题解 | 子数列求积

//就是一个前缀乘+逆元就OK了!!!
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1e9 + 7;
const int N = 1e5 + 9;
int a[N];
int fpow(int a, int b)
{
    a %= mod;
    int res = 1;
    while (b)
    {
        if (b & 1)
            res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

int ni(int a)
{
    return fpow(a, mod - 2) % mod;
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int n, q;
    cin >> n >> q;
    a[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        a[i] = a[i] * a[i - 1] % mod;
    }
    while (q--)
    {
        int l, r;
        cin >> l >> r;
        cout << a[r] * ni(a[l - 1]) % mod << ' ';
    }
    return 0;
}