注意事项：无。

代码：

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int inv[MAXn + 10];
void evainv(int n) {
    inv[1] = 1;
    for (int i = 2; i <= n; ++i){
        inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
    }
}

Luogu P3811 【模板】乘法逆元

代码：

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#include<bits/stdc++.h>
using namespace std;
#define int long long
const int MAXn = 3e6;

template <typename T>
inline void read(T &a) {
    char c;for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar());bool f = c == '-';T x = f ? 0 : (c ^ '0');for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) {x = x * 10 + (c ^ '0');}a = f ? -x : x;
}
template <typename T, typename ...Argv>
inline void read(T &a, Argv &...argv) {
    read(a), read(argv...);
}

int mod;

int inv[MAXn + 10];
void evainv(int n) {
    inv[1] = 1;
    for (int i = 2; i <= n; ++i){
        inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    }
}

int n;
signed main() {
    read(n, mod);
    evainv(n);
    for (int i = 1; i <= n; ++i) {
        printf("%lld\n", inv[i]);
    }
    return 0;
}