Please help What two Boolean operators can you use to narrow your search results?
1 answer:
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Answer:
Explanation:
The following program is written in Java and creates three levels with the different set of randomly generated arithmetic problems. As the player goes getting more points they advance every 5 points to the next level.
import java.util.Random;
import java.util.Scanner;
class Main {
private static Random random = new Random();
private static Scanner in = new Scanner(System.in);
static int points = 0;
public static void main(String[] args) {
while (true) {
System.out.println(points);
if (points < 5) {
System.out.println("Starting Level 1");
levelOne();
} else if (points < 10) {
System.out.println("Starting Level 2");
levelTwo();
} else if (points < 15) {
System.out.println("Starting Level 3");
levelThree();
}
}
}
public static boolean levelOne() {
int numberOne = random.nextInt(10);
int numberTwo = random.nextInt((10 - numberOne));
int answer = numberOne + numberTwo;
int count = 0;
while (points < 5 && count < 2) {
System.out.println("What is " + numberOne + " + " + numberTwo + "?: ");
int userChoice = in.nextInt();
if (userChoice == answer) {
points += 1;
levelOne();
} else {
count += 1;
}
}
return false;
}
public static boolean levelTwo() {
int numberOne = random.nextInt(10);
int numberTwo = random.nextInt(10);
int answer = numberOne + numberTwo;
int count = 0;
while (points < 10 && count < 2) {
System.out.println("What is " + numberOne + " + " + numberTwo + "?: ");
int userChoice = in.nextInt();
if (userChoice == answer) {
points += 1;
levelTwo();
} else {
count += 1;
}
}
return false;
}
public static boolean levelThree() {
int numberOne = random.nextInt(10);
int numberTwo = random.nextInt(10);
int answer = numberOne - numberTwo;
int count = 0;
if (answer <= 0) {
levelThree();
}
while (points < 15 && count < 2) {
System.out.println("What is " + numberOne + " - " + numberTwo + "?: ");
int userChoice = in.nextInt();
if (userChoice == answer) {
points += 1;
levelThree();
} else {
count += 1;
}
}
return false;
}
}
