注意事项：无。

代码：

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int mx[MAXn + 10][MAXlog2_n + 2];
void build(int n, int *a) {
    int log2_n = log(n) / ln2;
    for (int i = 1; i <= n; ++i) {
        mx[i][0] = a[i];
    }
    for (int i = 1; i <= log2_n; ++i) {
        for (int j = 1, topj = n - power2[i] + 1; j <= topj; ++j) {
            mx[j][i] = max(mx[j][i - 1], mx[j + power2[i - 1]][i - 1]);
        }
    }
}
int querymax(int l, int r) {
    int loglen = log(r - l + 1) / ln2;
    return max(mx[l][loglen], mx[r - power2[loglen] + 1][loglen]);
}

Luogu P3865 【模板】ST 表

代码：

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#include<bits/stdc++.h>
using namespace std;
const int MAXn = 1e5;
const int MAXlog2_n = 16;
const int MAXm = 2e6;

template <typename T>
inline void read(T &a) {
    char c;for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar());bool f = c == '-';T x = f ? 0 : (c ^ '0');for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) {x = x * 10 + (c ^ '0');}a = f ? -x : x;
}
template <typename T, typename ...Argv>
inline void read(T &a, Argv &...argv) {
    read(a), read(argv...);
}

double ln2 = log(2);
int power2[MAXlog2_n + 10];
void evapower2(int n) {
    power2[0] = 1;
    for (int i = 1; i <= n; ++i) {
        power2[i] = power2[i - 1] * 2;
    }
}

int mx[MAXn + 10][MAXlog2_n + 2];
void build(int n, int *a) {
    int log2_n = log(n) / ln2;
    for (int i = 1; i <= n; ++i) {
        mx[i][0] = a[i];
    }
    for (int i = 1; i <= log2_n; ++i) {
        for (int j = 1, topj = n - power2[i] + 1; j <= topj; ++j) {
            mx[j][i] = max(mx[j][i - 1], mx[j + power2[i - 1]][i - 1]);
        }
    }
}
int querymax(int l, int r) {
    int loglen = log(r - l + 1) / ln2;
    return max(mx[l][loglen], mx[r - power2[loglen] + 1][loglen]);
}

int n, m;
int a[MAXn + 10];
signed main() {
    read(n, m);
    evapower2(log(n) / ln2);
    for (int i = 1; i <= n; ++i) {
        read(a[i]);
    }
    build(n, a);
    for (int i = 1, l, r; i <= m; ++i) {
        read(l, r);
        printf("%d\n", querymax(l, r));
    }
    return 0;
}