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If control reaches the second condition (age < 65), then the first condition must be false so in fact 18 age < 65. Similarly, if control reaches the second else, then both conditions must be false so in fact age 65. An integer m is a multiple of an integer n if the remainder from the integer division of m by n is 0. So the compound condition m % n == 0 || n % m == 0 tests whether either is a multiple of the other: int main() { int m, n; cin >> m >> n; cout << (m % n == 0 || n % m == 0 "multiple" : "not") << endl; } 30 4 not 30 5 multiple The value of the conditional expression will be either "multiple" or "not", according to whether the compound condition is true. So sending the complete conditional expression to the output stream produces the desired result. The character representing the operation should be the control variable for the switch statement: int main() { int x, y; char op; cout << "Enter two integers: "; cin >> x >> y; cout << "Enter an operator: "; cin >> op; switch (op) { case '+': cout << x + y << endl; break; case '-': cout << x - y << endl; break; case '*': cout << x * y << endl; break; case '/': cout << x / y << endl; break; case '%': cout << x % y << endl; break; } } Enter two integers: 30 13 Enter an operator: % 4 In each of the five cases, we simply print the value of the corresponding arithmetic operation and then break. First define the two enum types Choice and Result. Then declare variables choice1, choice2, and result of these types, and use an integer n to get the required input and assign it to them: enum Choice {ROCK, PAPER, SCISSORS}; enum Winner {PLAYER1, PLAYER2, TIE}; int main() { int n; Choice choice1, choice2; Winner winner; cout << "Choose rock (0), paper (1), or scissors (2):" << endl; cout << "Player #1: "; cin >> n; choice1 = Choice(n);
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cout << "Player #2: "; cin >> n; choice2 = Choice(n); if (choice1 == choice2) winner = TIE; else if (choice1 == ROCK) if (choice2 == PAPER) winner = PLAYER2; else winner = PLAYER1; else if (choice1 == PAPER) if (choice2 == SCISSORS) winner = PLAYER2; else winner = PLAYER1; else // (choice1 == SCISSORS) if (choice2 == ROCK) winner = PLAYER2; else winner = PLAYER1; if (winner == TIE) cout << "\tYou tied.\n"; else if (winner == PLAYER1) cout << "\tPlayer #1 wins." <<endl; else cout << "\tPlayer #2 wins." << endl; } Choose rock (0), paper (1), or scissors (2): Player #1: 1 Player #2: 1 You tied. Choose rock (0), paper (1), or scissors (2): Player #1: 2 Player #2: 1 Player #1 wins. Choose rock (0), paper (1), or scissors (2): Player #1: 2 Player #2: 0 Player #2 wins. Through a series of nested if statements, we are able to cover all the possibilities. Using a switch statement: enum Winner {PLAYER1, PLAYER2, TIE}; int main() { int choice1, choice2; Winner winner; cout << "Choose rock (0), paper (1), or scissors (2):" << endl; cout << "Player #1: "; cin >> choice1; cout << "Player #2: "; cin >> choice2; switch (choice2 - choice1) { case 0: winner = TIE; break; case -1: case 2: winner = PLAYER1; break; case -2: case 1: winner = PLAYER2; }
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if (winner == TIE) cout << "\tYou tied.\n"; else if (winner == PLAYER1) cout << "\tPlayer #1 wins." << endl; else cout << "\tPlayer #2 wins." << endl; 3.11 } Using a switch statement and conditional expressions: enum Winner {PLAYER1, PLAYER2, TIE}; int main() { int choice1, choice2; cout << "Choose rock (0), paper (1), or scissors (2):" << endl; cout << "Player #1: "; cin >> choice1; cout << "Player #2: "; cin >> choice2; int n = (choice1 - choice2 + 3) % 3; Winner winner = ( n==0 TIE : (n==1 PLAYER1:PLAYER2) ); if (winner == TIE) cout << "\tYou tied.\n"; else if (winner == PLAYER1) cout << "\tPlayer #1 wins." << endl; else cout << "\tPlayer #2 wins." << endl; } The solution(s) to the quadratic equation is given by the quadratic formula:
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b b 2 4ac x = ------------------------------------2a But this will not apply if a is zero, so that condition must be checked separately. The formula also fails to work (for real numbers) if the expression under the square root is negative. That expression b2 + 4ac is called the discriminant of the quadratic. We define that as the separate variable d and check its sign. #include <iostream> #include <cmath> // defines the sqrt() function int main() { // solves the equation a*x*x + b*x + c == 0: float a, b, c; cout << "Enter coefficients of quadratic equation: "; cin >> a >> b >> c; if (a == 0) { cout << "This is not a quadratic equation: a == 0\n"; return 0; } cout << "The equation is: " << a << "x^2 + " << b << "x + " << c << " = 0\n"; double d, x1, x2; d = b*b - 4*a*c; // the discriminant if (d < 0) { cout << "This equation has no real solutions: d < 0\n"; return 0; } x1 = (-b + sqrt(d))/(2*a); x2 = (-b - sqrt(d))/(2*a); cout << "The solutions are: " << x1 << ", " << x2 << endl; }
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