cf 1216d

https://codeforc.es/problemset/problem/1216/D 

greedy:

Find the maximum of the a [i], so as t, i.e., t is assumed that the original x

Then b [i] = ta [i]; b [i] represents the number of each of the missing.  

 

Finally, all seeking a b [i] is the greatest common divisor. 

The answer is accumulated 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 }

View Code