Answers for "check armstrong number in c"

0

c program to check the number is armstrong or not

#include <stdio.h>
#include <math.h>
void main ()
{
    int n,t,r,c,s=0;
    printf("Enter the number: ");
    scanf("%d",&n);
    t=n;

        while(t!=0)
        {
            r=t%10;
            c=pow(r,3);
            s=s+c;
            t=t/10;
        }

        if(s==n)
            printf("Armstrong");
        else
            printf("Not Armstrong");
}
Posted by: Guest on June-24-2021
1

armstrong number in c

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

int cube(int a)
{
    int c;
    c =  a*a*a;
    return c;
}

int armnum(int *a)
{
    int x = *a, n = 0, rem, r = 0;
    while (x != 0) {
        x /= 10;
        n++;
    }
    x = *a;
    while (x != 0) {
        rem = x % 10;
        r += cube(rem);
        x /= 10;
    }
    if(r == *a){
        return 1;
    }
}

int main()
{
    int a, y;
    scanf("%d", &a);
    y = armnum(&a);
    if(y == 1){
        printf("It is an Armstrong number.");
    }
    else{
        printf("It is not an Armstrong number.");
    }
}
Posted by: Guest on August-26-2020
0

armstrong number in c

//Bismillahir Rahmanir Rahim
#include <stdio.h>
#include <math.h>

int digit(int n, int count){
    while( n != 0){
        n/=10;
        count++;
    }
    return count;
}

int main(){
	    int n, rem, num, sum = 0, count = 0, digits;
	    while(1){
	        printf("Enter a number (type 0 to exit) : ");
            scanf("%d", &n);
            
            if(n==0){
                break;
            }
            digits = digit(n,count);
           
    		num=n;
    		while(num != 0){
    			rem = num % 10;
    			sum += pow(rem,digits);
    			num /= 10;
    		
    			}
    		
    		if(sum == n){
    			printf("%d is an armstrong number!\n", n);
    			sum = 0;
    		}else{
    			printf("%d is not an armstrong number!\n", n);
    			sum = 0;
    		}
	
	    }
	    
	return 0;
}
Posted by: Guest on August-13-2021

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