Codeforces Round #587 (Div. 3) D - Swords

Posted 2022-11-18 xwl3109377858

原文链接：https://www.cnblogs.com/xwl3109377858/p/11564317.html

Codeforces Round #587 (Div. 3)

D - Swords

There were n types of swords in the theater basement which had been used during the plays. Moreover there were exactly x swords of each type. y people have broken into the theater basement and each of them has taken exactly z swords of some single type. Note that different people might have taken different types of swords. Note that the values x,y and z are unknown for you.

The next morning the director of the theater discovers the loss. He counts all swords — exactly ai swords of the i-th type are left untouched.

The director has no clue about the initial number of swords of each type in the basement, the number of people who have broken into the basement and how many swords each of them have taken.

For example, if n=3, a=[3,12,6] then one of the possible situations is x=12, y=5 and z=3. Then the first three people took swords of the first type and the other two people took swords of the third type. Note that you don‘t know values x,y and z beforehand but know values of n and a.

Thus he seeks for your help. Determine the minimum number of people y, which could have broken into the theater basement, and the number of swords z each of them has taken.

Input

The first line of the input contains one integer n (2≤n≤2⋅105) — the number of types of swords.

The second line of the input contains the sequence a1,a2,…,an (0≤ai≤109), where ai equals to the number of swords of the i-th type, which have remained in the basement after the theft. It is guaranteed that there exists at least one such pair of indices (j,k) that aj≠ak.

Output

Print two integers y and z — the minimum number of people which could have broken into the basement and the number of swords each of them has taken.

Examples

input

3

3 12 6

output

5 3

input

2

2 9

output

1 7

input

7

2 1000000000 4 6 8 4 2

output

2999999987 2

input

6

13 52 0 13 26 52

output

12 13

Note

In the first example the minimum value of y equals to 5, i.e. the minimum number of people who could have broken into the basement, is 5. Each of them has taken 3 swords: three of them have taken 3 swords of the first type, and two others have taken 3 swords of the third type.

In the second example the minimum value of y is 1, i.e. the minimum number of people who could have broken into the basement, equals to 1. He has taken 7 swords of the first type.

 

思路：用最大数减去所给的值，然后对前n-1个值求一个gcd，然后除掉gcd求和就是答案。

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<map>
 7 #include<set>
 8 #include<vector>
 9 #include<queue>
10 #include<list>
11 #include<stack>
12 using namespace std;
13 #define ll long long 
14 const int mod=1e9+7;
15 const int inf=1e9+7;
16  
17 const int maxn=2e5+10;
18  
19 ll int gcd(ll int a,ll int b)
20 
21     if(b==0)
22         return a;
23     else
24         return gcd(b,a%b);
25 
26  
27 int num[maxn];
28  
29 int main()
30 
31     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
32     
33     int n;
34     cin>>n;
35     
36     int maxx=-1;
37     
38     for(int i=0;i<n;i++)
39         cin>>num[i];
40     
41     sort(num,num+n);
42     
43     maxx=num[n-1];
44     
45     int now;
46     
47     for(int i=0;i<n-1;i++)
48     
49         if(i==0)
50             now=maxx-num[i];
51         else
52             now=gcd(now,maxx-num[i]);
53     
54     
55     ll int ans=0;
56     
57     for(int i=0;i<n-1;i++)
58     
59         ans+=(maxx-num[i])/now;
60     
61     
62     cout<<ans<<" "<<now<<endl;
63     
64     return 0;
65