P4245 模板任意模数多项式乘法(NTT)

Posted 2021-09-07 Jozky86

题意：

P4245 【模板】任意模数多项式乘法

题解：

NTT模板，记录一下

代码：

#include <bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for (int i= (a), _end_= (b); i < _end_; ++i)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

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

typedef long long LL;

const int oo= 0x3f3f3f3f;

int Mod;

const int max0= 262144;

struct comp
{
    double x, y;

    comp() : x(0), y(0)
    {
    }
    comp(const double& _x, const double& _y) : x(_x), y(_y)
    {
    }
};

inline comp operator+(const comp& a, const comp& b)
{
    return comp(a.x + b.x, a.y + b.y);
}
inline comp operator-(const comp& a, const comp& b)
{
    return comp(a.x - b.x, a.y - b.y);
}
inline comp operator*(const comp& a, const comp& b)
{
    return comp(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}
inline comp conj(const comp& a)
{
    return comp(a.x, -a.y);
}

const double PI= acos(-1);

int N, L;

comp w[max0 + 5];
int bitrev[max0 + 5];

void fft(comp* a, const int& n)
{
    REP(i, 0, n) if (i < bitrev[i]) swap(a[i], a[bitrev[i]]);
    for (int i= 2, lyc= n >> 1; i <= n; i<<= 1, lyc>>= 1)
        for (int j= 0; j < n; j+= i) {
            comp *l= a + j, *r= a + j + (i >> 1), *p= w;
            REP(k, 0, i >> 1)
            {
                comp tmp= *r * *p;
                *r= *l - tmp, *l= *l + tmp;
                ++l, ++r, p+= lyc;
            }
        }
}

inline void fft_prepare()
{
    REP(i, 0, N) bitrev[i]= bitrev[i >> 1] >> 1 | ((i & 1) << (L - 1));
    REP(i, 0, N) w[i]= comp(cos(2 * PI * i / N), sin(2 * PI * i / N));
}

inline void conv(int* x, int* y, int* z)
{
    REP(i, 0, N)(x[i]+= Mod)%= Mod, (y[i]+= Mod)%= Mod;
    static comp a[max0 + 5], b[max0 + 5];
    static comp dfta[max0 + 5], dftb[max0 + 5], dftc[max0 + 5], dftd[max0 + 5];

    REP(i, 0, N) a[i]= comp(x[i] & 32767, x[i] >> 15);
    REP(i, 0, N) b[i]= comp(y[i] & 32767, y[i] >> 15);
    fft(a, N), fft(b, N);
    REP(i, 0, N)
    {
        int j= (N - i) & (N - 1);
        static comp da, db, dc, dd;
        da= (a[i] + conj(a[j])) * comp(0.5, 0);
        db= (a[i] - conj(a[j])) * comp(0, -0.5);
        dc= (b[i] + conj(b[j])) * comp(0.5, 0);
        dd= (b[i] - conj(b[j])) * comp(0, -0.5);
        dfta[j]= da * dc;
        dftb[j]= da * dd;
        dftc[j]= db * dc;
        dftd[j]= db * dd;
    }
    REP(i, 0, N) a[i]= dfta[i] + dftb[i] * comp(0, 1);
    REP(i, 0, N) b[i]= dftc[i] + dftd[i] * comp(0, 1);
    fft(a, N), fft(b, N);
    REP(i, 0, N)
    {
        int da= (LL)(a[i].x / N + 0.5) % Mod;
        int db= (LL)(a[i].y / N + 0.5) % Mod;
        int dc= (LL)(b[i].x / N + 0.5) % Mod;
        int dd= (LL)(b[i].y / N + 0.5) % Mod;
        z[i]= (da + ((LL)(db + dc) << 15) + ((LL)dd << 30)) % Mod;
    }
}

int main()
{
    // #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    // #endif
    int n, m;
    static int a[max0 + 5], b[max0 + 5], c[max0 + 5];
    scanf("%d%d%d", &n, &m, &Mod), ++n, ++m;
    REP(i, 0, n) scanf("%d", a + i);
    REP(i, 0, m) scanf("%d", b + i);
    L= 0;
    for (; (1 << L) < n + m - 1; ++L)
        ;
    N= 1 << L;
    fft_prepare();
    conv(a, b, c);
    REP(i, 0, n + m - 1)(c[i]+= Mod)%= Mod, printf("%d ", c[i]);
    printf("\\n");
    return 0;
}