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P4721 【模版】分治FFT 题解

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P4721 【模版】分治FFT 题解

题面

题目链接

解法

由于$f_i$依赖于$f_j(1\le j\lt i)$，所以考虑分治计算，每次先计算左半段，将贡献加到右半段后继续递归计算右半段。

思考如何计算贡献。发现题目中$fi=\sum\limits{j=1}^{i}{f_{i-j}g_j}$的方式很像多项式乘法，区间$[l,mid]$对$f_i(mid \lt i \le r)$的贡献就是$F(x) = f(x) \times g(x)$的$F(i)$，用$NTT$进行计算即可。

时间复杂度$O(n \log^2 n)$（似乎比多项式求逆要慢不少）。

AC代码

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/**
 * @file:           P4721.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P4721
*/
// #pragma GCC optimize ("O2")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 1 << 18;
const int INFL = 0x3f3f3f3f3f3f3f3f;
const int MOD = 998244353;
const int G0 = 3;
namespace maths {
    int qpow(int x, int y) {
        int ret = 1;
        for (; y; y >>= 1, x = x * x % MOD)
            if (y & 1)
                ret = ret * x % MOD;
        return ret;
    }
    void change(int *f, int len) {
        static int rev[MAXN];
        for (int i = rev[0] = 0; i < len; ++i) {
            rev[i] = rev[i >> 1] >> 1;
            if (i & 1)
                rev[i] |= len >> 1;
        }
        for (int i = 0; i < len; ++i)
            if (i < rev[i])
                swap(f[i], f[rev[i]]);
    }
    void ntt(int *f, int len, int on) {
        change(f, len);
        for (int h = 2; h <= len; h <<= 1) {
            int gn = qpow(G0, (MOD - 1) / h);
            for (int j = 0; j < len; j += h) {
                int g = 1;
                for (int k = j; k < j + h / 2; ++k) {
                    int u = f[k], t = g * f[k + h / 2] % MOD;
                    f[k] = (u + t + MOD) % MOD;
                    f[k + h / 2] = (u - t + MOD) % MOD;
                    g = g * gn % MOD;
                }
            }
        }
        if (on == -1) {
            reverse(f + 1, f + len);
            int inv = qpow(len, MOD - 2);
            for (int i = 0; i < len; ++i)
                f[i] = f[i] * inv % MOD;
        }
    }
}
using maths::ntt;
int n, a[MAXN], b[MAXN], f[MAXN], g[MAXN];
void solve(int l, int r) {
    if (l >= r)
        return;
    int len = r - l + 1;
    int mid = (l + r) >> 1;
    solve(l, mid);
    copy(f + l, f + mid + 1, a);
    copy(g, g + len, b);
    fill(a + len / 2, a + len, 0);
    ntt(a, len, 1), ntt(b, len, 1);
    for (int i = 0; i < len; ++i)
        a[i] = a[i] * b[i] % MOD;
    ntt(a, len, -1);
    for (int i = mid + 1; i <= r; ++i)
        f[i] = (f[i] + a[i - l]) % MOD;
    solve(mid + 1, r);
}
signed main() {
    read(n);
    for (int i = 1; i < n; ++i)
        read(g[i]);
    int len = f[0] = 1;
    while (len < n)
        len <<= 1;
    solve(0, len - 1);
    for (int i = 0; i < n; ++i)
        write(f[i]), putchar(" \n"[i == n - 1]);
    return 0;
}