Codeforce Round #227 Div2 A

Codeforces Round #228 (Div. 2)
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Submission	Time	Verdict
5911354	02/06/2014 07:40PM	Accepted

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greedy

math

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Fox Ciel is playing a game with numbers now.

Ciel has n positive integers: x₁, x₂, ..., x_n. She can do the following operation as many times as needed: select two different indexes i and j such that x_i > x_j hold, and then apply assignment x_i = x_i - x_j. The goal is to make the sum of all numbers as small as possible.

Please help Ciel to find this minimal sum.

Input

The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x₁, x₂, ..., x_n (1 ≤ x_i ≤ 100).

Output

Output a single integer — the required minimal sum.

Sample test(s)

Input

2 1 2

Output

Input

3 2 4 6

Output

Input

2 12 18

Output

Input

5 45 12 27 30 18

Output

Note

In the first example the optimal way is to do the assignment: x₂ = x₂ - x₁.

In the second example the optimal sequence of operations is: x₃ = x₃ - x₂, x₂ = x₂ - x₁.

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <queue>
 6 #include <set>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define INF 0x7fffffff 
12 #define mod 1000000007
13 #define ll long long
14 #define maxn 20025
15 #define pi acos(-1.0)  
16 int n, m, k, c,t,b;
17 int a[maxn];
18 int gcd(int n, int m){ return m ? gcd(m, n%m) : n; }
19 int main(){
20     int x,s;
21     scanf("%d", &n);
22     for (int i = 0; i < n; i++){
23         scanf("%d", &a[i]);
24         if(i==0)s = gcd(a[i], a[i]);
25         else s = gcd(s, a[i]);
26     }
27     printf("%d\n", s*n);
28     return 0;
29 }

View Code

posted @ 2014-02-07 04:47 HaibaraAi 阅读(110) 评论(0) 收藏举报

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