Submission #1288522

#	Time	Username	Problem	Language	Result	Execution time	Memory
1288522		LIA	Tricks of the Trade (CEOI23_trade)	C++17	55 / 100	1271 ms	2162688 KiB

trade

#include <bits/stdc++.h>

using namespace std;
#define ll long long
#define vll vector<ll>
#define pll pair<ll, ll>
#define tp tuple<ll, ll, ll>
const ll inf = 1e18;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ll n, k, ans = -1e18;
  cin >> n >> k;
  vll b(n), s(n);
  for (ll i = 0; i < n; ++i)
    cin >> b[i];
  for (ll i = 0; i < n; ++i)
    cin >> s[i];

  vector<vll> dp(n + 1, vll(k + 1, -inf));
  for (ll i = 0; i <= n; ++i)
    dp[i][0] = 0;

  vll pref(n + 1, 0);
  for (ll i = 1; i <= n; ++i)
    pref[i] = pref[i - 1] + b[i - 1];

  for (ll j = 1; j <= k; ++j) {
    ll best = (j == 1 ? 0 : inf);
    for (ll i = 1; i <= n; ++i) {
      if (best != inf)
        dp[i][j] = s[i - 1] - pref[i] - best;
      if (dp[i][j - 1] > -inf / 2) {
        ll cand = -(dp[i][j - 1] + pref[i]);
        if (cand < best)
          best = cand;
      }
    }
  }
  for (ll i = 1; i <= n; ++i)
    ans = max(ans, dp[i][k]);

  set<ll> in;

  ll cur = ans;
  ll took = 0;

  vll br = b, sr = s;
  reverse(br.begin(), br.end());
  reverse(sr.begin(), sr.end());

  vll prefr(n + 1, 0);
  for (ll i = 1; i <= n; ++i)
    prefr[i] = prefr[i - 1] + br[i - 1];

  vector<vll> dpr(n + 1, vll(k + 1, -inf));
  for (ll i = 0; i <= n; ++i)
    dpr[i][0] = 0;

  for (ll j = 1; j <= k; ++j) {
    ll best = (j == 1 ? 0 : inf);
    for (ll i = 1; i <= n; ++i) {
      if (best != inf)
        dpr[i][j] = sr[i - 1] - prefr[i] - best;
      if (dpr[i][j - 1] > -inf / 2) {
        ll cand = -(dpr[i][j - 1] + prefr[i]);
        if (cand < best)
          best = cand;
      }
    }
  }

  vector<char> out(n, '0');
  for (ll i = 1; i <= n; ++i) {
    bool ok = false;
    for (ll j = 1; j <= k && !ok; ++j) {
      ll jr = k - j + 1;
      ll li = i;
      ll ri = n - i + 1;
      if (dp[li][j] <= -inf / 2)
        continue;
      if (dpr[ri][jr] <= -inf / 2)
        continue;
      ll val = dp[li][j] + dpr[ri][jr] - s[i - 1] + b[i - 1];
      if (val == ans)
        ok = true;
    }
    if (ok) {
      out[i - 1] = '1';
      in.insert(i);
    }
  }

  cout << ans << endl;
  for (ll i = 1; i <= n; ++i)
    cout << (in.count(i) ? '1' : '0');
  cout << "\n";
}

• Subtask 0: 0 / 0
• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 25 / 25
• Subtask 5: 0 / 45