Exercise P2.12.

Write a program that reads a number greater than or equal to 1,000 from the user and prints it with a comma separating the thousands. Here is a sample dialog; the user input is in boldface:

Please enter an integer >= 1000: 23456

23,456

Solution

#include < iostream>

using namespace std;

int main(){

  int n;
  cout << "type a number greater or equal to 1000 : ";
  cin >> n;

  // Goal: define n1 = 23 and n2 = 455

  int n1 = n / 1000; // 23455 / 1000 = 23.455 -> 23 because it is an integer vs. integer

  int n2 = n % 1000; // 455

//               2 3
//  1000 | 2 3 4 5 5
//  ________________
//         2 3 4 5
//         2 0 0 0
//  ________________
//           3 4 5 5
//           3 0 0 0
//    ________________
//             4 5 5

  // want to print out 23,455

  cout << n1 << "," << n2 << endl;

  return 0;
}

Exercise P2.13.

Write a program that reads a number between 1,000 and 999,999 from the user, where the user enters a comma in the input. Then print the number without a comma. Here is a sample dialog; the user input is in boldface:

Please enter an integer between 1,000 and 999,999: 23,456

23456

Hint: Read the input as a string. Measure the length of the string. Suppose it contains n characters. Then extract substrings consisting of the first n – 4 characters and the last three characters.

Solution

#include < iostream>
#include < string>

using namespace std;

int main(){

  string n;
  cout << "type a number between 1,000 and 999,999 : ";
  cin >> n;

  // suppose the input we get is 23,456
  // n = "3,456"

  int num_of_char = n.length(); // 6

  string n1 = n.substr(num_of_char-3); // 456
  string n2 = n.substr(0,num_of_char-4); // 23

  cout << n1 << n2 << endl; // will print out 23456

  return 0;
}

Exercise P3.1.

Write a program that prints all solutions to the quadratic equation $ax^2 + bx + c$. Read in a, b, c and use the quadratic formula. If the discriminant $b^2 - 4ac$ is negative, display a message stating that there are no solutions.

Solution

#include < iostream >

using namespace std;

int main()
{
  double a,b,c;

  cout << "type a : ";
  cin >> a;

  cout << "type b : ";
  cin >> b;

  cout << "type c : ";
  cin >> c;

  // b^2 - 4ac
  double discriminant = b*b - 4*a*c;

  if(discriminant < 0){
    cout << "no solution" << endl;
  }
  else if(discriminant == 0){
    cout << "repeating roots" << endl;
  }
  else{
    cout << "two distinct roots" << endl;
  }

}

Exercise P3.4.

Write a program that translates a letter grade into a number grade. Letter grades are A, B, C, D, and F, possibly followed by + or –. Their numeric values are 4, 3, 2, 1, and 0. There is no F+ or F–. A + increases the numeric value by 0.3, a – decreases it by 0.3. However, an A+ has value 4.0.

Enter a letter grade: B-

The numeric value is 2.7.

Solution 1

#include < iostream >
#include < string >

using namespace std;

int main(){

  string grade;

  cout << "Enter a letter grade: ";
  cin >> grade;

  if(grade == "A+"){
    cout << 4 << endl;
  }
  else if(grade == "A"){
    cout << 4 << endl;
  }
  else if(grade == "A-"){
    cout << 3.7 << endl;
  }
  else if(grade == "B+"){
    cout << 3.3 << endl;
  }
  else if(grade == "B"){
    cout << 3 << endl;
  }
  else if(grade == "B-"){
    cout << 2.7 << endl;
  }
  else if(grade == "C+"){
    cout << 2.3 << endl;
  }
  else if(grade == "C"){
    cout << 2 << endl;
  }
  else if(grade == "C-"){
    cout << 1.7 << endl;
  }
  else if(grade == "D+"){
    cout << 1.3 << endl;
  }
  else if(grade == "D"){
    cout << 1 << endl;
  }
  else if(grade == "D-"){
    cout << 0.7 << endl;
  }
  else if(grade == "F"){
    cout << 0 << endl;
  }
  return 0;
}

Solution 2

#include < iostream >
#include < string >

using namespace std;

int main(){

  string grade;

  cout << "Type grade : ";
  cin >> grade;

  string alphabet = grade.substr(0,1);
  double plus_minus = 0;

  if( grade.length() == 2 ){

    if( grade.subtr(1,1) == "+"){
      plus_minus = 0.3;
    }
    else{ // "-" case
      plus_minus = -0.3;
    }
  }

  if( alphabet == "A" ){
    cout << fmin(4,4 + plus_minus) << endl;
  }
  else if( alphabet == "B" ){
    cout << 3 + plus_minus << endl;
  }
  else if( alphabet == "C" ){
    cout << 2 + plus_minus << endl;
  }
  else if( alphabet == "D" ){
    cout << 1 + plus_minus << endl;
  }
  else if( alphabet == "F" ){
    cout << 0 << endl;
  }
  else{
    cout << "Wrong Input!" <