朴素版：
d最大的可能是a[i]-a[i-1]的最小间隔
求出之后 一直从大到小遍历
看哪一个符合

#include <iostream>
#include <algorithm>
using namespace std;
const int n = 100005;
int a[n];
int N;
int f(int d)
{
    int n = (a[N - 1] - a[0]) / d + 1;
    int x = a[0];
    int index = 0;
    while (x <= a[N - 1])
    {
        if (x == a[index])
            index++;
        x += d;
    }
    return (index == N ? n : 0);
}
int main()
{
    cin >> N;
    for (int i = 0; i < N; i++)
        scanf("%d", &a[i]);
    sort(a, a + N);
    int l = 1, r = 1e8;
    for (int i = 0; i < N - 1; i++)
        r = min(a[i + 1] - a[i], r);
    if (r == 1)
    {
        int n = (a[N - 1] - a[0]) / 1 + 1;
        cout << n;
        return 0;
    }
    int ans = 0;
    while (r--)
    {
        int t = f(r + 1);
        if (t)
        {
            ans = t;
            break;
        }
    }
    cout << ans;
    cout << " " << r + 1 << endl;
    return 0;
}