多项式板子·新

Posted 2021-02-04 oier

之前写过一个vector的多项式汇总。。。不过常数特别大

最近事儿比较少，准备写一个int[]的

目前没写完，先打了发多项式求逆，蛙了好几发。。。

最后发现递归下去回来之后忘了清系数了。。。

int[]版的主要是数组大小和高位系数比较恶心，还有数组清空啥的。。。

不过常数肯定比vector小很多，所以还是尝试写一下好了

#include <cstdio>
#include <algorithm>
using namespace std;

const int p = 998244353;

int qpow(int x, int y)
{
    int res = 1;
    while (y > 0)
    {
        if (y & 1) res = res * (long long)x % p;
        x = x * (long long)x % p, y >>= 1;
    }
    return res;
}

void ntt(int *a, int len, int flag)
{
    int *r = new int[len];
    r[0] = 0;
    for (int i = 1; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
    for (int i = 0; i < len; i++) if (i < r[i]) swap(a[i], a[r[i]]);
    for (int i = 1; i < len; i <<= 1)
    {
        int g1 = qpow(3, (p - 1) / (i * 2));
        for (int j = 0; j < len; j += i << 1)
            for (int g = 1, k = 0; k < i; k++, g = g * (long long)g1 % p)
            {
                int t = a[j + i + k] * (long long)g % p;
                a[j + i + k] = ((a[j + k] - t) % p + p) % p;
                a[j + k] = (a[j + k] + t) % p;
            }
    }
    if (flag == -1)
    {
        reverse(a + 1, a + len);
        for (int i = 0, inv = qpow(len, p - 2); i < len; i++) a[i] = a[i] * (long long)inv % p;
    }
    delete []r;
}

void poly_inv(int *a, int len)
{
    if (len == 1) { a[0] = qpow(a[0], p - 2); return; }
    int len1 = len / 2;
    int *f0 = new int[len * 2];
    for (int i = 0; i < len1; i++) f0[i] = a[i];
    for (int i = len1; i < len * 2; i++) f0[i] = 0;
    poly_inv(f0, len1);
    for (int i = len1; i < len * 2; i++) f0[i] = 0;
    ntt(f0, len * 2, 1), ntt(a, len * 2, 1);
    for (int i = 0; i < len * 2; i++) a[i] = ((2 * f0[i] % p - a[i] * (long long)f0[i] % p * f0[i] % p) % p + p) % p;
    ntt(a, len * 2, -1);
    delete []f0;
}

int a[500000], n, len = 1;

int main()
{
    scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    while (len < n) len *= 2;
    poly_inv(a, len);
    for (int i = 0; i < n; i++) printf("%d ", a[i]);
    return 0;
}