求逆元

(1)擴展歐幾里得

LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL r = exgcd(b, a % b, x, y);
    LL t = x % mod;
    x = y % mod;
    y = ((t - a / b * y) % mod + mod) % mod;
    return r;
}

求a對於b的逆元x

(2)費馬小定理 要求p爲素數

LL quick_mod(LL x, LL n, LL p)
{
    LL res = 1;
    while(n)
    {
        if (n & 1)
            res = res * x % p;
        x = x * x % p;
        n >>= 1;
    }
    return res;
}
LL inv(LL x, LL p)
{
    return quick_mod(x, p-2, p);
}

(3)初始化記錄逆元

void init()
{
    F[0] = 1; F[1] = 1;//階乘
    for (int i = 2; i < maxn; ++i)
        F[i] = F[i-1] * i % p;
    invF[maxn - 1] = quick_mod(F[maxn - 1], p-2);//階乘逆元
    for (int i = maxn - 2; i >= 0; --i)
        invF[i] = invF[i+1] * (i + 1) % p;
    inv[1] = 1;//1-n逆元
    for (int i = 2; i < max; ++i)
        inv[i] = (p - p / i) * inv[p % i] % p;
}