P5431 【模板】乘法逆元 2

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e6 + 10;
ll fac[N], sv[N], inv[N], a[N];
ll n, p, k;
void read(ll &x) {
    x = 0; int f = 1; char ch = getchar();
    while (ch < '0' || ch >'9') {if (ch == '-') f = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    x *= f;
}
ll quickpow(ll x, ll y) {
    ll ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % p;
        x = x * x % p;
        y >>= 1;
    }
    return ans % p;
}
signed main() {
    read(n), read(p), read(k);
    fac[0] = 1;
    for (int i = 1; i <= n; i ++) {
        read(a[i]);
        fac[i] = fac[i - 1] * a[i] % p;
    } 
    sv[n] = quickpow(fac[n], p - 2);
    for (int i = n; i >= 1; i --) sv[i - 1] = sv[i] * a[i] % p;
    ll tmp = 1, ans = 0;
    for (int i = 1; i <= n; i ++) {
        tmp = (tmp * k) % p;
        ans = (ans + ((sv[i] * fac[i - 1] % p)) * tmp) % p;
    }
    printf("%lld\n", ans);
    return 0;
}

 

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