1.递推

#include<bits/stdc++.h>
#include<unordered_map> 
#include<unordered_set>
using namespace std;
#define int long long //可能会超时 
#define PII pair<int,int>
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 2005;
int a, b,n;
int c[N][N];
signed main()
{
	ios::sync_with_stdio, cin.tie(0), cout.tie(0);
	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])%mod;
		}
	}
	cin >> n;
	while (n--) {
		cin >> a >> b;
		cout << c[a][b] << endl;
	}
	return 0;
}

2.定义出发，快速幂逆元

#include<bits/stdc++.h>
#include<unordered_map> 
#include<unordered_set>
using namespace std;
#define int long long //可能会超时 
#define PII pair<int,int>
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 1e5 + 5;
int fact[N], infact[N];
int n;
int pow(int a, int b, int p) {
	int ans = 1;
	while (b) {
		if (b & 1) ans = ans * a % p;
		b >>= 1;
		a = a * a % p;
	}
	return ans;
}
signed main()
{
	ios::sync_with_stdio, cin.tie(0), cout.tie(0);
	fact[0] = infact[0] = 1;
	for (int i = 1;i < N;i++) {
		fact[i] = fact[i - 1] * i % mod;
		infact[i] = infact[i - 1] * pow(i, mod - 2, mod) % mod;
	}
	cin >> n;
	while (n--) {
		int a, b;
		cin >> a >> b;
		cout << fact[a] * infact[b] % mod * infact[a - b] % mod;
		puts("");
	}
	return 0;
}

3.卢卡斯定理

#include<bits/stdc++.h>
#include<unordered_map> 
#include<unordered_set>
using namespace std;
#define int long long //可能会超时 
#define PII pair<int,int>
const int INF = 0x3f3f3f3f, mod = 998244353;
int a, b, p;
int n;
int pow(int a, int b, int p) {
	int ans = 1;
	while (b) {
		if (b & 1) ans = ans * a % p;
		a = a * a % p;
			b >>= 1;
	}
	return ans;
}
int C(int a, int b, int p) {
	if (b > a) return 0;
	int ans = 1;
	for (int i = 1,j = a;i <= b;i++, j--) {
		ans = ans * j % p;
		ans = ans * pow(i, p - 2, p) % p;
	}
	return ans;
}
int lucas(int a, int b, int p) {
	if (a < p && b < p) return C(a, b, p);
	return C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}
signed main()
{
	ios::sync_with_stdio, cin.tie(0), cout.tie(0);
	cin >> n;
	while (n--) {
		cin >> a >> b >> p;
		cout << lucas(a, b, p) << endl;
	}
	return 0;
}

4.无模数，高精度

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 5010;
int primes[N], cnt;
int sum[N];
bool st[N];
void get_primes(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] <= n / i; j++)
        {
            st[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)
{
    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()
{
    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--) printf("%d", res[i]);
    puts("");
    return 0;
}