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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");
}
} |
 |
| asked by Sankhya Gowda (taankkkkk) 7 years ago | |
2 | |
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