Codeforces Round #228 (Div. 2) A. Fox and Number Game

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, ..., xn. She can do the following operation as many times as needed: select two different indexes iand j such that xi > xj hold, and then apply assignment xi = xi - xj. 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, ..., xn (1 ≤ xi ≤ 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.

有史以来第一次冲到第四题，有点思路，但是做不下去了。

提前把代码贴上吧

很简单的一道题，刚开始的想法是循环排序并作a[i]-=a[i-1]处理，提交后超时了

想到可以用a[i]%=a[i-1]，但是这种情况要考虑到a[i]整除a[i-1]的情况，整除时另a[i]=a[i-1]

循环的条件是a[0]!=a[n-1]；

代码如下：

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#define maxn 200
using namespace std;

int a[maxn];
int main(void){
    
    int n;
    while(cin >> n){
     for(int i=0; i<n; ++i){
      cin >> a[i];
     }
     sort(a,a+n);
     
     while(a[n-1] != a[0]){
      for(int i=n-1; i>0; --i)
       if(a[i]%a[i-1] == 0)
        a[i] = a[i-1];
       else a[i] %= a[i-1];
      sort(a,a+n);
     }  
     int sum = 0;
     for(int i=0; i<n; ++i)
      sum += a[i];
     cout << sum << endl;
    }
    return 0;
}