#include <bits/stdc++.h>

using namespace std;

int main() {
    cin.tie(0)->sync_with_stdio(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int maxw = 0, maxh = 0;
        for (int i=0; i<n; i++) {
            int w, h;
            cin >> w >> h;
            maxw = max(maxw, w);
            maxh = max(maxh, h);
        }
        cout << 2 * (maxw + maxh) << "\n";
    }
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main() {
    cin.tie(0)->sync_with_stdio(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> A(n);
        for (int i=0; i<n; i++) cin >> A[i];
        int best = 0;
        for (int i=0; i<n; i++) {
            int curr = 0;
            for (int j=i; j<n; j++) {
                if (A[j] <= A[i]) {
                    curr += 1;
                }
            }
            best = max(best, curr);
        }
        cout << n - best << "\n";
    }
  return 0;
}

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
        int n;
        cin >> n;
        vector<ll> A(n);
        for (int i=0; i<n; i++) cin >> A[i];
        map<ll,vector<ll>> adj;
        for (int i=1; i<n; i++) {
            ll u = A[i] + i;
            ll v = u + i;
            adj[u].push_back(v);
        }
        set<ll> vis;
        function<void(ll)> dfs = [&](ll u) -> void {
            if (vis.count(u)) return;
            vis.insert(u);
            for (ll v : adj[u]) dfs(v);
        };
        dfs(n);
        cout << *vis.rbegin() << "\n";
  }
  return 0;
}

2027D1 - The Endspeaker (Easy Version)

Let's use dynamic programming. We will have $$$\operatorname{dp}_{i,j}$$$ be the minimum cost to remove the prefix of length $$$i$$$, where the current value of $$$k$$$ is $$$j$$$. By a type $$$1$$$ operation, we can transition from $$$\operatorname{dp}_{i,j}$$$ to $$$\operatorname{dp}_{i,j+1}$$$ at no cost. Otherwise, by a type $$$2$$$ operation, we need to remove some contiguous subarray $$$a_{i+1}, a_{i+2}, \dots, a_{x}$$$ (a prefix of the current array), to transition to $$$\operatorname{dp}_{x,j}$$$ with a cost of $$$m - k$$$.

Let $$$r$$$ be the largest value of $$$x$$$ possible. Given we're spending $$$m - k$$$ whatever value of $$$x$$$ we choose, it's clear to see that we only need to transition to $$$\operatorname{dp}_{r,j}$$$. To find $$$r$$$ for each value of $$$i$$$ and $$$j$$$, we can either binary search over the prefix sums or simply maintain $$$r$$$ as we increase $$$i$$$ for a fixed value of $$$k$$$. The answer is then $$$\operatorname{dp}_{n,k}$$$. The latter method solves the problem in $$$\mathcal{O}(nm)$$$.

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const int inf = 1 << 30;

void chmin(int &a, int b) {
  a = min(a, b);
}

int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<int> A(n);
    for (int i=0; i<n; i++) cin >> A[i];
    vector<int> B(m);
    for (int i=0; i<m; i++) cin >> B[i];
    vector nxt(n, vector<int>(m));
    for (int k=0; k<m; k++) {
      int r = -1, sum = 0;
      for (int i=0; i<n; i++) {
        while (r < n && sum <= B[k]) sum += A[++r];
        nxt[i][k] = r;
        sum -= A[i];
      }
    }
    vector dp(n+1, vector<int>(m, inf));
    dp[0][0] = 0;
    for (int k=0; k<m; k++) {
      for (int i=0; i<n; i++) {
        chmin(dp[nxt[i][k]][k], dp[i][k] + m - k - 1);
        if (k < m-1)
          chmin(dp[i][k+1], dp[i][k]);
      }
    }
    int ans = inf;
    for (int k=0; k<m; k++) {
      chmin(ans, dp[n][k]);
    }
    if (ans == inf) {
      cout << "-1\n";
    } else {
      cout << ans << "\n";
    }
  }
  return 0;
}

2027D2 - Финальный диалог (сложная версия)

Following on from the editorial for D1. Let's have the $$$\operatorname{dp}$$$ table store a pair of the minimum cost and the number of ways. Since we're now counting the ways, it's not enough just to consider the transition to $$$\operatorname{dp}_{r,j}$$$; we also need to transition to all $$$\operatorname{dp}_{x,j}$$$. Doing this naively is too slow, so let's instead find a way to perform range updates.

Let's say we want to range update $$$\operatorname{dp}_{l,j}, \operatorname{dp}_{l+1,j}, ..., \operatorname{dp}_{r,j}$$$. We'll store some updates in another table at either end of the range. Then, for a fixed value of $$$k$$$ as we iterate through increasing $$$i$$$-values, let's maintain a map of cost to ways. Whenever the number of ways falls to zero, we can remove it from the map. On each iteration, we can set $$$\operatorname{dp}_{i,j}$$$ to the smallest entry in the map, and then perform the transitions. This works in $$$\mathcal{O}(nm \log n)$$$.

Bonus: It's also possible to solve without the $$$\log n$$$ factor. We can use the fact that $$$\operatorname{dp}_{i,k}$$$ is non-increasing for a fixed value of $$$k$$$ to make the range updates non-intersecting by updating a range strictly after the previous iteration. Then we can just update a prefix sum array, instead of using a map.

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const int inf = 1 << 30;
const int MOD = 1000000007;

int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<int> A(n);
    for (int i=0; i<n; i++) cin >> A[i];
    vector<int> B(m);
    for (int i=0; i<m; i++) cin >> B[i];
    vector nxt(n, vector<int>(m));
    for (int k=0; k<m; k++) {
      int r = -1, sum = 0;
      for (int i=0; i<n; i++) {
        while (r < n && sum <= B[k]) sum += A[++r];
        nxt[i][k] = r;
        sum -= A[i];
      }
    }
    vector dp(n+1, vector<array<int,2>>(m, {inf, 0}));
    vector upd(n+1, vector(m, vector<array<int,3>>()));
    upd[0][0].push_back({0, 0, 1});
    upd[1][0].push_back({1, 0, 1});
    for (int k=0; k<m; k++) {
      map<int,array<int,2>> mp;
      for (int i=0; i<=n; i++) {
        for (auto [t, move, count] : upd[i][k]) {
          if (t == 0) {
            auto &[a, b] = mp[move];
            a += 1;
            (b += count) %= MOD;
          } else {
            auto &[a, b] = mp[move];
            a -= 1;
            (b += MOD - count) %= MOD;
            if (a == 0) mp.erase(move);
          }
        }
        if (mp.empty()) continue;
        auto &[move, info] = *mp.begin();
        dp[i][k] = {move, info[1]};
        if (i == n) continue;
        if (k < m-1) {
          upd[i][k+1].push_back({0, move, info[1]});
          upd[i+1][k+1].push_back({1, move, info[1]});
        }
        if (nxt[i][k] > i) {
          upd[i+1][k].push_back({0, move + (m - k - 1), info[1]});
          if (nxt[i][k] < n) {
            upd[nxt[i][k]+1][k].push_back({1, move + (m - k - 1), info[1]});
          }
        }
      }
    }
    map<int,int> mp;
    for (int k=0; k<m; k++) {
      auto &[move, count] = dp[n][k];
      (mp[move] += count) %= MOD;
    }
    auto &[move, count] = *mp.begin();
    if (move == inf) {
      cout << "-1\n";
    } else {
      cout << move << " " << count << "\n";
    }
  }
  return 0;
}

There exists an alternative solution for D1 & D2, using segment tree.
Continuing from the dynamic programming solution, let $$$\operatorname{dp}[i][j]$$$ represent the minimum score required to remove all elements after the $$$i$$$-th element, given that the current value of $$$k$$$ is $$$j$$$.

Instead of using a dp table, we maintain $$$m$$$ segment trees, each of length $$$n$$$. The $$$i$$$-th segment tree will represent the $$$i$$$-th column of the dp table.

We precalculate for each $$$i$$$ and $$$j$$$ the furthest position we can remove starting from $$$i$$$ — specifically, the maximum subarray starting from $$$i$$$ with a sum less than $$$b[j]$$$. We store this in $$$\operatorname{nxt}[i][j]$$$. This calculation can be done in $$$\mathcal{O}(nm)$$$ time using a sliding window.

To transition in the dp, we have:

This transition can be computed in $$$\mathcal{O}(\log n)$$$ time thanks to range querying on the segment tree, so our total complexity is $$$\mathcal{O}(nm \log n)$$$. For D2, we can store the count of minimums within each segment, and simply sum these counts to get the total number of ways.

#include <bits/stdc++.h>
using namespace std;

#define int long long
int modN = 1e9 + 7;

int mod(int n) {
    return (n + modN) % modN;
}

struct SegmentTree { 
    struct Node {
        int val = 1e18;
        int cnt = 1;
    }; 
 
    vector<Node> st;
    int n;
    SegmentTree(int n): n(n) {
        st.resize(4 * n + 1, Node());
    }
 
    SegmentTree(vector<int> a): n(a.size()) {
        st.resize(4 * n + 1, Node());
        build(a, 1, 0, n - 1);
    }
 
    void merge(Node& a, Node& b, Node& c) {
        a.val = min(b.val, c.val);
        if (b.val == c.val)
            a.cnt = mod(b.cnt + c.cnt);
        else if (b.val < c.val)
            a.cnt = b.cnt;
        else if (b.val > c.val)
            a.cnt = c.cnt;
    }
 
    void build(vector<int>& a, int id, int l, int r) {
        if (l == r) {
            st[id].val = a[l];
            return;
        }
        int mid = (l + r) / 2;
        build(a, id * 2, l, mid);
        build(a, id * 2 + 1, mid + 1, r);
        merge(st[id], st[id * 2], st[id * 2 + 1]);
    }
 
    void update(int id, int l, int r, int u, int val, int cnt) {
        if (l == r) {
            st[id].val = val; // or st[id].sum += val
            st[id].cnt = cnt;
            return;
        }
        int mid = (l + r) / 2;
        if (u <= mid) update(id * 2, l, mid, u, val, cnt);
        else update(id * 2 + 1, mid + 1, r, u, val, cnt); 
        merge(st[id], st[id * 2], st[id * 2 + 1]);
    }
 
    void update(int idx, int val, int cnt) { //wrapper
        update(1, 0, n - 1, idx, val, cnt);
    }
 
    Node query(int id, int l, int r, int u, int v) { //give 0, n - 1 as l and r and 1 as id
        if (v < l || r < u) return Node();
        if (u <= l && r <= v) {
            return st[id];
        }
        int mid = (l + r) / 2;
        auto a = query(id * 2, l, mid, u, v);
        auto b = query(id * 2 + 1, mid + 1, r, u, v);
        Node res;
        merge(res, a, b);
        return res;
    }
 
    Node query(int l, int r) { //wrapper
        return query(1, 0, n - 1, l, r);
    }
};

void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n), b(m);
    for (int &A : a) cin >> A;
    for (int &B : b) cin >> B;
    if (*max_element(a.begin(), a.end()) > b[0]) {
        cout << -1 << '\n';
        return;
    }
    vector<vector<int>> nxt(m, vector<int>(n));
    for (int i = 0; i < m; i++) {
        int curr = 0, r = -1;
        for (int j = 0; j < n; j++) {
            while (r + 1 < n && curr + a[r + 1] <= b[i]) 
                curr += a[r + 1], r += 1;
            nxt[i][j] = r + 1;
            if (j <= r) curr -= a[j];
            r = max(r, j);
        }
    }
    vector<SegmentTree> dp(m, SegmentTree(vector<int>(n + 1, 1e18)));
    for (int i = 0; i < m; i++)
        dp[i].update(n, 0, 1);
    for (int i = n - 1; i >= 0; i--) {
        for (int j = m - 1; j >= 0; j--) {
            auto q1 = dp[j].query(i + 1, nxt[j][i]);
            int v1 = q1.val + m - (j + 1), ps1 = q1.cnt;
            if (i + 1 <= nxt[j][i]) dp[j].update(i, v1, ps1); 
            if (j != m - 1) {
                auto q2 = dp[j + 1].query(i, i);
                int v2 = q2.val, ps2 = q2.cnt;
                auto q3 = dp[j].query(i, i);
                if (v2 < q3.val)
                    dp[j].update(i, v2, ps2);
                else if (v2 == q3.val)
                    dp[j].update(i, v2, mod(ps2 + q3.cnt));
            }
        }
    }
    cout << dp[0].query(0, 0).val << ' ' << dp[0].query(0, 0).cnt << '\n';
}

signed main() {
    cin.tie(0) -> sync_with_stdio(false);
    int t;  
    cin >> t;
    while (t--) 
       solve();
    return 0;
} 

#include <bits/stdc++.h>
 
using namespace std;
 
int nimber(int x, int a) {
  int aprime = 0;
  bool goodbit = false;
  for (int bit=30; bit>=0; bit--) {
    if (x & (1 << bit)) {
      aprime *= 2;
      if (goodbit || (a & (1 << bit))) {
        aprime += 1;
      }
    } else if (a & (1 << bit)) {
      goodbit = true;
    }
  }
 
  // g(2^k - 2) = 0, for all k >= 1.
  for (int k=1; k<=30; k++) {
    if (aprime == (1 << k) - 2) {
      return 0;
    }
  }
 
  // g(2^k - 1) = k, for all k >= 1.
  for (int k=1; k<=30; k++) {
    if (aprime == (1 << k) - 1) {
      return k;
    }
  }
 
  // g(2^k) = k + (-1)^k, for all k >= 0.
  for (int k=1; k<=30; k++) {
    if (aprime == (1 << k)) {
      if (k % 2) return k - 1;
      else return k + 1;
    }
  }
 
  // g(2^k+1) = g(2^k+2) = ... = g(2^{k+1} - 3) = k + 1, for all k >= 2.
  for (int k=2; k<=30; k++) {
    if ((1 << k) < aprime && aprime <= (2 << k) - 3) {
      return k + 1;
    }
  }
 
  // should never get to this point
  assert(false);
  return -1;
}
 
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> A(n);
    for (int i=0; i<n; i++) cin >> A[i];
    vector<int> X(n);
    for (int i=0; i<n; i++) cin >> X[i];
    int curr = 0;
    for (int i=0; i<n; i++) curr ^= nimber(X[i], A[i]);
    cout << (curr ? "Alice" : "Bob") << "\n";
  }
  return 0;
}

#include <bits/stdc++.h>
 
using namespace std;
        
int dp[32][32][6][2][2];
 
const int mod = 1000000007;
 
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> A(n);
    for (int i=0; i<n; i++) cin >> A[i];
    vector<int> B(n);
    for (int i=0; i<n; i++) cin >> B[i];
    vector<int> curr(32);
    curr[0] = 1; // identity
    for (int i=0; i<n; i++) {
      memset(dp, 0, sizeof dp);
      dp[0][0][0][0][0] = 1;
      for (int j=0; j<=29; j++) {
        int p = 29 - j; // place we are going to add a bit in
        for (int k=0; k<=29; k++) { // position of most significant one in a'
          for (int type=0; type<6; type++) { // 0, 1, 11111, 11110, 10000, else
            for (int good=0; good<2; good++) { // good=1 iff good bit has occured
              for (int low=0; low<2; low++) { // low=1 iff prefix below b
                for (int bit=0; bit<2; bit++) { // bit in x
                  if (dp[j][k][type][good][low] == 0) 
                    continue; // no point in transition since count is 0
                  if (!low && (B[i] & (1 << p)) == 0 && bit == 1) 
                    continue; // x can't go higher than B[i]
                  if (bit == 0) {
                    int good2 = good || (A[i] & (1 << p)) != 0; // check good bit
                    int low2 = low || (B[i] & (1 << p)) != 0; // check if low
                    // nothing added to a' so nothing else changes
                    (dp[j+1][k][type][good2][low2] += dp[j][k][type][good][low]) %= mod;
                  } else {
                    int bita = good || (A[i] & (1 << p)) != 0; // bit in a'
                    int k2 = type == 0 ? 0 : k + 1; // increase if MSOne exists
                    int type2 = bita ?
                      (
                        type == 0 ? 1 : // add first one
                        (type == 1 || type == 2) ? 2 : // 11111
                        5 // can't add 1 after a 0
                      ) : (
                        (type == 0) ? 0 : // 0
                        (type == 1 || type == 4) ? 4 : // 10000
                        (type == 2) ? 3 : // 11110
                        5 // can't have a zero in any other case
                      );
                    (dp[j+1][k2][type2][good][low] += dp[j][k][type][good][low]) %= mod;
                  }
                }
              }
            }
          }
        } 
      }
      vector<int> count(32); // number of x-values for each nimber
      for (int k=0; k<=29; k++) { // position of MSOne
        for (int good=0; good<2; good++) { // doesn't matter
          for (int low=0; low<2; low++) { // doesn't matter
            (count[0] += dp[30][k][0][good][low]) %= mod; // 0
            (count[1] += dp[30][k][1][good][low]) %= mod; // 1
            (count[k+1] += dp[30][k][2][good][low]) %= mod; // 11111
            (count[0] += dp[30][k][3][good][low]) %= mod; // 11110
            (count[k+(k%2?-1:1)] += dp[30][k][4][good][low]) %= mod; // 10000
            (count[k+1] += dp[30][k][5][good][low]) %= mod; // else
          }
        }
      }
      count[0] -= 1; // remove when x=0
      vector<int> next(32); // knapsack after adding this pile
      for (int j=0; j<32; j++)
        for (int k=0; k<32; k++)
          (next[j ^ k] += 1LL * curr[j] * count[k] % mod) %= mod;
      swap(curr, next);
    }
    cout << curr[0] << "\n";
  }
  return 0;
}