Answers for "armstrong example with even number in it"

2

armstrong number

temp=n;    
while(n>0)    
{    
r=n%10;    
sum=sum+(r*r*r);    
n=n/10;    
}    
if(temp==sum)    
printf("armstrong  number ");    
else    
printf("not armstrong number");    
return 0;
Posted by: Guest on January-09-2021
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

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