数论学习之卢卡斯定理(组合数取模)

利用逆元,当n和m很大,mod比较小的时候使用

C(n,m)%p=C(n/p,m/p)*C(n%p,m%p)%p,递归

const int N = 100000 + 5;
LL mul(LL a, LL b, LL p)
{//快速乘,计算a*b%p
    LL ret = 0;
    while(b)
    {
        if(b & 1)
        {
            ret = (ret + a) % p;
        }
        a = (a + a) % p;
        b >>= 1;
    }
    return ret;
}

LL fact(int n, LL p)
{//n的阶乘求余p
    LL ret = 1;
    for (int i = 1; i <= n ; i ++)
    {
        ret = ret * i % p ;
    }
    return ret ;
}

void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d)
{
    if (!b)
    {
        d = a, x = 1, y = 0;
    }
    else
    {
        ex_gcd(b, a % b, y, x, d);
        y -= x * (a / b);
    }
}

LL inv(LL t, LL p)
{//如果不存在,返回-1
    LL d, x, y;
    ex_gcd(t, p, x, y, d);
    return d == 1 ? (x % p + p) % p : -1;
}

LL comb(int n, int m, LL p)
{//C(n, m) % p
    if (m < 0 || m > n)
    {
        return 0;
    }
    return fact(n, p) * inv(fact(m, p), p) % p * inv(fact(n-m, p), p) % p;
}

LL Lucas(LL n, LL m, int p)
{
    return m ? Lucas(n/p, m/p, p) * comb(n%p, m%p, p) % p : 1;
}
需要注意一点,在乘法中,如果有可能爆long long记得用快速乘取模,防止爆精度

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转载自blog.csdn.net/neuq_zsmj/article/details/80155722