1.线性求 \(i\) 的逆元
for (int i = 2; i <= N; ++ i) {
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
2.费马小定理求 \(i\) 的逆元
inv[i] = QucikPower(i, mod - 2);
for (int i = 2; i <= N; ++ i) {
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
inv[i] = QucikPower(i, mod - 2);