armstrong numbers

1int main(){
2    
3    int a;
4    cin >> a;
5    int check = a;
6    int sum = 0;
7    while(a!=0){
8        int rem = a%10;
9        sum += pow(rem,3);  //include cmath
10        a/=10;
11    }
12
13    if(sum==check){
14        cout << "armstrong";
15    } else {
16        cout << "not armstrong";
17    }
18    return 0;
19}

1sum of cubes of the digits

1temp=n;    
2while(n>0)    
3{    
4r=n%10;    
5sum=sum+(r*r*r);    
6n=n/10;    
7}    
8if(temp==sum)    
9printf("armstrong  number ");    
10else    
11printf("not armstrong number");    
12return 0;  

1#include <stdio.h>
2#include <stdlib.h>
3#include <math.h>
4
5int cube(int a)
6{
7    int c;
8    c =  a*a*a;
9    return c;
10}
11
12int armnum(int *a)
13{
14    int x = *a, n = 0, rem, r = 0;
15    while (x != 0) {
16        x /= 10;
17        n++;
18    }
19    x = *a;
20    while (x != 0) {
21        rem = x % 10;
22        r += cube(rem);
23        x /= 10;
24    }
25    if(r == *a){
26        return 1;
27    }
28}
29
30int main()
31{
32    int a, y;
33    scanf("%d", &a);
34    y = armnum(&a);
35    if(y == 1){
36        printf("It is an Armstrong number.");
37    }
38    else{
39        printf("It is not an Armstrong number.");
40    }
41}

1Numbers -- Armstrong numbers
2Write a method that can check if a number is Armstrong  number
3 
4Solution:
5public  static  boolean ArmStrongNumber (int  num) {
6int a = 0,    b =0,    c= num;
7while(num>0){
8              a=num%10; 
9              num/=10; 
10              b=b+(a*a*a);
11}
12 
13if(c==b) {
14return true;
15}
16return false;
17}

1// Armstrong Numbers
21, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748...