[洛谷P4245]【模板】任意模数NTT-白红宇

[洛谷P4245]【模板】任意模数NTT

阅读量：6902 次

发布时间：2019-06-27

本文共 3844 字，大约阅读时间需要 12 分钟。

题目大意：给你两个多项式$f(x)$和$g(x)$以及一个模数$p(p\leqslant10^9)$，求$f*g\pmod p$

题解：任意模数$NTT$，最大的数为$p^2\times\max\{n,m\}\leqslant10^{23}$，所以一般选$3$个模数即可，求出这三个模数下的答案，然后中国剩余定理即可。

假设这一位的答案是$x$，三个模数分别为$A,B,C$，那么：

x\equiv x_1\pmod{A}\\

x\equiv x_2\pmod{B}\\

x\equiv x_3\pmod{C}

先把前两个合并：

x_1+k_1A=x_2+k_2B\\

x_1+k_1A\equiv x_2\pmod{B}\\

k_1\equiv \frac{x_2-x_1}A\pmod{B}

于是求出了$k_1$，也就求出了$x\equiv x_1+k_1A\pmod{AB}$，记$x_4=x_1+k_1A$

x_4+k_4AB=x_3+k_3C\\

x_4+k_4AB\equiv x_3\pmod{C}\\

k_4\equiv \dfrac{x_3-x_4}{AB}\pmod{C}

求出了$k_4$，$x\equiv x_4+k_4AB\pmod{ABC}$，因为$x<ABC$，所以$x=x_4+k_4AB$

卡点：$Wn$数组开小，中国剩余定理写错

C++ Code：

#include 
     
      #include 
      
       #include 
       
        int mod;namespace Math {	inline int pw(int base, int p, const int mod) {		static int res;		for (res = 1; p; p >>= 1, base = static_cast
        
          (base) * base % mod) if (p & 1) res = static_cast
         
           (res) * base % mod;		return res;	}	inline int inv(int x, const int mod) { return pw(x, mod - 2, mod); }}const int mod1 = 998244353, mod2 = 1004535809, mod3 = 469762049, G = 3;const long long mod_1_2 = static_cast
          
            (mod1) * mod2;const int inv_1 = Math::inv(mod1, mod2), inv_2 = Math::inv(mod_1_2 % mod3, mod3);struct Int { int A, B, C; explicit inline Int() { } explicit inline Int(int __num) : A(__num), B(__num), C(__num) { } explicit inline Int(int __A, int __B, int __C) : A(__A), B(__B), C(__C) { } static inline Int reduce(const Int &x) { return Int(x.A + (x.A >> 31 & mod1), x.B + (x.B >> 31 & mod2), x.C + (x.C >> 31 & mod3)); } inline friend Int operator + (const Int &lhs, const Int &rhs) { return reduce(Int(lhs.A + rhs.A - mod1, lhs.B + rhs.B - mod2, lhs.C + rhs.C - mod3)); } inline friend Int operator - (const Int &lhs, const Int &rhs) { return reduce(Int(lhs.A - rhs.A, lhs.B - rhs.B, lhs.C - rhs.C)); } inline friend Int operator * (const Int &lhs, const Int &rhs) { return Int(static_cast
           
             (lhs.A) * rhs.A % mod1, static_cast
            
              (lhs.B) * rhs.B % mod2, static_cast
             
               (lhs.C) * rhs.C % mod3); } inline int get() { long long x = static_cast
              
                (B - A + mod2) % mod2 * inv_1 % mod2 * mod1 + A; return (static_cast
               
                 (C - x % mod3 + mod3) % mod3 * inv_2 % mod3 * (mod_1_2 % mod) % mod + x) % mod; }} ;#define maxn 131072namespace Poly {#define N (maxn << 1) int lim, s, rev[N]; Int Wn[N | 1]; inline void init(int n) { s = -1, lim = 1; while (lim < n) lim <<= 1, ++s; for (register int i = 1; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s; const Int t(Math::pw(G, (mod1 - 1) / lim, mod1), Math::pw(G, (mod2 - 1) / lim, mod2), Math::pw(G, (mod3 - 1) / lim, mod3)); *Wn = Int(1); for (register Int *i = Wn; i != Wn + lim; ++i) *(i + 1) = *i * t; } inline void NTT(Int *A, const int op = 1) { for (register int i = 1; i < lim; ++i) if (i < rev[i]) std::swap(A[i], A[rev[i]]); for (register int mid = 1; mid < lim; mid <<= 1) { const int t = lim / mid >> 1; for (register int i = 0; i < lim; i += mid << 1) { for (register int j = 0; j < mid; ++j) { const Int W = op ? Wn[t * j] : Wn[lim - t * j]; const Int X = A[i + j], Y = A[i + j + mid] * W; A[i + j] = X + Y, A[i + j + mid] = X - Y; } } } if (!op) { const Int ilim(Math::inv(lim, mod1), Math::inv(lim, mod2), Math::inv(lim, mod3)); for (register Int *i = A; i != A + lim; ++i) *i = (*i) * ilim; } }#undef N}int n, m;Int A[maxn << 1], B[maxn << 1];int main() { scanf("%d%d%d", &n, &m, &mod); ++n, ++m; for (int i = 0, x; i < n; ++i) scanf("%d", &x), A[i] = Int(x % mod); for (int i = 0, x; i < m; ++i) scanf("%d", &x), B[i] = Int(x % mod); Poly::init(n + m); Poly::NTT(A), Poly::NTT(B); for (int i = 0; i < Poly::lim; ++i) A[i] = A[i] * B[i]; Poly::NTT(A, 0); for (int i = 0; i < n + m - 1; ++i) { printf("%d", A[i].get()); putchar(i == n + m - 2 ? '\n' : ' '); } return 0;}