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Sagot :
The only way without a while loop and no method I can think of is use switch with every possible variation of the four digit binary which is 15.
Here is the starter code:
import java.util.Scanner;
public class MyClass {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int digit = sc.nextInt();
switch(digit) {
case 0000:
System.out.println("0");
break;
case 0001:
System.out.println("1");
break;
case 0010:
System.out.println("2");
break;
case 0011:
System.out.println("3");
break;
case 0100:
System.out.println("4");
break;
… (fill in other cases)
}
}
}
Use this link: https://www.electronics-tutorials.ws/binary/bin_3.html
There might be a better way, but without loops or methods this is all I got.
Here is the starter code:
import java.util.Scanner;
public class MyClass {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int digit = sc.nextInt();
switch(digit) {
case 0000:
System.out.println("0");
break;
case 0001:
System.out.println("1");
break;
case 0010:
System.out.println("2");
break;
case 0011:
System.out.println("3");
break;
case 0100:
System.out.println("4");
break;
… (fill in other cases)
}
}
}
Use this link: https://www.electronics-tutorials.ws/binary/bin_3.html
There might be a better way, but without loops or methods this is all I got.
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