傻傻地敲了好多遍ST表。

10119. 「一本通 4.2 例 1」数列区间最大值

#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
struct st {
    int a[N][21];
    void build(int *src, int n) {
        for (int i = 1; i <= n; i++) a[i][0] = src[i];
        for (int i = 1; i <= 20; i++)
            for (int j = 1; j <= n - (1 << i) + 1; j++)
                a[j][i] = max(a[j][i - 1], a[j + (1 << (i - 1))][i - 1]);
    }
    int query(int l, int r) {
        int j = log2(r - l + 1);
        return max(a[l][j], a[r - (1 << j) + 1][j]);
    }
};
int n, m, a[N];
st s;
int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    s.build(a, n);
    for (int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        cout << s.query(u, v) << endl;
    }
}

10120. 「一本通 4.2 例 2」最敏捷的机器人

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

struct Monoqueue {
    deque<pair<int, int> > q;
    void push(int val, int tim) {
        while (q.size() && q.front().first >= val) q.pop_front();
        q.push_front(make_pair(val, tim));
    }
    int get(int tim) {
        while (q.size() && q.back().second < tim) q.pop_back();
        return q.back().first;
    }
};

Monoqueue a, b;
int n, k, t;

int main() {
    ios::sync_with_stdio(false);
    cin >> n >> k;
    for (int i = 1; i < k; i++) {
        cin >> t;
        a.push(t, i + k - 1);
        b.push(-t, i + k - 1);
    }
    for (int i = k; i <= n; i++) {
        cin >> t;
        a.push(t, i + k - 1);
        b.push(-t, i + k - 1);
        cout << -b.get(i) << " " << a.get(i) << endl;
    }
}

10121. 「一本通 4.2 例 3」与众不同

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

int n, m, a[3000005], L, R;

namespace Monoqueue {
int head = -1, tail = 0, q[2000005], buck[2000005], jmp[2000005];
void solve() {
    for (int i = 1; i <= n; i++) {
        q[++head] = a[i];
        while (buck[a[i]]) buck[q[tail++]]--;
        buck[a[i]]++;
        jmp[i] = (head - tail + 1);
    }
}
}  // namespace Monoqueue

namespace ST {
int s[2000005][21];
void build(int n, int *src) {
    for (int i = 1; i <= n; i++) s[i][0] = src[i];
    for (int j = 1; j < 20; j++)
        for (int i = 1; i <= n - (1 << j >> 1); i++) s[i][j] = max(s[i][j - 1], s[i + (1 << j >> 1)][j - 1]);
}
int query(int l, int r) {
    int lg = log2(r - l + 1);
    return max(s[l][lg], s[r - (1 << lg) + 1][lg]);
}
}  // namespace ST

int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i], a[i] += 1000000;
    Monoqueue::solve();
    ST::build(n, Monoqueue::jmp);
    for (int i = 1; i <= m; i++) {
        int l, r;
        cin >> l >> r;
        ++l;
        ++r;
        int L = 1, R = r - l + 2;
        while (R > L) {
            int M = (L + R) >> 1;
            int x = ST::query(l + M - 1, r);
            // cout<<L<<" "<<R<<" "<<M<<" "<<x<<endl;
            if (x >= M)
                L = M + 1;
            else
                R = M;
        }
        cout << L - 1 << endl;
    }
}

10122. 「一本通 4.2 练习 1」天才的记忆

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

int n, m, a[1000005][21];

void build() {
    for (int i = 1; i <= 20; i++)
        for (int j = 1; j <= n - (1 << i >> 1) + 1; j++)
            a[j][i] = max(a[j][i - 1], a[j + (1 << i >> 1)][i - 1]);
}

int query(int l, int r) {
    int lg = log2(r - l + 1);
    return max(a[l][lg], a[r - (1 << lg) + 1][lg]);
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i][0];
    build();
    cin >> m;
    for (int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        cout << query(u, v) << endl;
    }
}

10123. 「一本通 4.2 练习 2」Balanced Lineup

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

int n, m, a[1000005][21], b[1000005][21], u, v;

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i][0], b[i][0] = a[i][0];
    for (int i = 1; i <= 20; i++)
        for (int j = 1; j + (1 << i >> 1) - 1 <= n; j++)
            a[j][i] = max(a[j][i - 1], a[j + (1 << i >> 1)][i - 1]),
            b[j][i] = min(b[j][i - 1], b[j + (1 << i >> 1)][i - 1]);
    for (int i = 1; i <= m; i++) {
        cin >> u >> v;
        int lg = log2(v - u + 1);
        cout << max(a[u][lg], a[v - (1 << lg) + 1][lg]) - min(b[u][lg], b[v - (1 << lg) + 1][lg]) << endl;
    }
}