思路：

使用ST表。ST表求区间最大值改为按位或即可。

ST模板可参考MT3024 max=min

代码：

1.暴力6/10

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int M = 5e5 + 10;
int n, m;
int num[N];
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> num[i];
    }
    int l, r;
    for (int i = 1; i <= m; i++)
    {
        cin >> l >> r;
        int ans = num[l];
        for (int j = l + 1; j <= r; j++)
        {
            ans = ans | num[j];
        }
        cout << ans << endl;
    }
}

 2.ST表：

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int M = 5e5 + 10;
int n, m;
int mn[N][50], Lg[N], a[N];
void pre()
{
    Lg[1] = 0;
    for (int i = 2; i <= n; i++)
    {
        Lg[i] = Lg[i >> 1] + 1;
    }
}
void ST_create()
{ // 创建ST表
    for (int i = 1; i <= n; i++)
    {
        mn[i][0] = a[i];
    }
    for (int j = 1; j <= Lg[n]; j++)
    {
        for (int i = 1; i <= n - (1 << j) + 1; i++)
        {
            mn[i][j] = (mn[i][j - 1] | mn[i + (1 << (j - 1))][j - 1]); // 改成或运算
        }
    }
}
int ST_q(int l, int r)
{ // ST表求区间或
    int k = Lg[r - l + 1];
    return (mn[l][k] | mn[r - (1 << k) + 1][k]); // 改成或运算
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    pre();
    ST_create();
    int l, r;
    for (int i = 1; i <= m; i++)
    {
        cin >> l >> r;
        cout << ST_q(l, r) << endl;
    }
}