JavaScript Question

Question

How do I put the output on the second line?

Intended output(example)

Enter·the·number·up·to·which·you·would·like·to·look·for·perfect·numbers:

500↵

Looking·for·perfect·numbers·from·1·to·500↵

6·is·a·perfect·number·it's·factors·are:·1·2·3·↵ 28·is·a·perfect·number·it's·factors·are:·1·2·4·7·14·↵ 496·is·a·perfect·number·it's·factors·are:·1·2·4·8·16·31·62·124·248·↵

What I have:

Enter·the·number·up·to·which·you·would·like·to·look·for·perfect·numbers:

500↵

Looking·for·perfect·numbers·from·1·to·↵

6·is·a·perfect·number·it's·factors·are:·1·2·3·↵ 28·is·a·perfect·number·it's·factors·are:·1·2·4·7·14·↵ 496·is·a·perfect·number·it's·factors·are:·1·2·4·8·16·31·62·124·248·↵

My code:

import java.util.Scanner;

class PerfectNumbers {

public static boolean isPerfect(int N)

{

int sum = 0,j=0;

//sum the factors of the number N if they are equal , then it's perfect number

for (int i = 1; i < N - 1; i++)

{

if (N%i == 0)

{

sum += i;

}

}

if (N == sum)

return true;

else

return false;

}

public static String factors(int N)

{

String s="";

int first=1;

for(int i =1; i< N; i++)

{

if(N%i == 0)

{

if(first == 1)

{

//System.out.println("s"+s);

s = Integer.toString(i);

first=0;

}

else

{

s+=" ";

s+=Integer.toString(i);

//System.out.println("s"+s);

}

}

}

//System.out.println("s"+s);

return s;

}

public static void main(String[] args) {

Scanner sc=new Scanner(System.in);

System.out.print("Enter the number up to which you would like to look for perfect numbers:");

System.out.println("Looking for perfect numbers from 1 to ");

int n=sc.nextInt();

String []str=new String[20];

for(int i = 1 ; i < n;i++)

{

if(isPerfect(i))

{

System.out.println(""+i+" is a perfect number it's factors are: "+factors(i)+" " );

//System.out.println("It's factors are: "+factors(i));

}

}

}

}

Answer 1

import java.util.ArrayList;

import java.util.Scanner;

class PerfectNumber

{

private Scanner scanner;

PerfectNumber() { scanner = new Scanner(System.in); }

//returns the arrayList of factors of N, including 1 but not N;

// ex. GetFactors(32,factors); ---> factors = {1,2,4,8,16}

public void GetFactors(long N, ArrayList<Long> factors)

{

factors.add(1L);

for (long iLoop=2; iLoop<N/2+1; iLoop++)

{

if (N % iLoop==0)

{

factors.add(iLoop);

}

}

}

boolean IsPerfectNumber( long nLongIntNum, ArrayList<Long> factors)

{

GetFactors(nLongIntNum,factors);

int N = factors.size();

boolean booleanReturn=false;

long sumFactors = 0;

for (int iLoop=0; iLoop<N; iLoop++)

{

sumFactors += (long) factors.get(iLoop);

}

return (booleanReturn = (sumFactors==nLongIntNum));

}

void DisplayPerfectNumber( long nLongIntNum, ArrayList<Long> factors)

{

System.out.print( nLongIntNum + " is a perfect number. It's factors are :");

int N = factors.size();

for (int iLoop=0; iLoop<N; iLoop++)

{

System.out.print(factors.get(iLoop)+ " ");

}

System.out.println("");

}

public void Go()

{

long N=-1;

ArrayList<Long> factors = new ArrayList<Long>();

// 2 and 3 are not perfect numbers

while (N<3)

{

System.out.print(" Enter the number up to which you would like to look for perfect numbers :>");

N = scanner.nextLong();

}

System.out.println("Scanning for perfect numbers between 2 and " + N);

for (long iLoop=3; iLoop<=N; iLoop++)

{

if (iLoop%10000==0) { System.out.println(" checking " + iLoop + "...."); }

if (IsPerfectNumber(iLoop,factors))

{

DisplayPerfectNumber(iLoop,factors);

}

factors.clear();

}

}

public static void main (String args[])

{

new PerfectNumber().Go();

}

}