P5431 【模板】乘法逆元 2

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N = 5e6 + 10;
 5 ll fac[N], sv[N], inv[N], a[N];
 6 ll n, p, k;
 7 void read(ll &x) {
 8     x = 0; int f = 1; char ch = getchar();
 9     while (ch < '0' || ch >'9') {if (ch == '-') f = -1; ch = getchar();}
10     while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
11     x *= f;
12 }
13 ll quickpow(ll x, ll y) {
14     ll ans = 1;
15     while (y) {
16         if (y & 1) ans = ans * x % p;
17         x = x * x % p;
18         y >>= 1;
19     }
20     return ans % p;
21 }
22 signed main() {
23     read(n), read(p), read(k);
24     fac[0] = 1;
25     for (int i = 1; i <= n; i ++) {
26         read(a[i]);
27         fac[i] = fac[i - 1] * a[i] % p;
28     } 
29     sv[n] = quickpow(fac[n], p - 2);
30     for (int i = n; i >= 1; i --) sv[i - 1] = sv[i] * a[i] % p;
31     ll tmp = 1, ans = 0;
32     for (int i = 1; i <= n; i ++) {
33         tmp = (tmp * k) % p;
34         ans = (ans + ((sv[i] * fac[i - 1] % p)) * tmp) % p;
35     }
36     printf("%lld\n", ans);
37     return 0;
38 }