EECE 231 Lab 2 Solution, Fall 2014

/* * course: EECE 231 * Lab: 2 * Problem: 1 * author: Ameen Jaber */ #include <iostream> using namespace std; int main() { // Declare variables int i,j,k; bool expA, expB, expC, expD, expE; // Read values for i,j, and k from user cout << "Please enter values for i,j, and k: "; cin >> i >> j >> k; // Compute the expressions expA = i<5; expB = (i>= 5) && (i<=10); expC = (i>0) || (j%5 == 0); expD = (i>0) || (j>=0 && (j%5 == 0)); expE = (i == j && j == k) || (i != j && j != k && i != k); // Print the values of the expressions cout << "The value of expression A is: " << expA << endl; cout << "The value of expression B is: " << expB << endl; cout << "The value of expression C is: " << expC << endl; cout << "The value of expression D is: " << expD << endl; cout << "The value of expression E is: " << expE << endl; return 0; } //Problem Assignment 2, Fall 14/15 //Problem 2 //Author: Maya Safieddine #include <iostream> #include <cmath> //to use the pow function using namespace std; int main() { double a, b, c;//declare the coefficient variables of the quadratic equation cout<<"Please enter the values of a, b, and c"<<endl; cin>>a >> b>> c;//prompt the user to enter the values of variables a, b, and c double discriminant = pow(b,2.0)-4*a*c;//compute the discriminant value into variable of type double (since expected value of the equation is decimal) //we use 2.0 in pow function to remove ambiguity, i.e. selcting the pow function that expects double inputs. The pow function is defined in the cmath library cout<<"discriminant: "<<discriminant<<endl; if(discriminant ==0) { cout<<"single repeated root:"<<endl; double root = -b/(2*a);//compute the single root cout<<root<<endl; } else if(discriminant < 0) { cout<<"two complex roots"<<endl; } else { cout<<"two real roots:"<<endl; double root_1 = (-b + pow(discriminant,0.5))/(2*a);//compute the two real roots double root_2 = (-b - pow(discriminant,0.5))/(2*a); cout<<root_1<<" and "<< root_2 <<endl; } //test example1: 1 1 1 two complex roots, discrimint: -3, two complex roots //test example2: 1, 2, 1 single real, discrimint:0, root: -1 //test example3: 1, 5, 1 two real roots: -4.79, -0.208, discrimint:21 return 0; } //Problem Assignment 2, Fall 14/15 //Problem 3 //Author: Maya Safieddine #include <iostream> using namespace std; int main() { cout << "Enter the number of integers in the list: "; int n; cin >> n; int count = 1;// counter to keep track of the number items int number; // to store the next user input int max, min; // to store the max and the min cout << "Enter "<< n<<" integers: "; while ( count<=n) //this condition indicates when to stop looping over the loop body. //Since we are reading a number at every loop, exit when all numbers are read and processed i.e. n numbers. //count tracks the number of read numbers. Since count is initially set to 1, stop when count reaches n -> loop while count< n+1 (or equivalently count<=n) { cin >> number; // read number from the user if(count == 1) { // if count ==1, i.e., we are looking at the first number, we initialize max and min to this number, as it's the only number max = number; min = number; } else { //count>1, a new number is added other than first if (number<min)//if this number is less than min min = number; //set min to the number if (number>max)//if this number is greater than max max=number; //set max to this number } count++;//increment count } if (n != 0) {//if the list is not empty cout << "The largest number: " << max <<endl; cout << "The smallest number: "<< min <<endl; } else cout << "No input." << endl; return 0; } ///Assignment 2-Problem 4 Solution //Ali Yassin #include <iostream> using namespace std; int main() { int n; int numOfEven=0; int sumOfEven=0; int x; cout<<"Enter the number of integers in the list please."<<endl; cin>>n; cout<<"Enter an integer"<<endl; for(int i=0;i<n;i++) { cin>>x; if(x%2==0){ numOfEven++; sumOfEven=sumOfEven+x; } } cout<<"The number of the even integers in the list is: "<<numOfEven<<endl; cout<<"The sum of the even integers in the list is: "<<sumOfEven<<endl; return 0; } #include <iostream> #include <cmath> using namespace std; int main() { int n; cout <<"This is a primality test. Please enter a positive integer:"<<endl; cin >> n; bool isPrime = true; if (n==1) isPrime = false; // 1 is not prime by convention else { int i = 2; while(i<= n-1&& isPrime==true) { if( n%i == 0) isPrime =false; i++; } } if(isPrime) cout <<endl<<n<<" is prime"<<endl; else cout <<endl<<n<<" is not prime"<<endl; return 0; } #include <iostream> #include <cmath> using namespace std; int main() { int n; cout <<"This is a primality test. Please enter a positive integer:"<<endl; cin >> n; bool isPrime = true; if (n==1) isPrime = false; // 1 is not prime by convention else { int i = 2; while(i*i<=n && isPrime==true ) { // "i*i<=n" is equivalent to "i<=square_root(n)" if( n%i == 0) isPrime =false; i++; } } if(isPrime) cout <<endl<<n<<" is prime"<<endl; else cout <<endl<<n<<" is not prime"<<endl; return 0; } #include <iostream> #include <cmath> using namespace std; int main() { int x; cout<<"Please enter a positve integer: "; cin>> x; int numberOfPrimes = 0; int n=2; while(n<=x) { //////////// code from part (b) ///////////// bool isPrime = true; int i = 2; while(i*i<=n && isPrime == true) { // "i*i<=n" is equivalent to "i<=square_root(n)" if( n%i == 0) isPrime =false; i++; } ////////// end of code from part (b) ///////// if(isPrime) numberOfPrimes ++; n++; } cout<<endl; cout<<"f("<<x<<")= "<<(numberOfPrimes+0.0)/x<<endl; return 0; }