Question

C program--write a program to find whether the given number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 371 is since 3**3 + 7**3 + 1**3 = 371.
Output:

Enter a number, or * to quit : 432

4**3 + 3**3 + 2**3 is not = 432
Enter a number, or * to quit : 371

3**3 + 7**3 + 1**3 is = 371
Enter a number, or * to quit : *

Please modify my code but not using atoi...

#include <stdio.h>
int main()
{
    int number, oriNumber, rem, result = 0;

    printf("Enter a number, or * to quit: ");
    scanf("%d", &number);

    oriNumber = number;

    while (oriNumber != 0)
    {
        rem = oriNumber%10;
        result += rem*rem*rem;
        oriNumber /= 10;
    }

    if(result == number)
        printf("%d**3 + %d**3 + %d**3 is not = %d ",number);
    else
        printf("%d**3 + %d**3 + %d**3 is = %d.",number);

    return 0;
}

Answer 1

#include <stdio.h>

int main() {
    int number, oriNumber, rem, result = 0, d1, d2, d3;

    printf("Enter a number, or * to quit: ");
    while (scanf("%d", &number) == 1) {
        oriNumber = number;
        result = 0;
        while (oriNumber != 0) {
            rem = oriNumber % 10;
            result += rem * rem * rem;
            oriNumber /= 10;
        }

        oriNumber = number;
        d1 = oriNumber / 100;
        oriNumber %= 100;
        d2 = oriNumber / 10;
        oriNumber %= 10;
        d3 = oriNumber;

        if (result == number)
            printf("%d**3 + %d**3 + %d**3 is = %d\n", d1, d2, d3, number);
        else
            printf("%d**3 + %d**3 + %d**3 is not = %d\n", d1, d2, d3, number);
        printf("Enter a number, or * to quit: ");
    }
    return 0;
}