注意到一定是购买按照排序后前一半，以后一半售出最佳，如果一共有偶数件，直接按照一半一半的购买即可，但是如果是奇数件，需要从前往后买 $n/2$ 向下取整，从后往前卖$n/2$ 向下取整，会空出中间一个。

#include <iostream>
#include <vector>
#include <algorithm>

#define int long long

using namespace std;

typedef pair<int, int> PII;

void solve()
{
    int n;
    cin >> n;
    vector<PII> a(n);
    for (int i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
    sort(a.begin(), a.end());
    int sum = 0;
    for (auto t : a) sum += t.second;
    int cnt = sum / 2;
    int sum1 = 0, sum2 = 0;
    for (auto t : a)
    {
        if (t.second <= cnt)
        {
            sum1 += t.first * t.second;
            cnt -= t.second;
        }
        else if (cnt > 0)
        {
            sum1 += cnt * t.first;
            cnt = 0;
            break;
        }
    }
    cnt = sum / 2;
    for (int i = n - 1; i >= 0; i--)
    {
        if (a[i].second <= cnt)
        {
            sum2 += a[i].first * a[i].second;
            cnt -= a[i].second;
        }
        else if (cnt > 0)
        {
            sum2 += cnt * a[i].first;
            cnt = 0;
            break;
        }
    }
    cout << sum2 - sum1 << '\n';
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int t = 1;
    while (t--)
    {
        solve();
    }
}