题解 | 子数列求积

#include <bits/stdc++.h>
using namespace std;
const int N=3e5+10;
const int mod = 1e9+7;
typedef long long ll;
typedef unsigned long long ull;
int n,q;

ll qpow(ll a,ll b,ll p)
{
    ll res = 1;
    for(;b;b>>=1)
    {
        if(b&1)
        {
            res = res*a%p;
        }
        a = a*a%p;
    }
    return res;

}
ll a[N];
ll pres[N];

void solve()
{
    cin>>n>>q;
    pres[0]=1;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        pres[i] = pres[i-1]*a[i]%mod;
    }



    while(q--)
    {
        int l,r;
        cin>>l>>r;
        ll ans = pres[r]*qpow(pres[l-1],mod-2,mod)%mod;
        cout<<ans<<' ';
    }
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t=1;
    // cin>>t;
    while(t--)
    {
        solve();

    }


    return 0;
}

简单的数学题，会逆元即可