题解 | #小紫的劣势博弈#

#include <bits/stdc++.h>

using namespace std;

using LL = long long;
vector<LL> v;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n;
    if(!(cin >> n)) return 0;
    v.resize(n);
    for(int i = 0;i < n;i++)
    {
        cin >> v[0];
    }
    
    v.resort();
    
    LL x = 0;
    
    for(int i = 1;i <= n;i++)
    {
        if(i / 2 != 0) x += v[i-1];
        else if(i / 2 == 0) x -= v[i-1];
    }
    
    cout << x << "\n";
    return 0;
}