Codeforces 389A (最大公约数)

Fox and Number Game

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

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 jsuch 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 Input

Input

2
1 2

Output

2

Input

3
2 4 6

Output

6

Input

2
12 18

Output

12

Input

5
45 12 27 30 18

Output

15

Hint

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.

Source

Codeforces Round #228 (Div. 2)

题解：将题目转化一下，就是求n个数的最大公约数再乘以n。

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
    if (b==0) return a;
    else return gcd(b,a%b);
}
int main()
{
    int i,n,ans,a[105];
    while (cin>>n)
    {
        cin>>a[0];
        ans=a[0];
        for (i=1;i<n;i++)
        {
            cin>>a[i];
            ans=gcd(ans,a[i]);
        }
        cout<<ans*n<<endl;
    }
    return 0;
}