9069 等差数列

思路：先通过数列第1、2小的值求最大公约数得到公差，然后再推算出项数公式算有多少项

 

#include<bits/stdc++.h>
#define f(i,s,e) for(int i = s; i <= e; i++)
#define ll long long
using namespace std;
const int N = 1e6+10,inf = 0x3f3f3f3f;
int a[N],n; //存储给的n个数
int minn1 = inf,minn2 = inf; 
int gcd(int a,int b)
{
    if(b == 0) return a;
    else return gcd(b,a % b);
}
int main()
{
    cin >> n;
    f(i,1,n) scanf("%d",&a[i]);
    sort(a + 1,a + 1 + n); //从小到大排序
    f(i,2,n) //比较a[i] 和 a[i-1]来获取第一小和第二小差值
    {
        if(a[i] - a[i - 1] <= minn1)
        {
            minn2 = minn1;
            minn1 = a[i] - a[i - 1];
        }
        else if(a[i] - a[i - 1] <= minn2)
            minn2 = a[i] - a[i - 1];
    } 
    int d = gcd(minn1,minn2); //gcd：辗转相除法求最大公约数 
    int ans = (a[n] - a[1]) / d + 1;
    cout << ans;
    return 0;
}