P5431 【模板】乘法逆元 2

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#include<cstdio>
#define re register
const int MAXn = 5e6;
#define int long long

template <class T> 
inline void read(T &a) {
    register char c;while (c = getchar(), c < '0' || c >'9');register T x(c - '0');while (c = getchar(), c >= '0' && c <= '9')x = (x << 1) + (x << 3) + c - '0';a = x;  
}

int power(int x, int y, int mod) {
    int ans = 1;
    while (y) {
        if (y & 1) {
            ans = ans * x % mod;
        }
        x = x * x % mod;
        y >>= 1;
    }
    return ans;
}
int inv(int n, int mod) {
    int inv = power(n, mod - 2, mod);
    return (inv % mod + mod) % mod;
}

int n, p, k, a[MAXn + 10], pi[MAXn + 10], mi[MAXn + 10], invv, ans;
signed main() {
    read(n), read(p), read(k);
    pi[0] = 1;
    for (re int i = 1; i <= n; ++i) {
        read(a[i]);
        pi[i] = (pi[i - 1] * a[i]) % p;
    }
    mi[0] = 1;
    for (re int i = 1; i <= n; ++i) {
        mi[i] = (mi[i - 1] * k) % p;
    }
    invv = inv(pi[n], p);
    for (re int i = n; i; --i) {
        ans = (ans + mi[i] * (invv * pi[i - 1] % p)) % p;
        invv = (invv * a[i]) % p;
    }
    printf("%lld\n", ans);
}