组合计数 （四）1.递归 2. 快速幂求逆元 3.卢卡斯定理 4.阶乘分解加大数据乘法运算

a,b较小时 2000

#include<iostream>
using namespace std;
const int N = 2010, mod = 1e9 + 7;
int c[N][N];
int n;
int main()
{
    cin >> n;
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j <= i; j++)
        {
            if(j == 0) c[i][j] = 1;
            else if(j == i) c[i][j] = 1;
            else c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
        }
    }
    
    while(n--)
    {
        int x, y;
        cin >> x >> y;
        cout << c[x][y] << endl;
    }
    return 0;
}

a,b较大时 利用乘法逆元求 a,b为10^5

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
const int N = 100010, mod = 1e9 + 7;
int n;
int fact[N], infact[N];

int quick_mul(int a, int b, int mod)
{
    int ans = 1 % mod;
    while (b)
    {
        if (b & 1)
            ans = (ll)ans * a % mod;
        a = (ll)a * a % mod;
        b >>= 1;
    }
    return ans;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n;
    fact[0] = infact[0] = 1;
    for (int i = 1; i < N; i++)
    {
        fact[i] = (ll)fact[i - 1] * i % mod;
        infact[i] = (ll)infact[i - 1] * quick_mul(i, mod - 2, mod) % mod;
    }

    while (n--)
    {
        int a, b;
        cin >> a >> b;
        cout << 1ll * fact[a] * infact[b] % mod * infact[a - b] % mod << endl;
    }
    return 0;
}

a, b 10^18， 利用卢卡斯定理

原始定义，实现C函数

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int n;
ll a, b, p;

int quick_mul(ll a, ll b, ll mod)
{
    ll ans = 1 % mod;
    while (b)
    {
        if (b & 1)
            ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

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

ll lucas(ll a, ll b, ll p)
{
    if (a < b)
        return 0;
    if (a < p && b < p)
        return C(a, b, p);
    return C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n;

    while (n--)
    {
        cin >> a >> b >> p;
        cout << lucas(a, b, p) << endl;
    }
    return 0;
}

当不用取模时，用阶乘分解加大数据乘法

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 5010;
int n, a, b;
int sum[N];
int prime[N], cnt;
bool vis[N];

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

int get_num(int n, int p)
{
    int sum = 0;
    while (n)
    {
        sum += n / p;
        n /= p;
    }
    return sum;
}

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

    while (t)
    {
        ans.push_back(t % 10);
        t /= 10;
    }

    return ans;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> a >> b;

    get_primes(a);

    for (int i = 0; i < cnt; i++)
    {
        int p = prime[i];
        sum[i] = get_num(a, p) - get_num(b, p) - get_num(a - b, p);
    }

    vector<int> ans;
    ans.push_back(1);

    for (int i = 0; i < cnt; i++)
        for (int j = 0; j < sum[i]; j++)
            ans = mul(ans, prime[i]);

    for (int i = ans.size() - 1; i >= 0; i--)
        cout << ans[i];
    cout << endl;
    return 0;
}