Codeforces 1355 E. Restorer Distance

分段枚举。

AC代码：

const int N = 2e5 + 50;
ll n, m, a, r;
ll h[N], sum[N], suf[N];

int main()
{
  cin >> n >> a >> r >> m;
  rep(i, 1, n)
    sld(h[i]);
  sort(h + 1, h + 1 + n);
  sum[0] = 0;
  rep(i, 1, n)
    sum[i] = sum[i - 1] + h[i];//前缀和
  suf[n + 1] = 0;
  per(i, n, 1)
    suf[i] = suf[i + 1] + h[i];//后缀和
  ll ans = INF;
  h[n + 1] = h[n];
  rep(i, 1, n)
  {
    ll p = i, s = n - i, tar;
    if (p * a <= s * r)
      tar = h[i + 1];
    else
      tar = h[i];
    ll add = p * tar - sum[i], remove = suf[i + 1] - s * tar;
    ll res = 0;
    if (a + r >= m)
    {
      ll tmp = min(add, remove);
      res += tmp * m, add -= tmp, remove -= tmp;
    }
    res += a * add + r * remove;
    ans = min(ans, res);
  }
  ll ave = sum[n] / n;
  rep(i, max(0ll, ave - 2), ave + 2)
  {
    ll add = 0, remove = 0;
    rep(j, 1, n)
    {
      if (h[j] < i)
        add += (i - h[j]);
      else if (h[j] > i)
        remove += (h[j] - i);
    }
    ll res = 0;
    if (a + r >= m)
    {
      ll tmp = min(add, remove);
      res += tmp * m, add -= tmp, remove -= tmp;
    }
    res += a * add + r * remove;
    ans = min(ans, res);
  }
  ans = min(ans, r * sum[n]);
  pld(ans);
  return 0;
}