Answers for "int c=0,a,temp; int n=153;//It is the number to check armstrong temp=n; while(n>0) { a=n%10; n=n/10; c=c+(a*a*a); } if(temp==c) System.out.println("armstrong number"); else System.out.println("Not armstrong number"); }"

armstrong number in java

int c=0,a,temp;  
    int n=153;//It is the number to check armstrong  
    temp=n;  
    while(n>0)  
    {  
    a=n%10;  
    n=n/10;  
    c=c+(a*a*a);  
    }  
    if(temp==c)  
    System.out.println("armstrong number");   
    else  
        System.out.println("Not armstrong number");

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.");
    }
}