（a/b)%mod=a*pow(b，mod-2);
例题:给定一个数组,多次问询,求区间乘积取余
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
long long a[N], b[N], f[N];
long long ans;
long long mod = 1e9 + 7;
int n, m;
long long qmi(long long x, long long y) {
long long ans = 1;
long long base = x;
while (y) {
if (y & 1)
ans = ans * base % mod;
y >>= 1;
base = base * base % mod;
}
return ans;
}
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++)
cin >> a[i];
b[0] = 1;
f[0]=1;
for (int i = 1; i <= n; i++) {
b[i] = b[i - 1] * a[i] % mod;
f[i] = qmi(b[i], mod - 2);
}
while (m--) {
long long l, r;
cin >> l >> r;
cout << b[r] * f[l - 1] %mod<< endl;
}
}