shibaji.paul

 Profile Answers by shibaji.paul 

• Apr 20th, 2006

 

Check Armstrong Number
1.start
2. input n.
3. n1:=n.
4. let s := 0.
5. r := n1%10.
6. s:=s+ r*r*r.
7. n1 :=n1/10.
8. if n1<>0 then goto step 5.
9.  if n=s then
10. print "ARMSTRONG NO"
11. else
12.print "NOT ARMSTRONG NO"
13. end

The above algorithm would check only the 3 digit numbers, if u need it for all digit numbers then u need to request me again. CHAO.

vvsureshreddy

 Profile Answers by vvsureshreddy 

• Aug 29th, 2006

 

Would appreciate the Generic algoirthm for the problem :-)

saurabh

• Sep 16th, 2006

 

could you please send armstrong algo for the all digits

thnking you

saurabh

jyoti

• Sep 27th, 2006

 

simply add a do-while loop to it for (n>0) and it whould wor for all number of digits

 

jyoti

www.jyoticlub.jaijyoti.com

JonBruse

 Profile Answers by JonBruse 

• Nov 14th, 2006

 

in C#:

public static bool isArmstrong(int intInteger)

{

// Declarations

int Length, Total1, Temp;

string strString = intInteger.ToString();

Total1 = 0;

//get length of number

Length = strString.Length;

//iterate through each digit, get n power, add to sum

for (int i = 0; i <= Length - 1; i++)

{

Temp = Convert.ToInt32(strString.Substring(i, 1).ToString());

Total1 += Convert.ToInt32(System.Math.Pow(Convert.ToDouble(Temp), Convert.ToDouble(Length)));

}

if (Total1 == intInteger)

{

return true;

}

else

{

return false;

}

}

Sukumar Paul

 Profile Answers by Sukumar Paul 

• Aug 1st, 2010

 

First of all I would like to introduce to you with amstrong number.

Lets take an example: 153 which is amstrong number because 153 = (1*1*1) + (5*5*5) + (3*3*3) i.e.
153 = 1^3 + 5^3 + 3^3(where ^ indicates power i.e. 3^3 is 3*3*3)

Now logic behind writing the program to check the amstrong number:

1: initialize the number i.e. num
y <-- num, sum <-- 0, temp <-- 0
2: repeat step untill y !=0
i: calculate y%10 and store it in temp
ii: calculate temp*temp*temp and add it with the previous value of sum
iii: find y <-- y/10
3: check whether sum is equal to num , if so num is amstrong number
4: otherwise num is not an amstrong number

Here is the actual program which implements the above logic and find out whether a number is amstrong number or not:

<-------------------------------------------------------------------------------------------->
#include
<stdio.h>
void main()
{
int num=153;
int y=num,sum=0,temp;
while(y != 0)
{
temp = y%10;
sum = sum+temp*temp*temp;
y = y/10;
}
if(sum == num)
printf("amstrong number");
else
printf("Not amstrong number");
}
<------------------------------------------------------------------------------------>

Kapil S Shinde

• Feb 25th, 2014

 

Code
start


Accept num


 


noOfDigit=calculate number of digits   //use counting digit function


 


while(num!=0)


{


remainder=inum%10;


sum=sum+pow(remainder,noOfDigit);     //use power function from math.h header file


num=num/10;


}


if(sum==num)


 return true


else


 return false


stop

jbode

 Profile Answers by jbode Questions by jbode

• Mar 7th, 2014

 

First, count up the number of digits in the candidate value; an easy way to do this is take the floor value of the log of the number and add 1 (for example, floor(log10(1234)) = 3, so 3 + 1 = 4 digits).

Then extract each digit (an easy way to do this is to use a remainder or modulus operation), raise it to the power of the number of digits, and sum the results together.

Compare this sum to your candidate value; if they're equal, then the candidate value is narcissistic. Attached is an example written in C that examines a single number or range of numbers using base 10, along with some sample output.

Code
/**


 * This program analyzes a range of numbers to see if they are


 * narcissistic or Armstrong numbers (sum of its own digits, each


 * raised to the power of the number of digits) in base 10.


 *


 * The output is simply the number if it meets the necessary


 * criteria.


 */


#include <stdio.h>


#include <stdlib.h>


#include <ctype.h>


#include <math.h>


 


/**


 * Test a candidate value to see if it is narcissisitc.


 * Returns true (1) or false (0).


 */


int test( long cand )


{


  int digits = (int) log10( (double) cand ) + 1;


  long sum = 0;


  long temp = cand;


 


  for ( int i = 0; i < digits; i++ )


  {


    int curdigit = temp % 10;


    temp /= 10;


    sum += (long) pow( (double) curdigit, digits );


  }


 


  return sum == cand;


}


 


/**


 * Process the command line arguments; must provide


 * a valid integer or range of integers.


 */


int getargs( int argc, char **argv, long *start, long *end )


{


  int result = 1;


  if ( argc < 2 )


  {


    result = 0;


  }


  else if ( argc >= 2 )


  {


    char *chk;


    long val = strtol( argv[1], &chk, 0 );


    if ( isspace( *chk ) || *chk == 0 )


      *start = *end = val;


    else


      result = 0;


 


    if ( argc > 2 )


    {


      val = (int) strtol( argv[2], &chk, 0 );


      if ( isspace( *chk ) || *chk == 0 )


        *end = val;


      else


        result = 0;


    }


  }


 return result;


}


 


/**


 * Main program; loop through the specified range


 * of values and test each for narcissism; if


 * the number is narcissistic, write it to


 * standard output.


 */


int main( int argc, char **argv )


{


  long start=0, end=0;


  const char *usage = "USAGE: %s start [end]n";


  if ( !getargs (argc, argv, &start, &end ))


  {


    fprintf( stderr, usage, argv[0] );


    return EXIT_FAILURE;


  }


 


  for( long t = start; t <= end; t++ )


    if ( test( t ))


      fprintf( stdout, "%ld


", t );


 


  return EXIT_SUCCESS;


}


 


[fbgo448@n9dvap997]~/prototypes/armstrong: gcc -o armstrong -std=c99 -pedantic -g -Wall -Werror armstrong.c -lm


[fbgo448@n9dvap997]~/prototypes/armstrong: ./armstrong 153


153


[fbgo448@n9dvap997]~/prototypes/armstrong: ./armstrong 154


[fbgo448@n9dvap997]~/prototypes/armstrong: ./armstrong 0 10000


0


1


2


3


4


5


6


7


8


9


153


370


371


407


1634


8208


9474