O(n²)

int c[N][N];//存储c(n,m)%P的组合数

for (int i = 0; i < N; i++)
{
    for (int j = 0; j <= i; j++)
    {
        if (!j)
        {
            c[i][j] = 1;
        }
        else
        {
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % P;
        }
    }
}

O(nlogn)

int fact[N], infact[N];//fact存储阶乘取模余数,infact存储阶乘取模逆元

int qmi(int m, int k, int p) //求m^k mod p,注意数据范围
{
    int res = 1;
    while (k)
    {
        if (k & 1)
        {
            res = (ll)res * m % p;
        }
        m = (ll)m * m % p;
        k >>= 1;
    }
    return res;
}

int C(int a, int b, int p) //C(a,b)%P的值
{
    return (ll)fact[a] * infact[b] % p * infact[a - b] % p;
}

fact[0] = infact[0] = 1;
for (int i = 1; i < N; i++) //预处理
{
    fact[i] = (ll)fact[i - 1] * i % P;
    infact[i] = (ll)infact[i - 1] * qmi(i, P - 2, P) % P;
}

O(plogn)

int qmi(int m, int k, int p) //求m^k mod p,注意数据范围
{
    int res = 1;
    while (k)
    {
        if (k & 1)
        {
            res = (ll)res * m % p;
        }
        m = (ll)m * m % p;
        k >>= 1;
    }
    return res;
}

int C(int a, int b, int p)
{
    if (a < b)
    {
        return 0;
    }
    int res = 1, inv = 1;
    for (int i = 1, j = a; i <= b; i++, j--)
    {
        res = (ll)res * j % p;
        inv = (ll)inv * i % p;
    }
    res = (ll)res * qmi(inv, p - 2, p) % p;
    return res;
}

int lucas(ll a, ll b, int p)//求c(a,b)%p,p为质数
{
    if (a < p && b < p)
    {
        return C(a, b, p);
    }
    return (ll)C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}

O(n)求C(a,b)真实值

int primes[N], cnt;
int sum[N];
bool s[N];

void get_primes(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!s[i])
        {
            primes[cnt++] = i;
        }
        for (int j = 0; primes[j] <= n / i; j++)
        {
            s[primes[j] * i] = true;
            if (i % primes[j] == 0)
            {
                break;
            }
        }
    }
}

int get(int n, int p)
{
    int res = 0;
    while (n)
    {
        res += n / p;
        n /= p;
    }
    return res;
}

vector<int> mul(vector<int> a, int b)//高精乘
{
    vector<int> c;
    int t = 0;
    for (int i = 0; i < a.size(); i++)
    {
        t += a[i] * b;
        c.push_back(t % 10);
        t /= 10;
    }
    while (t)
    {
        c.push_back(t % 10);
        t /= 10;
    }
    return c;
}

int main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int a, b;
    cin >> a >> b;
    get_primes(a);
    for (int i = 0; i < cnt; i++)
    {
        int p = primes[i];
        sum[i] = get(a, p) - get(a - b, p) - get(b, p);
    }
    vector<int> res;
    res.push_back(1);
    for (int i = 0; i < cnt; i++)
    {
        for (int j = 0; j < sum[i]; j++)
        {
            res = mul(res, primes[i]);
        }
    }
    for (int i = res.size() - 1; i >= 0; i--)
    {
        cout << res[i];
    }
    return 0;
}