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Armstrong Number

/* Check Armstrong Number */

#include<stdio.h>
int main()
{
    int n, sum = 0, t, r;

    printf("Please enter a number: ");       
    scanf("%d",&n);

    for(t=n;t>0;t=t/10)
    {
        r = t%10;
        sum = sum + r*r*r;
    }
     if ( n == sum )
        printf("%d is an armstrong number.",n);
    else
        printf("%d is not an armstrong number.",n);

    return 0;
}

Related Links:
Print Armstrong Numbers Upto N
http://cbasicprogram.blogspot.in/2012/05/print-armstrong-numbers-upto-n.html

5 comments:

  1. Replies
    1. for which number you have tested it ??

      Delete
    2. abe its correct

      Delete
  2. #include

    using namespace std;

    int main()
    {
    int an= 371;
    int t=0,t1=an, sum=0;

    while(an>0){
    t = an%10;
    sum = sum + t*t*t;
    t = 0;
    an = an/10;

    }
    if(t1 == sum){
    cout<<"armstrong number";
    }else{
    cout<<"not armstrong number";
    }

    return 0;
    }

    ReplyDelete
  3. //using C#
    int num, sum = 0;
    num = Convert.ToInt32(Console.ReadLine());
    for (int i = num; i > 0;)
    {
    sum += (i % 10) * (i % 10) * (i % 10);
    i = i / 10;

    }
    if(sum==num)
    {
    Console.WriteLine("No is armstrong");
    }
    else
    {
    Console.WriteLine("No is not armstrong ");
    }
    Console.ReadLine();
    }

    ReplyDelete