cf228 div2 A. Fox and Number Game (模拟）

题意大致就是在给出的数列中不断取出 i，j 两个数，ai>aj , 然后把 ai-aj重新放入，直到不能做为止

我用了个优先队列模拟。 因为可能同时有几个最大值。。所以不能单纯的比较最大值和后面一个最大值相等就停止

题目：

A. Fox and Number Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

2

Input

3
2 4 6

Output

6

Input

2
12 18

Output

12

Input

5
45 12 27 30 18

Output

15

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.

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;
#define LL lolng long
const double pi = acos(0.) * 2;
priority_queue<int >q;
vector<int> ans;
int n;
int num[200];
int main()
{
    scanf("%d", &n);
    for(int i=0;i<n;i++)
    {
        int d;
        scanf("%d",&d);
        q.push(d);
    }
    long long ret =0;
    while(1)
    {
        int f,s;
        f = q.top();q.pop();
        if( f==q.top())
        {
            ans.push_back(f);
            while(!q.empty()&&f == q.top())
            {
                f= q.top();
                ans.push_back(q.top());q.pop();
            }
            if(!q.empty())
            {
                int p = ans.front();
                ans.pop_back();
                q.push( p-q.top());
                for(int i=0;i<ans.size();i++)
                    q.push(p);
                ans.clear();
            }
            else
                break;
        }
        else
            q.push(f-q.top());
    }

    while(!ans.empty())
    {
        ret+= ans.back();
        ans.pop_back();
    }
    printf("%d
",ret);
    return 0;
}