ICSE − Q & A

Anyone from Mumbai

I'm so glad I found ResPaper! Nice to have mates from over the country(well mainly Bangalore and Mumbai but yeah :P)

Well, I should be saying all this on the last board exam but well passion, mistress of desires and it sways it to the mood of what it likes or loathes.

+ 1 more questions by shauns  

Guys cant we use string functions and stuff for Disarium number, Other types of numbers?
asked by Aquaman- King of the Seven Seas (shauns) 5 years ago
2

Why it is not working. It is a program for Duck Number

import java.util.Scanner;
public class magic
{
    public static void main (String args[])
    {
        Scanner sc = new Scanner(System.in);
        System.out.println("ENTER NO TO CHECK");
        int num = sc.nextInt();
        String str = Integer.toString(num);
        int len = str.length();
        int count = 0;
        int i;
        for(i=0 ; i <len ; i++)
        {
            if ( str.charAt(i) == '0')
            count ++;
        }
        
        
        if(count > 0 && str.charAt(0) != '0')
        System.out.println("It is a Duck Number");
        else
        System.out.println("It is not a Duck Number");
            
    }
}

+ 2 more questions by teleansh  

what is use of if (num == 0) { num = sum; sum = 0; in magic number code
asked by Ansh Anand (teleansh) 5 years ago
0

Why it is not working. This code is for Disarium Number public class sum { public static void main (int num) { int i = 0; int sum = 0; String str = Integer.toString(num); int len = str.length(); while (num > 0) { i = num%10; sum = sum + (int)Math.pow( i , len); len -- ; // or len = len -1; num = num / 10; } if ( sum == num) System.out.println("It is a Disarium Number"); else System.out.println("It is not a Disarium Number"); } }
asked by Ansh Anand (teleansh) 5 years ago
2

any one here commercial applications....

If there is difference between two things asked for 2 marks and we have only one point are examples accepted?

What is the scope of local,instance & class variable?

Hi

Accept a no. from the user and find if it is a fascinating number.( FASCINATING NUMBER : When a number( 3 digit or more) is multiplied by 2 and 3 ,and when both these products are concatenated with the original number all digits from 1 to 9 are present exactly once, regardless of the number of zeroes)

hi guys these are two answers to the disarmium question everyone has been bothered about- without using string functions
import java.util.Scanner;
public class disarmy1
{
    void hello()
    {  Scanner in = new Scanner(System.in);
        System.out.println("enter the number ahole");
        int s = in.nextInt();
        int a = s;
        int b=s;
        int d=0;//to find the digits
        int c=0;//to count the number of digits
        int i,e=0,sum=0;
        while(s>0)
        {  d= s%10;
            c++;
            s=s/10;
        }
        for(i=c;i>=1;i--)
        { 
            
            
          e= a%10;
           sum = sum+ (int)(Math.pow(e,i));//since i is equal to number of digits
            a=a/10;
        
       }
       if(sum==b)
       System.out.println("disarmium number");
       else
       System.out.println("not");
    }
}
//THIS IS THE SECOND ANSWER TO THE QUESTION WHICH IS A LITTLE HARDER TO COPREHEND 
import java.util.Scanner;
public class disarmy2
{
    void hello()
    {
        //this is using arrays
         Scanner in = new Scanner(System.in);
        System.out.println("enter the number ahole");
        int s = in.nextInt();
        int n =s;
        int q=s;
        int d=0;
        int i;
        int c=0;
        int x=0;
        int sum=0;
        while(s>0)//to find the number of digits the number has
        {d = s%10;
            c++;
            s=s/10;
        }
        int a[]= new int[c];
        for(i=0;i<c;i++)//to put all these digits into an array
        {
            x=n%10;
            a[i]= x;
            n=n/10;
        }
        for(i=c-1;i>=1;i--)//taking array from the last element and adding the power of index+1 of that number to sum
        {
            sum = sum+ (int)Math.pow(a[i],i+1);
        }
        if(sum==q)
        System.out.println("disarmium");
        else
        System.out.println(" not");
    }
    
}

This is for all.

Can everyone please decribe how you he/she divide his/her time for completing the paper.

and what would you do additional in your copy to fetch good marks from the examiner.