【模板】組合數取模

\(N \le 2000, M \le 2000\)
直接利用遞推式預處理便可。
代碼以下

#include <bits/stdc++.h>

using namespace std;

const int mod = 1e9 + 7;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    static int f[2010][2010];
    for (int i = 0; i <= 2000; i++) {
        f[i][0] = 1;
    }
    for (int i = 1; i <= 2000; i++) {
        for (int j = 1; j <= 2000; j++) {
            f[i][j] = (f[i - 1][j] + f[i - 1][j - 1]) % mod;
        }
    }
    while (n--) {
        int a, b;
        cin >> a >> b;
        cout << f[a][b] << endl;
    }
    return 0;
}

\(N, M \le 1e5\)
預處理出階乘和階乘的逆元，利用組合數的定義直接回答。

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;

LL fpow(LL a, LL b) {
    LL ret = 1 % mod;
    for (; b; b >>= 1, a = a * a % mod) {
        if (b & 1) {
            ret = ret * a % mod;
        }
    }
    return ret;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    vector<LL> inv(maxn), fac(maxn);
    auto prework = [&]() {
        fac[0] = 1;
        for (int i = 1; i <= 1e5; i++) {
            fac[i] = fac[i - 1] * i % mod;
        }
        inv[1e5] = fpow(fac[1e5], mod - 2);
        for (int i = 1e5 - 1; i >= 0; i--) {
            inv[i] = inv[i + 1] * (i + 1) % mod;
        }
    };
    prework();
    auto get = [&](int a, int b) {
        return fac[a] * inv[a - b] % mod * inv[b] % mod;
    };
    while (n--) {
        int a, b;
        cin >> a >> b;
        cout << get(a, b) << endl;
    }
    return 0;
}
/*
(a, b) = a! / ((a - b)! * b!)
a! = a * (a - 1)!
a!' * a = (a - 1)!'
*/

\(N, M \le 1e18, p \le 1e5,p \in prime\)
預處理出 \(1...p\) 的階乘和階乘的逆元，用盧卡斯定理進行回答。
代碼以下

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

LL fpow(LL a, LL b, LL c) {
    LL ret = 1 % c;
    for (; b; b >>= 1, a = a * a % c) {
        if (b & 1) {
            ret = ret * a % c;
        }
    }
    return ret;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    while (n--) {
        LL a, b, p;
        cin >> a >> b >> p;
        vector<LL> inv(p + 1), fac(p + 1);
        fac[0] = 1;
        for (int i = 1; i <= p; i++) {
            fac[i] = fac[i - 1] * i % p;
        }
        inv[p - 1] = fpow(fac[p - 1], p - 2, p);
        for (int i = p - 2; i >= 0; i--) {
            inv[i] = inv[i + 1] * (i + 1) % p;
        }
        auto C = [&](LL x, LL y) -> LL {
            if (x < y) {
                return 0;
            }
            return fac[x] * inv[x - y] % p * inv[y] % p;
        };
        function<LL(LL, LL)> Lucas = [&](LL x, LL y) -> LL {
            if (y == 0) {
                return 1;
            }
            return C(x % p, y % p) * Lucas(x / p, y / p) % p;
        };
        cout << Lucas(a, b) << endl;
    }
    return 0;
}