent + incr - maint_per_unit)*(n-1); while (profit_for_n_units < profit_for_n_minus_1_units) { n = n-1; rent = rent+incr; profit_for_n_units = profit_for_n_minus_1_units; profit_for_n_minus_1_units = (rent + incr - maint_per_unit)*(n-1); cout << "\n\nmaximum profit is when rent is " << rent << " which gives "; cout << n << " units occupied, and a profit of " << profit_for_n_units << endl; Question 4 This question was way too easy. The supplemental exam recursion question will be harder. It will involve array recursion. (a) (i) 6 marks f(2,2) = 0 f(5,2) = 3

6 f(7,3) = 4 f(5,0) = 0 (ii) 3 marks subtraction (iii) 3 marks it causes an infinite recursion, with x going towards minus infinity. (b) 8 marks; 3 for the base case, 3 for the recursive call, 2 for adding m! int multiply (int m, int n) { if (n == 0) else return multiply (m, n-1) + m; Question 5 (a) 6 marks int search (string s, string sarray[], int size) { //returns the position of s in sarray or -1 if not found int i=0; while (i < size && sarray[i]!= s) i++; if (i < size) return i; else return -1; There is this alternative but note that if your return -1 line was part of the if statement, as an else clause, and therefore inside the loop, you lost 2 marks!! Be careful! int search (string s, string sarray[], int size) { //returns the position of s in sarray or -1 if not found for (int i=0; i < size; i++) if (sarray[i] == s) return i; //if we get here we didn't find s in the array return -1;

7 (b) 14 marks; This was an arrays question so most of the marks go for array declaration and processing. 3 for declaring the arrays correctly; 4 for the reading in of the names and grades (the for loop); 2 for calling search correctly and 1 for handling the not found case; 1 for looking up the score in the correct slot; 3 for the sentinel controlled loop; int main () { string name; const int ASIZE = 500; string names[asize]; int scores[asize]; //read in the names and grades for (int i=0; i<asize; i++) cin >> names[i] >> scores[i]; cout << "Enter a name (XXX to terminate):"; cin >> name; while (name!= "XXX") { int slot = search(name, names, ASIZE); if (slot!= -1) //if found cout << "Grade: " << scores[slot] << endl; else cout << "No score for " << name << ". Try again.\n"; cout << "Enter a name (XXX to terminate):"; cin >> name; cout << "Goodbye.\n"; Other Miscellaneous Comments that I noted while grading: Sentinel Controlled Loops Template is cin >> x; while (x!= ) { ***

8 cin > x; Where the *** are you put all processing of the current valid x. The processing of x must all be inside the loop. I had many examples of cin >> x; while (x!= ) { cin > x; tot = tot+x; //wrong more stuff on x.//wrong But those last two lines must be inside the loop, before the cin > x statement. Obviously instead of x, you might have studentno or code or n or whatever it is that you are reading that will eventually be the sentinel instead of a valid value. Many people used the above template (with ***s in it or in it!!!!) every time they thought a while loop was needed. The template is ONLY for SENTINEL CONTROLLED input. Functions Many of these comments or similar appear in Notes for 2005 exam. Local variables and accessing a function s value: If a function looks like this int f(int x) { int ans; ans = x; return ans; you can t do this z = 4; f(z); cout << ans; // wrong ans here is not related to f s ans and expect it to print the value of f(z) ans only has meaning inside f. Even if you declared ans in the calling program, that would be a separate variable from the one f uses, which is a local variable whose scope in the function f. The calling program has no access to f s ans. The correct way to call f and use the value it returns is:

9 cout << f(z); or int x = f(z); //now use or print x Accessing parameters: With the following declaration of f int f(int x, int y) the body of f has access to variables called x and y, and the values in there are the ones provided by the calling function. Therefore, f should not redeclare these variables read values for these variable initialize these variables So more than likely these are all wrong: int f(int x, int y) { int x, y; //creates new local variables x and y which //effectively hides the ones passed in as parameters int f(int x, int y) { cin >> x >> y; //if it succeeds in reading data it will replace // the values the calling function provided. int f(int x, int y) { y=0; //if the calling function provides f with a value for y //f should use it!! Not replace it. And, unrelated to parameters, but the f declared above should not print any values, but must return its answer. So NOT int f(int x, int y) { cout << (x*y)/2; but probably something like int f(int x, int y) { return (x*y)/2;

10 Good luck!