Find the perfect number between two numbers-CodePudding

I'm new to Java, and in order to practice I found a task on the Internet:

"Find all the perfect numbers between the two numbers you enter."

By the way - a perfect number is a natural number equal to the sum of all its own divisors. So I got to work and ran into such a problem that when I enter two numbers.

For example: 1 100, I get the correct answer in the console: 6,28. But if the first number is, for example, 100 and the second is 500 I get this output to the console: 6,28,496,, while I should get only 496. That is, for some reason, the minimum value does not shorten the search for a perfect number.

Here's where I ended up:

import java.util.Scanner;

public class IsPerfect {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int min = in.nextInt();
        int max = in.nextInt();

        System.out.println(min  " to " max " perfect numbers:");
        for (min = 1; min <= max; min  ) {
            int sum = 0;
            for (int e = 1; e < min; e  ) {
                if ((min % e) == 0) {
                    sum  = e;
                }
            }

            if (sum == min) {
                System.out.print(sum  ",");
            }
        }
    }
}

If anyone has any advice on how to fix this error, I would sincerely appreciate it.

CodePudding user response：

You are assigning the minimum value to 1, no matter what the user enters. Change your outer for loop to start from the minimum value the user enters:

import java.util.Scanner;

public class IsPerfect {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int min = in.nextInt();
        int max = in.nextInt();

        System.out.println(min  " to " max " perfect numbers:");
        for (; min <= max; min  ) {
            int sum = 0;
            for (int e = 1; e < min; e  ) {
                if ((min % e) == 0) {
                    sum  = e;
                }
            }

            if (sum == min) {
                System.out.print(sum  ",");
            }
        }
    }
}

CodePudding user response：

I think that was the point.

import java.util.Scanner;

public class IsPerfect {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in); 
        
           int min = in.nextInt();
           
           int max = in.nextInt();
        
           System.out.println(min  " to " max " perfect numbers:");
        for (int i = min; i <= max; i  )  
            {  
                int sum = 0;  
                for (int j = 1; j < i; j  )  
                {  
                    if (i % j == 0)  
                        sum = sum   j;  
                }  
                if (sum == i)  
                    System.out.println(i ", ");  
            }

    }

}

CodePudding user response：

Let me give you a piece of advise: integers in Java are defined as:

int 4 bytes: from -2,147,483,648 to 2,147,483,647

As a perfect number is positive, you only need to consider the ones from 0 to 2,147,483,647, and they are not that common:
6, 28, 496, 8128, 33550336 and the next one is already larger than the upper limit. So you just put those numbers in a basic collection (like an array) and you verify against that.

CodePudding user response：

Once entered, min and max should remain constant to make the code easier to read.

You need two loops, which conventionally use i and j as their loop values.

Change your loops to:

for (int i = min; i <= max; i  ) {  
    int sum = 0;  
    for (int j = 1; j <= i/2; j  ) {  
        if (i % j == 0) {
            sum  = j;  
        }
    }
    if (sum == i) {
        System.out.println(i   ", ");  
    }
}

Note the slight performance gain by only iterating j up to i/2

A further and greater improvement to performance is this for the inner loop, which only iterates up to the square root of i and finds factors 2 at a time (eg for 28, if 2 is a factor then so is 14):

int root = (int)Math.sqrt(i);
for (int j = 1; j <= root; j  ) {  
    if (i % j == 0) {
        sum  = j;
        sum  = i / j;
    }
}