a  a + d  a + 2d .....  a + kd
每一项与第一项的差一定是d的倍数
len = (a[max] - a[min]) / d(所有d的最大公约数) + 1
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
typedef long long LL;
const int N = 100010;
int n;
int a[N];

int gcd(int a,int b){
    return b ? gcd(b,a % b) : a;
}

int main(){
    scanf("%d",&n);
    for(int i = 0;i < n;++i)    scanf("%d",&a[i]);
    sort(a,a + n);
    int d = 0;
    for(int i = 1;i < n;++i)    d = gcd(d,a[i] - a[0]);
    if(!d)  printf("%d\n",n);
    else{
        LL len = (LL)(a[n - 1] - a[0]) / d + 1;
        printf("%lld\n",len);
    }   
    return 0;
}