題目描述

源代碼

思路
依次比較每個位置公差的最大公約數

#include <iostream>
#include <algorithm>
using namespace std;long gcd(long a, long b){  //求兩個數最大公約數if(b == 0){return a;}return gcd(b, a % b);
}int main() {long n;cin >> n;long a[n + 10];for (long i = 0; i < n; i++) {cin >> a[i];}sort(a, a + n);long d = a[1] - a[0];if (d == 0) {  //如果公差等于0cout << n;return 0;}for (long i = 2; i < n; i++) {d = gcd(a[i] - a[i - 1], d);}cout << (a[n - 1] - a[0]) / d + 1;return 0;
}