2. (5 Points) When calculating a combination (n choose k) the recursive definition is n k =1 if

Transcription

1 COSC 320 Exam 1 Key Spring 2011 Part 1: Recursion 1. (5 Points) Write a recursive function for calculating the factorial of a non-negative integer. long factorial(long n) if (n == 0) else return n * factorial(n-1); 2. (5 Points) When calculating a combination (n choose k) the recursive definition is n k =1 if k = 0 or if k = n, n k =0 if k > n and in the case where none of these are true we have n k = n 1 k 1 n 1 k Write a recursive function long comb(long n, long k) that will calculate n choose k. long comb(long n, long k) if (k == 0) else if (k == n) else if (k > n) return 0; else return comb(n-1, k-1) + comb(n-1, k);

2 3. (15 Points)Given the following function and main determine the output of the program. Also draw the calling tree for each of the calls from the main. long f(long n, long m) if (n < 0) else if (m > 5) return 5; else return f(n-2, m+1) + f(n-1, m+2); #include <iostream> int main() using namespace std; cout << "Result: " << f(1, 1) << endl; cout << "Result: " << f(2, 2) << endl; cout << "Result: " << f(5, 3) << endl; return 0; Result: 3 Result: 8 Result: 21 Call List Result: Result: Result: 21

3 Part 2: Binary Trees 1. (5 Points Each) Assuming that we have a the definition of a standard tnode class given below, write the following functions. class tnode public: T nodevalue; tnode<t> *left, *right; tnode() ; tnode (const T& item, tnode<t> *lptr = NULL, tnode<t> *rptr = NULL): nodevalue(item), left(lptr), right(rptr) (a) Write a function that will output the tree using an inorder scan. void inorderoutput(tnode<t> *t, const string& separator = " ") inorderoutput(t->left, separator); cout << t->nodevalue << separator; inorderoutput(t->right, separator); (b) Write a function that will output the tree using a postorder scan. void postorderoutput(tnode<t> *t, const string& separator = " ") postorderoutput(t->left, separator); postorderoutput(t->right, separator); cout << t->nodevalue << separator;

4 (c) Write a function that will count the number of leaves in the tree. void countleaf (tnode<t> *t, int& count) if (t->left == NULL && t->right == NULL) count++; countleaf(t->left, count); countleaf(t->right, count); (d) Write a function that will count the number of only children. void countonlychildren (tnode<t> *t, int& count) if ((t->left == NULL && t->right!= NULL) (t->left!= NULL && t->right == NULL)) count++; countonlychildren(t->left, count); countonlychildren(t->right, count); 2. (5 Points) Write a main that will create the following tree. C L K I J F B H G E D tnode<char> *root; tnode<char> *a, *b, *c, *d, *e; a = new tnode<char> ('H'); b = new tnode<char> ('G'); c = new tnode<char> ('F',a,b); d = new tnode<char> ('D'); e = new tnode<char> ('E'); b = new tnode<char> ('B',e,d); d = new tnode<char> ('K',c,b); a = new tnode<char> ('I'); c = new tnode<char> ('J'); e = new tnode<char> ('L',a,c); root = new tnode<char> ('C',e,d);

5 Part 3: BST Trees 1. (5 Points) Draw the Binary Search Tree that is created by the following block of code. int list[] = 6,, 5, -1, 10, 8, 4, 2, 0; int listsize = sizeof(list)/sizeof(int); BinSTree<int> tree; for (int i = 0; i < listsize; i++) tree.insertnode(list[i]); (5 Points Each) Say we have the following binary search tree. For each question start with the full tree (a) Draw the tree after the node 6 is removed (b) Draw the tree after the node 8 is removed (c) Draw the tree after the node -1 is removed (d) Draw the tree after the node is removed (e) Draw the tree after the node 4 is removed

6 Part 4: STL Sets and Maps 1. (5 Points Each) Assuming that we have three sets with the declaration set<int> A, B, C, D; and assume that we have overloaded the operations + (union), * (intersection) and (set difference). Write the statement that will produce the following and store in in D. (A (B+C)) + ((B*C) A) ((A*B) C) + ((A*C) B) + ((B*C) A) (A+B+C) (A*B+A*C+B*C)

7 2. (5 Points) Say that we have the following class and block of code from the main. Use map iterators to interchange the IDs of Don Spickler and Sue Me. class DumbClass private: std::string name; std::string id; int age; public: DumbClass(); DumbClass (const std::string& inputname, const std::string& inputid, const int& inputage); friend std::ostream& operator<< (std::ostream& ostr, const DumbClass& dc); friend bool operator< (const DumbClass& a, const DumbClass& b); friend bool operator== (const DumbClass& a, const DumbClass& b); std::string getname(); std::string getid(); int getage(); ; void setname(std::string& inputname); void setid(std::string& inputid); void setage(int inputage); // From the main map<string, DumbClass> staff; map<string, DumbClass>::iterator iter; string oldname, newname; int newhours; DumbClass emp1("don Spickler", " ", 45); DumbClass emp2("jack A. Beanstalk", " ", 25); DumbClass emp3("sue Me", " ", 21); staff["don"] = emp1; staff["jack"] = emp2; staff["sue"] = emp3; map<string, DumbClass>::iterator iter2; string id, id2; iter = staff.find("don"); if (iter!= staff.end()) id = (*iter).second.getid(); iter2 = staff.find("sue"); if (iter2!= staff.end()) id2 = (*iter2).second.getid(); (*iter2).second.setid(id); (*iter).second.setid(id2);