找出n个数字的公因子数（不包括“ 1”作为公因子）

Question

Here is my code that i have written in gcc. Don't know if my logic is incorrect or i'm making some other mistake.The output is coming as 0 every time.

int main()
{
int n,a[20],count=0;
cin>>n;

for(int i=0;i<n;i++)
{
    cin>>a[i];
}

for(int k=0;k<n;k++)
{
    int c=0;
    for(int j=2;j<n;j++)
    {
        if(a[k]%j==0)
        {
            c++;
        }
        else
        {
            continue;
        }
    }
    if(c==n)
    {
      count++;
    }
}
cout<<count;
}

Answer 1

Your approach isn't correct from what I gather. You should take every number from 2 to max(a) [1] and check if it is a factor/divisor of every number in a . If it is, you can increment count .

---

[1] or even better max(sqrt(a[i]) for a[i] in a) in pseudo-code-ish syntax.

Answer 2

You got the loops in the wrong order. You are checking if there are n dividers of any of the numbers.

Try swapping the loops. Then the count will be of the number of numbers that divide the input.

You also have the wrong upper bound. You need the highest possible divider.

int max_divider = 2;
for (i = 0; i < n; i++) {
    if (a[i] > max_divider) 
    {
        // Naive approach
        max_divider = a[i];
    }
}
// For each number 2..max_divider
for(int j=2;j <= max_divider;j++)
{
    int c=0;
    for(int k=0;k<n;k++)
    {
        if(a[k]%j==0)
        {
            c++;
        }
        else
        {
            continue;
        }
    }
    if(c==n)
    {
      count++;
    }
}

Answer 3

Replace this loop

for(int j=2; j<n; j++)

with

for(int j=2; (j < a[k]); j++)

Note that you can get rid of else part, it is having no effect.