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);// #endifint 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;
}