思路：

ST表，ST表模板可参考MT3024 max=min

注意，这里使用快读快写避免超时

代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, m, a[N], mn[N][20], Lg[N], l, r, ans;
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] = gcd(mn[i][j - 1], mn[i + (1 << (j - 1))][j - 1]);
        }
    }
}
int ST_qgcd(int l, int r)
{ // ST表求gcd
    int k = Lg[r - l + 1];
    return gcd(mn[l][k], mn[r - (1 << k) + 1][k]);
}
// 快读快写：
int read()
{
    int ret = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9')
        ch = getchar();
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret;
}
void write(int x)
{
    if (x >= 10)
        write(x / 10);
    putchar(x % 10 + '0');
}
int main()
{
    n = read(), m = read();
    for (int i = 1; i <= n; i++)
        a[i] = read();
    pre();
    ST_create();
    while (m--)
    {
        l = read(), r = read();
        ans = ST_qgcd(l, r);
        write(ans);
        putchar('\n');
    }
    return 0;
}