CODEWITHSUNDEEP
c
plaguedoctor24
plaguedoctor24

Reputation: 13

How to check if array elements are perfect numbers?

I'm trying to make a program where the array size is not entered by the user,but the elements are, until 0 is entered.Now I want to check for each element which one is a perfect number,for that I have to do the sum of the divisors.Problem is I can't manage to do the sum of divisors for each element in the array,instead it adds all the divisors of all the elements in the array.

#include <stdio.h>

int main()
{
    int n = 1000, i, j, sum = 0;
    int v[n];

    for (i = 1; i < n; i++)
    {
        scanf("%d", &v[i]);
        if (v[i] == 0)
        {
            break;
        }

        for (j = 1; j < v[i]; j++)
        {
            if (v[i] % j == 0)
            {
                printf("%d", j);
                sum = sum + j;
            }
        }
    }

    printf("\n%d",sum);

    return 0;
}

OUTPUT

Upvotes: 1

Views: 627

Answers (1)

0___________
0___________

Reputation: 67805

Brut force check can be very expensive. It is faster to build the table of perfect numbers using Euclides formula and then simple check if the number is perfect.

static unsigned long long getp(int x)
{
    return (2ULL << (x - 2)) * ((2ULL << (x - 1)) - 1);
}

int isperfect(unsigned long long x)
{
    const int primes[] = {2, 3, 5, 7, 13, 17, 19, 31};
    static unsigned long long array[sizeof(primes) / sizeof(primes[0])];
    int result = 0;

    if(!array[0])
    {
        for(size_t index = 0; index < sizeof(primes) / sizeof(primes[0]); index++)
        {
            array[index] = getp(primes[index]);
        }
    }
    for(size_t index = 0; index < sizeof(primes) / sizeof(primes[0]); index++)
    {
        if(x == array[index])
        {
            result = 1;
            break;
        }
    }
    return result;
}

The array of perfect numbers is build only one time on the first function call.

And some usage (your code a bit modified)

int main(void)
{
    size_t n = 1000, i;
    unsigned long long v[n];

    for (i = 1; i < n; i++)
    {
        scanf("%llu", &v[i]);
        if (v[i] == 0)
        {
            break;
        }
        printf("%llu is %s perfect number\n", v[i], isperfect(v[i]) ? "" : "not");
    }

    return 0;
}

https://godbolt.org/z/exMs345xb

Upvotes: 1

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