求组合数

1：如果a、b小于2000：可以预处理

#include<iostream>
using namespace std;
const int maxn = 2010;
const int mod = 1e9 + 7;
int c[maxn][maxn];

void init() {
    int n = 2000;
    for (int i = 0; i <= n; i++) c[i][0] = 1;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
}

int main() {
    init();
    int T, a, b;
    cin >> T;
    while (T--) {
        cin >> a >> b;
        cout << c[a][b] << endl;
    }
    return 0;
}

2.如果a、b达到1e5 mod p
可以求逆元(费马小定理）

#include<iostream>
using namespace std;
const long long maxn = 1e5 + 10;
const long long mod = 1e9 + 7;


typedef long long ll;
ll f[maxn], invf[maxn];

ll qpow(ll a, ll b) {
    ll ans = 1;
    for (; b; b >>= 1) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
    }
    return ans;
}

void init() {
    f[0] = 1;
    invf[0] = 1;
    for (int i = 1; i <= 100001; i++) f[i] = (f[i - 1] * i) % mod;
    for (int i = 1; i <= 100001; i++) invf[i] = invf[i - 1] * qpow(i, mod - 2) % mod;
}

int main() {
    init();
    int T, n, m;
    cin >> T;
    while (T--) {
        cin >> n >> m;
        cout << f[n] * invf[n - m] % mod * invf[m] % mod<< endl;
    }
    return 0;
}

3.如果a、b达到1e18,但模数p比较小（1e5)、
利用Lucas定理递归求解,转换成1e5的情况，转化为问题2（这里的p每次不一样，直接用定义算即可）

#include <iostream>
using namespace std;

typedef long long LL;

int p;

int qmi(int a, int b) {
    int res = 1;
    for (; b; b >>= 1) {
        if (b & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
    }
    return res;
}

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

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

int main() {
    int T;
    cin >> T;
    while (T--) {
        LL a, b;
        cin >> a >> b >> p;
        cout << lucas(a, b) << endl;
    }
    return 0;
}

4.a、b达到5000，不取模
换个支持高精度的语言（bushi）
预处理a以内的质数，对组合数进行巧妙的因式分解(详见get()函数），然后把所有质因子暴力相乘（

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 5010;
bool st[N];
int prime[N], sum[N], cnt;

void eulor(int n) {
    for (int i = 2; i <= n; i++) {
        if (!st[i]) prime[++cnt] = i;
        for (int j = 1; prime[j] <= n / i; j++) {
            st[prime[j] * i] = true;
            if (i % prime[j]) 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;
    eulor(a);

    for (int i = 1; i <= cnt; i++) {
        int p = prime[i];
        sum[i] = get(a, p) - get(b, p) - get(a - b, p);
    }

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

    for (int i = 1; i <= cnt; i++)
        for (int j = 1; j <= sum[i]; j++)
            res = mul(res, prime[i]);

    for (int i = res.size() - 1; i >= 0; i--) cout << res[i];

    return 0;
}

应用：求卡特兰数
经验：maxn开成两倍最大范围比较好QAQ

#include<iostream>
using namespace std;
const long long maxn = 1e5 + 10;
const long long mod = 1e9 + 7;

typedef long long ll;
ll f[maxn], invf[maxn];

ll qpow(ll a, ll b) {
    ll ans = 1;
    for (; b; b >>= 1) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
    }
    return ans;
}

void init() {
    f[0] = 1;
    invf[0] = 1;
    for (int i = 1; i <= 100001; i++) f[i] = (f[i - 1] * i) % mod;
    for (int i = 1; i <= 100001; i++) invf[i] = invf[i - 1] * qpow(i, mod - 2) % mod;
}

signed main() {
    init();
    int n;
    cin >> n;
    cout << f[n * 2] * invf[n] % mod * invf[n] % mod  << endl;
    return 0;
}