How do you write a C program to check for Armstrong numbers in three digit values from 100 to 999 using do while loop?

Answer 1

import java.io.*;

public class Arms

{

public static void main(String args[]) throws Exception

{

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter the number to be printed: ");

int n=Integer.parseInt(br.readLine());

int num,sum,r,temp;

System.out.println("Armstrong numbers upto given limit are:\n");

for(num=1;num<n;num++)

{

temp=num;

sum =0;

while (temp!=0)

{

r=temp%10;

sum=sum+(r*r*r);

temp=temp/10;

}

if(sum==num)

{

System.out.println(num);

}

}

}

}

Regards

Ramakrishna Nallapati

Answer 2

#include<limits.h> // defines CHAR_BIT

#include<math.h> // defines pow()

#include<stdio.h> // defines printf()

bool is_armstrong (const int n, const int b=10) {

// Returns true if n base b is narcissistic.

// Preconditions: n>=0, b>=2

int digit[sizeof (int) * CHAR_BIT]; // caters for all digits in all bases

int i, j, k; // local variables

i = 0; // count of digits (so far)

j = n;

while (j) {

digit[i] = j%b; // store least significant digit

j/=b; // shift-right by one digit

++i; // increment count

}

k = 0; // sum of all digits raised to the power of i

for (j=0; j<i; ++j) k+=pow (digit[j], i);

return k==n;

}

int main (void) {

for (int n=100; n<=999; ++n)

if (is_armstrong (n)) // base 10

printf ("%d ", n);

return 0;

}

Output:

153 370 371 407