E
维护每个字符串是否连续 \(k\) 次出现，第一次出现的位置，和出现总次数即可。比较签到。
void solve() {int n, k, m, q;cin >> n >> k >> m >> q;vector <int> ans;vector<string> s(n);set <string> S;map <string, int> idx;for (int i = 0; i < n; i++) {cin >> s[i];S.insert(s[i]);}int o = 0;for (auto &x : S) {idx[x] = o++;}vector <int> first(o, -1), cnt(o), can(o);auto check1 = [&](const string &t, int i, int x) -> bool {if (i < k) return 0;if (can[x]) return 1;for (int j = i - 1; j >= i - k; j--) {if (t != s[j]) {return 0;}}can[x] = 1;return 1;};auto update = [&](const string &t, int i, int x) -> void {if (i - k + 1 < 0) return;if (can[x]) return;for (int j = i; j >= i - k + 1; j--) {if (t != s[j]) {return;}}can[x] = 1;};auto check2 = [&](int x, int i) -> bool {if (first[x] == -1) return 0;return i - first[x] > m;};for (int i = 0; i < n; i++) {int cur = idx[s[i]];if (cnt[cur] == q) continue;if (check1(s[i], i, cur) && check2(cur, i)) ans.push_back(i + 1);if (cnt[cur] == 0) first[cur] = i;cnt[cur]++;update(s[i], i, cur);}if (ans.empty()) cout << "empty" << '\n';else {for (auto &x: ans) cout << x << ' ';cout << '\n';}
}
H
经典二维数点。我们把 \(L\) 看作横坐标轴，\(R\) 看作纵坐标轴，每一段区间 \([l_i,r_i]\) 看作平面上的点，点的权值为 \(r_i-l_i+1\)。则一次询问就是要找到在矩形 \(L\le l_i,r_i\le R\) 中的最大点权。
离线操作，把询问和点按左端点从大到小排序，在左端点 \(l_i \ge L\) 的情况下，将点插入树状数组。维护右端点 \(r_i\le R\) 的最大权值。由于值域很大，需要离散化。时间复杂度 \(\mathcal{O}(n\log n + q\log(n+q))\)
struct Segment {int l, r;int len() const { return r - l + 1; }bool operator < (const Segment &other) const {return l > other.l;}
};
struct Query {int l, r;int id;bool operator < (const Query &other) const {return l > other.l;}
};
struct BIT {int n;vector<int> t;BIT(int n) : n(n), t(n + 1, 0) {}void update(int x, int v) {for (int i = x; i <= n; i += i & -i) {t[i] = max(t[i], v);}}int query(int x) const {int res = 0;for (int i = x; i > 0; i -= i & -i) {res = max(res, t[i]);}return res;}
};void solve() {int n, q;cin >> n >> q;vector <Segment> points(n);vector <int> Rs(n);for (int i = 0; i < n; i++) {cin >> points[i].l >> points[i].r;Rs[i] = points[i].r;}sort(Rs.begin(), Rs.end());Rs.erase(unique(Rs.begin(), Rs.end()), Rs.end());vector <Query> queries(q);for (int i = 0; i < q; i++) {cin >> queries[i].l >> queries[i].r;queries[i].id = i;}sort(points.begin(), points.end());sort(queries.begin(), queries.end());vector <int> ans(q);BIT bit(Rs.size());int ptr = 0;for (auto &[L, R, id] : queries) {while (ptr < n && points[ptr].l >= L) {int len = points[ptr].len();int x = lower_bound(Rs.begin(), Rs.end(), points[ptr].r) - Rs.begin() + 1;bit.update(x, len);ptr++;}int x = upper_bound(Rs.begin(), Rs.end(), R) - Rs.begin();ans[id] = bit.query(x);}for (auto &x : ans) cout << x << '\n';
}
J
把一个局面映射成 100 位的二进制数字。把所有局面都放到异或线性基里，对应的权值也同时异或处理。看最终询问的局面能否被线性基所表示。时间复杂度 \(\mathcal{O}(100 \cdot 100 \cdot n)\)
struct Node {array <int, 100> num;int S;Node() {fill(num.begin(), num.end(), 0); S = 0;}
};array <int, 100> Zero;struct XorBasis {array<Node, 100> basis;bool insert(Node x) {for (int i = 99; i >= 0; i--) {if (x.num[i] == 1) {if (basis[i].num == Zero) {basis[i].num = x.num;basis[i].S = x.S;return 1;}for (int j = 0; j <= 99; j++) {x.num[j] ^= basis[i].num[j]; }x.S ^= basis[i].S;}}return 0;}int query(Node x) {for (int i = 99; i >= 0; i--) {if (x.num[i] == 1) {for (int j = 0; j <= 99; j++) {x.num[j] ^= basis[i].num[j]; }x.S ^= basis[i].S;}}if (x.num == Zero) return x.S;return -1;}
};void solve() {int K;cin >> K;vector <Node> a(K);XorBasis B;for (int i = 0; i < K; i++) {int C, S;cin >> C >> S;a[i].S = S;while (C--) {int L;cin >> L;L--;a[i].num[L] ^= 1;}B.insert(a[i]);}int Q;cin >> Q;while (Q--) {int D;cin >> D;Node x;while (D--) {int R;cin >> R;R--;x.num[R] ^= 1;}cout << B.query(x) << '\n';}
}
好像只能做出这么多了，希望这些简单题下次能做快一些。