C Armstrong Number and multiple coding questions

main()

{

int amm=0,temp=0,n,x;

printf("enter thealue of n");

scanf("%d",&n);

x=n;

for(;n>0;)

{

temp=n%10;

amm=amm+(temp*temp*temp);

n=n/10;

}

if(x==n)

{

printf("THE GIVEN NUMBER IS AMSTRONG NUMBER");

}

else

printf("NOT AMSTRONG NUMBER");

}

 

q6

 show(int i)
      3 {
      4 i&&show(i-1);
      5 printf("%d", i);
      6 }
      7  main()
      8 {
      9 show (10);
     10 }

    #include<stdio.h>
int main()
{
int i,j,k,l,m;
printf("give the number to find Amswrong (three digit)");
scanf("%d",&i);
j=i/100;
k=i%100;
l=k/10;
m=k%10;
j=j*j*j;
l=l*l*l;
m=m*m*m;
if(i==j+l+m)
{
printf("yes %d is an Amswrong numbern",i);
}
else
printf("not an Amswrong number");
}

Armstrong number is a number that in a given base is the sum of its own digits to the power of the number of digits.For example:1 + 5 + 3 = 153 To put it algebraically, let be an integer with representation dkdk − 1...d1 in base-b notation. If for some m it happens that then n is a narcissistic number (or an m-narcissistic number).In "A Mathematician's Apology", G. H. Hardy wrote:"There are just four numbers, after unity, which are the sums of the cubes of their digits:153 = 13 + 53 + 33370 = 33 + 73 + 03371 = 33 + 73 + 13and 407 = 43 + 03 + 73. These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals to the mathematician."However, it is not known if the only base 10 numbers equal to the sum of the cubes of their digits are 1, 153, 370, 371, and 407.Some "base ten" Armstrong numbers are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, ... (sequence A005188 in OEIS)Some "base three" Armstrong numbers are: 0,1,2,12,122Some "base four" Armstrong numbers are: 0,1,2,3,313

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static void reverse_file(char *, unsigned int);

int main(int argc, char **argv)
{
        char *ptr;
        unsigned int size, count = 0;
        FILE *fp;

        if(argc < 2)
        {
                printf("%s n", argv[0]);
                return EXIT_FAILURE;
        }

        if(!(fp = fopen(argv[1], "r")))
                return EXIT_FAILURE;

        fseek( fp, 0L, SEEK_END );
        count = ftell( fp );
        rewind(fp);

        if(!(ptr = malloc(count+1)))
                return EXIT_FAILURE;

        ptr[count] = '';

        fread(ptr, count, sizeof(char), fp);

        fclose(fp);

        reverse_file(ptr, count);

return EXIT_SUCCESS;
}

static void reverse_file(char *base, unsigned int size)
{
        char *end = &base[size-1];
        if(*end == 'n')
        {
                *end = '';
                --end;
        }

        for( ; end >= base; --end)
                if(*end == 'n')
                {
                        *end = ''; 
                        printf("%sn", end+1); /* if you want read into file, using fputs u can */
                }
                printf("%sn", base);
                free(base);
}


     regards,
G. Siva Prakash Reddy

Keep a table of names and their function pointers:

int myfunc1(), myfunc2();

struct
{
  char *name;
  int (*func_ptr)();
} func_table[] = {"myfunc1", myfunc1,
                  "myfunc2", myfunc2,};


Search the table for the name, and call via the associated function pointer.
         
       regards,
G. Siva Prakash Reddy.

this program reverse the content of file that is first character becomes the last character of the file so on.. previous posted answer  for the same question, it just reverse the file not the content...


#include<stdio.h>
#include<string.h>

int main( int argc, char **argv )
{
char *fileName;
FILE *fp;
int pos,i = 0, switchCase = 0;

        if( argc == 1 )
        {
                printf("n INCORRECT USAGE n");
                return 0;
        }
fileName = argv[1];

fp = fopen( fileName, "r" );
        if( fp )
        {
                pos = ftell(fp);
                fseek(fp, 0L, SEEK_END );
                pos = ftell(fp);
                char buffer[pos];
                memset( buffer, 0, pos );
                pos = pos - 1;
                        fseek( fp, pos, SEEK_SET );
                while ( pos >= 0 )
                {
                        fseek( fp, pos, SEEK_SET );
                        buffer[i] = fgetc( fp );
                        pos--;
                        i++;
                }
                printf("buffer :: %s n", buffer );

        }

        else
        {
                printf("n give file name doesn't exist n");
        }
return 0;
}

import java.util.*;

public class ArmstrongNos
{
    public astatic void main (String [] args)
    {
       int no = 0;
       int noOfDigits = 0;
       int tmp = 0;
      
       System.out.println ("Enter number");
       Scanner kb = new Scanner (System.in);
       no = kb.nextInt();
       
       tmp = no;
       do
          {
             if (tmp > o)
             tmp = tmp / 10;
             noOfDigits = noOfDigiis + 1;
           }while (tmp > 0);
         
          int quotient = 0;
          int sum = 0;
          int digit = 0;

          do
             {
                digit = quotient % 10;
                sum = sum + digit^(noOfDigits);
          quotient = quotient / 10;
              }while (quotient >0);

          if (sum == no)
             System.out.println(no+ " is an armstrong no");
          else
                System.out.println (no+" is not an armstrong no");
    }
}

1. import java.util*.;
2. public class armstrong number
3.     {
4.        public static void main(strings args);
5.        {
6.         keyboard scanner = new key board scanner;
7.         screen wirter = new screen writer;
   
8.         int kb writer.next=new kb writer
   
9.          {
10.          int counter =0;
11.          int sum;
12.          }       
    
13.          system.out.println(please Enter a number);
    
14.          }
    
15.           whilecounter(i<=o;i++);
16.         {
17.          for (i=no;&&no^;i++)
18.            {
19.            if (no^base++sum)
20.             system.out.println(sum+"number is an  armstrong number");
21.             else
22.             system.out.println(sum+"number is not an armtrong number");
    
23.           }
24.       }
    
25.   }
    

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static void rev_file(char *, unsigned int);

int main(int argc, char **argv)
{
        char *ptr;
        unsigned int size, count = 0;
        FILE *fp;

        if(argc < 2)
        {
                printf("%s n", argv[0]);
                return EXIT_FAILURE;
        }

        if(!(fp = fopen(argv[1], "r")))
                return EXIT_FAILURE;

        fseek( fp, 0L, SEEK_END );
        count = ftell( fp );  rewind(fp);

        if(!(ptr = malloc(count+1)))
                return EXIT_FAILURE;

        ptr[count] = '';fread(ptr, count, sizeof(char), fp);

        fclose(fp); reverse_file(ptr, count);

return EXIT_SUCCESS;
}

static void rev_file(char *base, unsigned int size)
{
        char *end = &base[size-1];
        if(*end == 'n')
        {
                *end = '';
                --end;
        }

        for( ; end >= base; --end)
                if(*end == 'n')
                {
                        *end = '';printf("%sn", end+1);
                }
                printf("%sn", base);
                free(base);
}

with regards,
 
P.Nirmal

#include<conio.h>
#include<math.h>
#include<stdio.h>
main()
{
int a,b,c,d,i;
for(i=500;i>=1;i--)
{
a=(i-(i%100))/100;
b=((i%100)-(i%10))/10;
c=(i%10);
d= (a*a*a)+(b*b*b)+(c*c*c);
if(d==i)
printf("%dn",i);
}
getch();
}

void main()
{
int n,r,k,s=0;
clrscr();
printf("n enter the value=n");
scanf("%d",&n);
k=n;
while(n>0)
{
r=n%10;
s=s+r*r*r;
n=n/10;
}
if(k==s)
printf("n the number is armstrong=%d",k);
else
printf("n the number is not armstrong=%d",k);
getch();
}

answer 1.
class anisha
{
    void main(int n)
    {
       int a,r,s;
           for(a=n;a>0;a=a/10)      
           {         
                b=a%10;
                s=s=(b*b*b);     
           }
           if(s==n)
            System.out.println(No. is armstrong no);
            else
            System.out.println(No. is not Armstrong);  
        }
}      
           
      

Answer 6

class abc
{
void main()
{
int a;
for(a=1;a>=100;a++)

{
System.out.println("how r  u");
}

#include<stdio.h>

#include<conio.h>

void main()

{

int a;

clrscr();

while (printf ("%dt",a++)-4);

getch();

}

posted by:

Rakesh Kumar

KIIT University,

Bhubaneswar,

Orissa,

India

ph: 09776131845

#include<stdio.h>
#include<conio.h>
void main()
{
 long int a,b=0,c,d;
 clrscr();
 printf("Enter the value");
 scanf("%ld",&a);
 d=a;
  while(a/10!=0)
  {
  printf("%ld",a%10);
  c=a%10;
  b=b+(c*c*c);
  a=a/10;
  }
  printf("%ld",a);
  b=b+(a*a*a);
  if(b==d)
  printf("nnnumber is armstrong %ld",b);
  else
  printf("nnnumber is not armstrong %ld",b);
 getch();
}

#include<stdio.h>
int armstrong(int);
main()
{
int i ,num;
printf("Enetr the
number to which you want to checkn");
scanf("%d", &i);
num=armstrong(i);
if( i
== num)
printf(" the number is armstrongn");
else
printf("The number is not an
armstrong numbern");
}
int armstrong(int i)
{
int r,sum=0;
while(i > 0)
{
r=i %
10;
sum = sum + r * r * r;
i = i / 10;
}
return(sum);
}

#include<stdio.h>
#include<conio.h>
void main()
{ int x,a,n,s=0;
clrscr();
printf(“Enter a number”);
scan f(“%d”,&n);
x=n;
while(n>0)
{a=n%10;
S=s+(a*a*a);
n=n/10;
}
If (s==x)
printf(“number is Armstrong”);
Else
printf(“number is not Armstrong”);
getch();
}
Output
Enter a number
1 5 3
Number is Armstrong

// C program to check whether a given number is Armstrong or not

#include<stdio.h>
#include<math.h>
int
main ()
{
  int i, n, m, k, cnt = 0, sum = 0, rem;
  printf ("Enter the numbert");
  scanf ("%d", &n);
  m = k = n;
  while (n > 0)
    {
      n = n / 10;
      cnt++;
    }
//printf("number of digits in the number are %d ",cnt);
  while (m > 0)
    {
      rem = m % 10;
      sum = sum + pow (rem, cnt);
      m /= 10;
    }
  if (k == sum)
    printf (" %d is a armstrong number n", sum);

  return 0;
}

This Program will Print the Armstrong Numbers between 1 to 500

Code
 class Program
    {
       
        static void Main(string[] args)
        {
            int[] number=new int[3];
            int digit1, digit2, digit3,temp;
            string a=null,b=null,c=null;
            int i;
            for (i = 1; i <= 500; i++)
            {
                string[] k = new string[3];
                string value= Convert.ToString(i);
 
                for (int j = 0; j < value.Length; j++)
                {
                    k[j] = value[j].ToString();
                }
 
                       if (k[0] != null)
                        {
                            a = k[0];
                        }
                   
                        if (k[1] != null)
                        {
                            b = k[1];
                        }
                   
                        if (k[2] != null)
                        {
                            c = k[2];
                        }
                 
           
                number[0] = Convert.ToInt32(a);
                number[1] = Convert.ToInt32(b);
                number[2] = Convert.ToInt32(c);
                
                digit1 = number[0];
                digit2 = number[1];
                digit3 = number[2];
 
                temp = (digit1 * digit1 * digit1) + (digit2 * digit2 * digit2) + (digit3 * digit3 * digit3);
                if (temp == i)
                {
                    Console.WriteLine("The Arm Strong Numbers are : "+i);
                }
 
            }
        }
    }

a program to accept a number and find how many digits it contain

//
declare
a number:=&a;
v_le number;
begin
select length (a) into v_le from dual;
dbms_output.put_line('the no of digits is '||v_le);
end;
//