模拟冒泡

这道题虽然用模拟和冒泡一起上，但是思路很简单，用冒泡排序的想法相邻的做减法即可

# include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int a[n], s = 0;
    for (int i = 0; i < n; i ++) cin >> a[i];
    for (int i = 0; i < n; i ++)
    {
        for (int j = 0; j < n - 1; j ++)
        {
            if (a[j] > a[j + 1]) a[j] -= a[j + 1];
            else if (a[j] < a[j + 1]) a[j + 1] -= a[j];
        }
    }
    for (int i = 0; i < n; i ++) s += a[i];
    cout << s << endl;
    return 0;
}