[生成函數] HackerRank Count Fox Sequences

Solution$Solution$

這是個有序的序列，只要計算一個數出現了最多次，最後乘m$m$ 就好了。
枚舉出現了k$k$ 次。
那剩下的就是

∏i=1m−1(∑j=0k−1xj)=(1−xk)m−1(1−x)1−m$$\prod _{i=1}^{m-1}(\sum _{j=0}^{k-1}{x}^j)=(1-x^k{)}^{m-1}(1-x{)}^{1-m}$$

負指數冪的話：

(1+x)−n=∑k=0∞(−1)k(n+k−1k)xk$$(1+x{)}^{-n}=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k(\genfrac{}{}{0ex}{}{n+k-1}{k})x^k$$

枚舉兩邊的冪指數，是個調和級數。複雜度O(Tnlogn)$\mathcal{O}(Tn\log n)$

#include <bits/stdc++.h>
using namespace std;

const int MOD = 1000000007;
const int N = 202020;
typedef long long ll;

int fac[N], ifac[N];
int n, lo, hi, test, ans;

inline void pre(int n) {
    ifac[1] = 1;
    for (int i = 2; i <= n; i++)
        ifac[i] = (ll)(MOD - MOD / i) * ifac[MOD % i] % MOD;
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= n; i++) {
        fac[i] = (ll)fac[i - 1] * i % MOD;
        ifac[i] = (ll)ifac[i - 1] * ifac[i] % MOD;
    }
}
inline int C(int n, int m) {
    if (n < 0 && m >= 0)
        return (m & 1) ? MOD - C(-n + m - 1, m) : C(-n + m - 1, m);
    if (n < m) return 0;
    return (ll)fac[n] * ifac[m] % MOD * ifac[n - m] % MOD;
}
inline void add(int &x, int a) {
    x += a; while (x >= MOD) x -= MOD;
}
inline int pow(int a, int b) {
    int c = 1;
    while (b) {
        if (b & 1) c = (ll)c * a % MOD;
        b >>= 1; a = (ll)a * a % MOD;
    }
    return c;
}
inline int inv(int x) {
    return pow(x, MOD - 2);
}

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    cin >> test; pre(200000);
    while (test--) {
        cin >> n >> lo >> hi;
        int m = hi - lo + 1;
        ans = 0;
        for (int k = 1; k <= n; k++)
            for (int j = 0; n - k - j >= 0 && j / k <= m - 1; j += k) {
                int up = C(m - 1, j / k);
                if ((j / k) & 1) up = MOD - up;
                int down = C(m + n - k - j - 2, n - k - j);
                add(ans, (ll)up * down % MOD);
            }
        ans = (ll)ans * m % MOD;
        cout << ans << endl;
    }
    return 0;
}