https://codeforc.es/problemset/problem/1216/D
贪心：
找出最大的a[i]，令为t，即假设这个t就是原来的x
然后b[i]=t-a[i]; b[i]表示每个缺了的数量。
最后求一个所有b[i]的最大公约数。
答案就是累加b[i]/gcd
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 int const N = 200000 + 10;
5 int a[N], b[N], n, ta;
6 int gcd(int a, int b)
7 {
8 return b ? gcd(b, a % b) : a;
9 }
10 int main()
11 {
12 scanf("%d", &n);
13 for (int i = 1; i <= n; i++)
14 {
15 scanf("%d", &a[i]);
16 ta = max(ta, a[i]);
17 }
18 for (int i = 1; i <= n; i++)
19 b[i] = ta - a[i];
20 int g = b[1];
21 for (int i = 2; i <= n; i++)
22 g = gcd(g, b[i]);
23 ll ans = 0;
24 for (int i = 1; i <= n; i++)
25 ans += b[i] / g;
26 printf("%lld %d\n", ans, g);
27 return 0;
28 }