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[NOI2020统一省选 A] 组合数问题 (推式子)

news 来源：原创 2024/4/25 12:42:35

题意

给定四个整数 $n, x, p, m$ ，求
$\sum_{i=0}^{n}f(i)\times x^i\times \binom{n}{i}$
对 $p$ 取模，其中 $a_0 + a_1x + a_2x ^ 2 + \cdots + a_mx ^ m$

$\le n,x,p \le 10 ^ 9, 0 \le a_i \le 10 ^ 9, 0 \le m \le \min(n, 10 ^ 3)$

分析：

首先把 $f (i)$ 带入原式
$\sum_{i=0}^{n} x^i\times \binom{n}{i} \sum_{j = 0} ^ {m} a_j \times i ^ {j}$
看到 $i ^ j$ ，故想到展开 $\sum\limits_{k = 0} ^ {j} {j \brace k} i ^ {\underline k}$

$\sum_{i=0}^{n} x^i\times \binom{n}{i} \sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times \frac{i!}{(i - k)!}$

把前面的 $\dbinom{n}{i}$ 放到最后面化简

$\sum_{i=0}^{n} x^i \sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times\dfrac{n!}{i! \times (n - i)!} \times \frac{i!}{(i - k)!} \\\\ = \sum_{i=0}^{n} x^i \sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times\dfrac{n!}{(n - i)! \times (i - k)!}$

考虑凑组合数 $\dbinom{n - k}{n - i} = \dfrac{(n - k)!}{(n - i)! \times (i - k)!}$ ，所以分式上下同乘 $(n - k)!$ ，即
$\sum_{i=0}^{n} x^i \sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times \binom{n - k}{n - i} \times n ^ {\underline k}$
交换求和次序，将 $i$ 放到最后求和

$\sum_{j = 0} ^ {m} a_{j} \sum_{k = 0} ^ {j} {j \brace k} \times n ^ {\underline k} \sum_{i=0}^{n} x^i \times \binom{n - k}{n - i} \\\\ = \sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times n ^ {\underline k} \sum_{i=0}^{n} x^i \times \binom{n - k}{i - k} \\\\ = \sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times n ^ {\underline k} \sum_{i=k}^{n} x^i \times \binom{n - k}{i - k}$

做变换 $\rightarrow i$

$\sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times n ^ {\underline k} \sum_{i=0}^{n - k} x^{i + k} \times \binom{n - k}{i} \\\\ = \sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times n ^ {\underline k} \times x ^ {k} \sum_{i=0}^{n - k} x^{i} \times \binom{n - k}{i}$

考虑二项式展开 $\sum\limits_{i = 0} ^ {n} \dbinom{n}{i} a ^ {i} b ^ {n - i}$ ，所以 $\sum\limits_{i=0}^{n - k} x^{i} \times \dbinom{n - k}{i} = (1 + x) ^ {n - k}$ ，故式子变为
$\sum_{j = 0} ^ {m} a_j \sum_{k = 0} ^ {j} {j \brace k} \times n ^ {\underline k} \times x ^ {k} \times (1 + x) ^ {n - k}$
这样式子就变为 $O(m ^ 2)$ 了，第二类斯特林数可以预处理，下降幂可以线性维护。

代码：

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
constexpr int N = 1e3;
int mod;
int norm(int x) {
    if (x < 0) {
        x += mod;
    }
    if (x >= mod) {
        x -= mod;
    }
    return x;
}
template<class T>
T power(T a, int b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}
struct Z {
    int x;
    Z(int x = 0) : x(norm(x)) {}
    int val() const {
        return x;
    }
    Z operator-() const {
        return Z(norm(mod - x));
    }
    Z inv() const {
        assert(x != 0);
        return power(*this, mod - 2);
    }
    Z &operator*=(const Z &rhs) {
        x = i64(x) * rhs.x % mod;
        return *this;
    }
    Z &operator+=(const Z &rhs) {
        x = norm(x + rhs.x);
        return *this;
    }
    Z &operator-=(const Z &rhs) {
        x = norm(x - rhs.x);
        return *this;
    }
    Z &operator/=(const Z &rhs) {
        return *this *= rhs.inv();
    }
    friend Z operator*(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res *= rhs;
        return res;
    }
    friend Z operator+(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res += rhs;
        return res;
    }
    friend Z operator-(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res -= rhs;
        return res;
    }
    friend Z operator/(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res /= rhs;
        return res;
    }
    friend istream &operator>>(istream &is, Z &a) {
        i64 v;
        is >> v;
        a = Z(v);
        return is;
    }
    friend ostream &operator<<(ostream &os, const Z &a) {
        return os << a.val();
    }
};
vector<vector<Z>> stirling(N + 1, vector<Z>(N + 1));
void init() {
    stirling[0][0] = 1;
    for (int i = 1; i <= N; i ++) {
        for (int j = 1; j <= i; j ++) {
            stirling[i][j] = stirling[i - 1][j - 1] + j * stirling[i - 1][j];
        }
    }
}
signed main() {
    cin.tie(0) -> sync_with_stdio(0);
    int n, x, m;
    cin >> n >> x >> mod >> m;
    init();
    vector<Z> a(m + 1);
    for (int i = 0; i <= m; i ++) {
        cin >> a[i];
    }
    Z res;
    for (int j = 0; j <= m; j ++) {
        Z sum = 1;
        for (int k = 0, cnt = n; k <= j; k ++, cnt --) {
            res += a[j] * stirling[j][k] * power(Z(x), k) * sum * power(Z(1 + x), n - k);
            sum *= cnt;
        }
    }
    cout << res << "\n";
}