#2409. 「THUPC 2017」小 L 的計算題 / Sum（生成函數）

題意

給定一個長爲 \(n\) 的序列 \(\{a_i\}\) 對於 \(k \in [1, n]\) 求
\[ f_k = \sum_{i = 1}^{n} a_i^k \pmod {998244353} \]

\(n \le 2 \times 10^5\)

題解

不會牛頓恆等式TAT，參考了這位大佬的博客。

我們令 \(F(x)\) 爲 \(f_k\) 的生成函數，我們有
\[ \begin{aligned} F(x) &= \sum_{k} (\sum_{i = 1}^{n} a_i^k) x^k\\ &=\sum_{i = 1}^n \sum_k a_i^kx^k\\ &=\sum_{i = 1}^{n} \frac{1}{1 - a_ix}\\ \end{aligned} \]

這幾步都比較基礎，但看起來還是挺不可做的，我們用一些神奇的操作。
\[ \begin{aligned} F(x) &= \sum_{i = 1}^{n} (1 + \frac{a_ix}{1 - a_ix})\\ &= n - x \sum_{i= 1}^{n} \frac{-a_i}{1 - a_ix}\\ &= n - x \sum_{i = 1}^{n} \ln'(1 - a_ix)\\ &= n - x (\ln (\prod_{i = 1}^n(1 - a_ix)))' \end{aligned} \]

我們先分治求出 \(\prod_{i = 1}^{n}(1 - a_ix)\) 然後套 \(\ln\) 的板子即可。\(\mathcal O(n \log^2 n)\)

代碼

#include <bits/stdc++.h>

#define For(i, l, r) for (register int i = (l), i##end = int(r); i <= i##end; ++i)
#define Fordown(i, r, l) for (register int i = (r), i##end = (int)(l); i >= i##end; --i)
#define Rep(i, r) for (register int i = (0), i##end = int(r); i < i##end; ++i)
#define Set(a, v) memset(a, v, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define debug(x) cout << #x << ": " << (x) << endl

using namespace std;

typedef vector<int> VI;

template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return b > a ? a = b, 1 : 0; }

inline int read() {
    int x(0), sgn(1); char ch(getchar());
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') sgn = -1;
    for (; isdigit(ch); ch = getchar()) x = (x * 10) + (ch ^ 48);
    return x * sgn;
}

void File() {
#ifdef zjp_shadow
    freopen ("2409.in", "r", stdin);
    freopen ("2409.out", "w", stdout);
#endif
}

const int Mod = 998244353;

inline int fpm(int x, int power) {
    int res(1);
    for (; power; power >>= 1, x = 1ll * x * x % Mod)
        if (power & 1) res = 1ll * res * x % Mod;
    return res;
}

int findlen(int l) {
    int res = 1; while (res <= l) res <<= 1; return res;
}

template<int Maxn>
struct Poly {

    const int g = 3, invg = fpm(g, Mod - 2);

    int rev[Maxn], W[Maxn], len;

    void NTT(int *P, int opt) {
        Rep (i, len) if (i < rev[i]) swap(P[i], P[rev[i]]);
        for (int i = 2, p; p = i >> 1, i <= len; i <<= 1) {
            W[0] = 1; W[1] = fpm(~opt ? g : invg, (Mod - 1) / i);
            For (k, 2, p - 1) W[k] = 1ll * W[k - 1] * W[1] % Mod;
            for (int j = 0; j < len; j += i) Rep (k, p) {
                int u = P[j + k], v = 1ll * P[j + k + p] * W[k] % Mod;
                P[j + k] = (u + v) % Mod; P[j + k + p] = (u - v + Mod) % Mod;
            }
        }
        if (!~opt) {
            int invn = fpm(len, Mod - 2);
            Rep (i, len) P[i] = 1ll * P[i] * invn % Mod;
        }
    }

    void Prepare(int lc) {
        int cnt = -1;
        for (len = 1; len <= lc; len <<= 1) ++ cnt;
        Rep (i, len) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << cnt);
    }

    int A[Maxn], B[Maxn], C[Maxn];
    void Mult(int *a, int *b, int *c, int la, int lb) {
        int lc = la + lb; Prepare(lc);
        if (lc <= 400) {
            static int tmp[410] = {0};
            For (i, 0, la) if (a[i]) For (j, 0, lb)
                tmp[i + j] = (tmp[i + j] + 1ll * a[i] * b[j]) % Mod;
            For (i, 0, lc) c[i] = tmp[i], tmp[i] = 0; return;
        }
        Rep (i, len) A[i] = i <= la ? a[i] : 0; NTT(A, 1);
        Rep (i, len) B[i] = i <= lb ? b[i] : 0; NTT(B, 1);
        Rep (i, len) C[i] = 1ll * A[i] * B[i] % Mod; NTT(C, -1);
        For (i, 0, lc) c[i] = C[i];
    }

    VI Mult(VI a, VI b) {
        static int ta[Maxn], tb[Maxn], tc[Maxn]; VI c;
        Rep (i, a.size()) ta[i] = a[i];
        Rep (i, b.size()) tb[i] = b[i];
        Mult(ta, tb, tc, a.size() - 1, b.size() - 1);
        Rep (i, a.size() + b.size() - 1) c.push_back(tc[i]); return c;
    }

    void Inv(int *f, int *g, int lf) {
        if (lf == 1) return void(g[0] = fpm(f[0], Mod - 2));
        Inv(f, g, lf >> 1); Prepare(lf << 1);
        Rep (i, len) A[i] = i < lf ? f[i] : 0; NTT(A, 1); 
        Rep (i, len) B[i] = i < lf ? g[i] : 0; NTT(B, 1);
        Rep (i, len) C[i] = 1ll * A[i] * B[i] % Mod * B[i] % Mod; NTT(C, -1);
        Rep (i, lf) g[i] = (g[i] * 2ll + Mod - C[i]) % Mod;
    }

    int inv[Maxn];
    Poly() {
        inv[1] = 1;
        For (i, 2, Maxn - 1)
            inv[i] = 1ll * inv[Mod % i] * (Mod - Mod / i) % Mod;
    }

    void Der(int *f, int *g, int lf) {
        For (i, 1, lf) g[i - 1] = 1ll * i * f[i] % Mod; g[lf] = 0;
    }

    void Int(int *f, int *g, int lf) {
        g[0] = 0;
        For (i, 1, lf + 1)
            g[i] = 1ll * f[i - 1] * inv[i] % Mod;
    }

    void Ln(int *f, int *g, int lf) {
        static int der[Maxn], tmp[Maxn];
        Der(f, der, lf); Inv(f, tmp, lf);
        Mult(der, tmp, tmp, lf, lf); Int(tmp, g, lf);
    }

};

const int N = 1 << 20;

Poly<N> T;

int n, a[N], f[N], g[N], ans[N];

VI Solve(int l, int r) {
    if (l == r) return {1, Mod - a[l]};
    int mid = (l + r) >> 1;
    return T.Mult(Solve(l, mid), Solve(mid + 1, r));
}

int main () {

    File();

    int cases = read();

    while (cases --) {
        For (i, 1, n = read()) a[i] = read() % Mod;

        VI res = Solve(1, n);
        For (i, 0, n) f[i] = res[i];

        T.Ln(f, g, findlen(n)); T.Der(g, f, n);
        ans[0] = n; For (i, 1, n) ans[i] = Mod - f[i - 1];

        int Ans = 0;
        For (i, 1, n) Ans ^= ans[i];
        printf ("%d\n", Ans);
    }

    return 0;

}