2333

\(\color{red}{\text{多项式全集之二 任长任模的FFT:}}\)

三模NTT实现任模FFT：

\(1.\)为什么要用MTT：当\(p\)不是NTT模数或者多项式长度大于模数限制时，就要使用MTT。

\(2.\)MTT的使用原理：我们对初始多项式取模，那么如果在不取模卷积情况下，答案\(x\)不会超过\(N\times p^2\)。我们取三个NTT模数\(p_1,p_2,p_3\)，分别做多项式乘法，得到\(x\)分别\(mod~p_1,p_2,p_3\)的答案，通过CRT合并可以得到\(x~mod~p_1p_2p_3\)的答案，如果\(x<p_1p_2p_3\)那么就可以得到准确的答案，再对\(p\)取模即可。

\(3.\)CRT合并的小优化：

\(step~0:\)初始式子

\[{\begin{cases}x\equiv c_1(mod~p_1)\\x\equiv c_2(mod~p_2)\\x\equiv c_3(mod~p_3)\end{cases}} \]

\(step~1:\)把一式二式合并(LL范围内)。

\[{\begin{cases}x\equiv a(mod~p_1p_2)\\x\equiv c_3(mod~p_3)\end{cases}} \]

\(step~2:\)再次合并(不需要\(long~double\) 快速乘)。

\(4.\)常用NTT模数：

以下模数的共同\(g=3189\)

\(p=r\times 2^k+1\) \(k\) \(g\)
\(104857601\) \(22\) \(3\)
\(167772161\) \(25\) \(3\)
\(469762049\) \(26\) \(3\)
\(950009857\) \(21\) \(7\)
\(998244353\) \(23\) \(3\)
\(1004535809\) \(21\) \(3\)
\(2013265921\) \(27\) \(31\)
\(2281701377\) \(27\) \(3\)
\(3221225473\) \(30\) \(5\)
拆系数FFT(CFFT)实现任模FFT：

\(1.\)实现原理：运用实数FFT不取模做乘法，然后取模回归到整数。但是由于误差较大（值域是\(10^{23}\)），我们令\(t=\sqrt{m}\)把系数\(a_i=k_it+b_i\)，对\(k_i,t_i\)交叉做四遍卷积，求出答案按系数贡献取模加入。

\(2.\)可按合并DFT的方法优化DFT次数。

\(bluestein\)算法实现任长FFT：

当\(m\)不是\(2\)的幂次的时候，我们从式子入手：
令\(X_i=a_iw_m^{\frac {i^2} 2},Y_i=w_m^{\frac{-i^2}2}\)

\(\begin{align*} A_k & = \sum_{j=0}^{m-1}a_jw_m^{jk}\\ & = \sum_{j=0}^{m-1}a_jw_m^{\frac{j^2+k^2-{(k-j)}^2}{2}}\\ &=w_m^{\frac {k^2} 2}\sum_{j=0}^{m-1}a_jw_m^{\frac{j^2} 2}w_m^{\frac{-{(k-j)}^2}{2}}\\ &=w_m^{\frac {k^2} 2}\sum_{j=0}^{m-1}X_jY_{k-j} \end{align*}\)

喜闻乐见的模板：

三模NTT模板（注意：不可以MTT回来，因为系数会取模）

namespace MTT{
	typedef long long LL;
	int n, m;
	LL p, mod;
	const LL p1 = 998244353;
	const LL p2 = 1004535809;
	const LL p3 = 104857601;
	const int g = 3189;
	LL a[300005], b[300005], c[300005], cpa[300005], cpb[300005];
	LL c3[300005], c1[300005], c2[300005];
	LL qpow(LL a, LL b, LL mod) {
		LL ans = 1;
		while(b) {
			if(b & 1)	ans = ans * a % mod;
			a = a * a % mod;
			b >>= 1;
		}
		return ans;
	}
	const LL inv12 = qpow(p1, p2 - 2, p2);
	const LL inv123 = qpow(p1 * p2 % p3, p3 - 2, p3);
	struct p_l_e{
		int wz[300005];
		void MTT(LL *a, int N, int op) {
			for(int i = 0; i < N; i++)
				if(i < wz[i]) swap(a[i], a[wz[i]]);
			for(int le = 2; le <= N; le <<= 1) {
				int mid = le >> 1;
				LL wn = qpow(g, (mod - 1) / le, mod);
				if(op == -1) wn = qpow(wn, mod - 2, mod);
				for(int i = 0; i < N ;i += le) {
					LL w = 1, x, y;
					for(int j = 0; j < mid; j++) {
						x = a[i + j];
						y = a[i + j + mid] * w % mod;
						a[i + j] = (x + y) % mod;
						a[i + j + mid] = (x - y + mod) % mod;
						w = w * wn % mod;
					}
				}
			}
		}
		void mult(LL *a, LL *b, LL *c, int M) {
			int N = 1, len = 0;
			while(N < M) N <<= 1, len++;
			for(int i = 0; i < N; i++)
				wz[i] = (wz[i >> 1] >> 1) | ((i & 1) << (len - 1));
			MTT(a, N, 1); MTT(b, N, 1);
			for(int i = 0; i < N; i++)	c[i] = a[i] * b[i] % mod;
			MTT(c, N, -1);
			LL t = qpow(N, mod - 2, mod);
			for(int i = 0; i < N; i++)	c[i] = c[i] * t % mod;
		}

	}PLE;
	LL CRT(LL c1, LL c2, LL c3) {
		LL x = (c1 + p1 * ((c2 - c1 + p2) % p2 * inv12 % p2));
		LL y = (x % p + p1 * p2 % p * ((c3 - x % p3 + p3) % p3 * inv123 % p3) % p) % p;
		return y;
	}
	void merge(LL *c1, LL *c2, LL *c3, LL *c, int N) {
		for(int i = 0; i < N; i++)
			c[i] = CRT(c1[i], c2[i], c3[i]);
		return;
	}
	void main() {
		scanf("%d%d%lld", &n, &m, &p); n++; m++;
		for(int i = 0; i < n; i++)	scanf("%lld", &a[i]);
		for(int i = 0; i < m; i++)	scanf("%lld", &b[i]);
		mod = p1; memcpy(cpa, a, sizeof(a)); memcpy(cpb, b, sizeof(b)); PLE.mult(cpa, cpb, c1, n + m - 1);
		mod = p2; memcpy(cpa, a, sizeof(a)); memcpy(cpb, b, sizeof(b)); PLE.mult(cpa, cpb, c2, n + m - 1);
		mod = p3; memcpy(cpa, a, sizeof(a)); memcpy(cpb, b, sizeof(b)); PLE.mult(cpa, cpb, c3, n + m - 1);
		merge(c1, c2, c3, c, n + m - 1);
		for(int i = 0; i < n + m - 1; i++)	printf("%lld ", (c[i] % p + p) % p);
		return;
	}
}

拆系数FFT模板（注意：相同系数的两项可以合并一起IDFT。采用共轭优化法，只进行四次DFT）

namespace CFFT{
	typedef long long LL;
	int n, m, p ,sqrp; 
	int a[300005], b[300005];
	const long double pi = acos(-1);
	struct cp{
		long double x, y;
		cp() {x = y = 0;}
		cp(long double X,long double Y) {x = X; y = Y; }
		cp conj() {return (cp) {x, -y};}
	}ka[300005], kb[300005], ta[300005], tb[300005], kk[300005], kt[300005], tt[300005], c[300005], I(0, 1), d[300005];
	 
	cp operator+ (const cp &a, const cp &b) {return (cp){a.x + b.x, a.y + b.y}; }
	cp operator- (const cp &a, const cp &b) {return (cp){a.x - b.x, a.y - b.y}; }
	cp operator* (const cp &a, const cp &b) {return (cp){a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};}
	cp operator* (const cp &a, long double b) {return (cp){a.x * b, a.y * b};}
	cp operator/ (const cp &a, long double b) {return (cp){a.x / b, a.y / b};}	
	struct p_l_e{
		int wz[300005];
		void FFT(cp *a, int N, int op){
			for(int i = 0; i < N; i++)
				if(i < wz[i])	swap(a[i], a[wz[i]]);
			for(int le = 2; le <= N; le <<= 1){
				int mid = le >> 1;
				cp x, y, w, wn = (cp){cos(op * 2 * pi / le), sin(op * 2 * pi / le)};
				for(int i = 0; i < N; i += le){
					w = (cp){1, 0};
					for(int j = 0; j < mid; j++){
						x = a[i + j];
						y = a[i + j + mid] * w;
						a[i + j] = x + y;
						a[i + j + mid] = x - y;
						w = w * wn;
					}
				}
			} 
		}
		void D_FFT(cp *a, cp *b, int N, int op){
			for(int i = 0; i < N; i++)	d[i] = a[i] + I * b[i];
			FFT(d, N, op);
			d[N] = d[0];
			if(op == 1){
				for(int i = 0; i < N; i++){
					a[i] = (d[i] + d[N - i].conj()) / 2;
					b[i] = I * (-1) * (d[i] - d[N - i].conj()) / 2;
				}
			} else {
				for(int i = 0; i < N; i++){
					a[i] = cp(d[i].x, 0);
					b[i] = cp(d[i].y, 0);
				}
			}
			d[N] = cp(0, 0);
		}
		void mult(int *a, int *b, int M){
			int N = 1, len = 0;
			while(N < M) N <<= 1, len++;
			for(int i = 0; i < N; i++)
				wz[i] = (wz[i >> 1] >> 1) | ((i & 1) << (len - 1));
			for(int i = 0; i < N; i++){
				ka[i].x = a[i] >> 15;
				kb[i].x = b[i] >> 15;
				ta[i].x = a[i] & 32767;
				tb[i].x = b[i] & 32767;
			}
			D_FFT(ta, ka, N, 1); D_FFT(tb, kb, N, 1);
			for(int i = 0; i < N; i++){
				kk[i] = ka[i] * kb[i];
				kt[i] = ka[i] * tb[i] + ta[i] * kb[i];
				tt[i] = ta[i] * tb[i];
			}
			D_FFT(tt, kk, N, -1); FFT(kt, N, -1);
			for(int i = 0; i < N; i++){
				tt[i] = tt[i] / N;
				kt[i] = kt[i] / N;
				kk[i] = kk[i] / N;
			}
		}
	}PLE;
	void main() {
		scanf("%d%d%d", &n, &m, &p); n++; m++;
		for(int i = 0; i < n; i++)	scanf("%d", &a[i]),a[i] = a[i] % p;
		for(int i = 0; i < m; i++)	scanf("%d", &b[i]),b[i] = b[i] % p;
		PLE.mult(a, b, n + m - 1);
		for(int i = 0; i < n + m - 1; i++)
			printf("%lld ",(((((LL)round(kk[i].x)) % p) << 30) + ((((LL)round(kt[i].x)) % p) << 15) + ((LL)round(tt[i].x)) % p) % p);
	}
}

\(blue\_stein\)模板：

struct polynie {
	CP getw(int m, int k, int op) {
		return CP(cos(2 * pi * k / m), op * sin(2 * pi * k / m));
	}
	int wz[MAXN];
	CP A[MAXN], B[MAXN], C[MAXN];
	void FFT(CP *a, int N, int op) {
		rop(i, 0, N) if(i < wz[i]) swap(a[i], a[wz[i]]);
		for(int l = 2; l <= N; l <<= 1) {
			int mid = l >> 1;
			CP x, y, w, wn = CP(cos(pi / mid), sin(op * pi / mid));
			for(int i = 0; i < N; i += l) {
				w = CP(1, 0);
				rop(j, 0, mid) {
					x = a[i + j];
					y = w * a[i + j + mid];
					a[i + j] = x + y;
					a[i + j + mid] = x - y;
					w = w * wn;
				}
			}
		}
	}
	void mult(CP *a, CP *b, CP *c, int M) {
		int N = 1, len = 0;
		while(N < M) N <<= 1, len++;
		rop(i, 0, N) wz[i] = (wz[i >> 1] >> 1) | ((i & 1) << (len - 1));
		FFT(a, N, 1); FFT(b, N, 1);
		rop(i, 0, N) c[i] = a[i] * b[i];
		FFT(c, N, -1);
		rop(i, 0, N) c[i].x = c[i].x / N, c[i].y = c[i].y / N;
	}
	void blue_stein(CP *a, int M, int op) {
		int M2 = M << 1;
		memset(A, 0, sizeof(A));
		memset(B, 0, sizeof(B));
		memset(C, 0, sizeof(C));
		rop(i, 0, M) A[i] = a[i] * getw(M2, 1ll * i * i % M2, op);
		rop(i, 0, M2) B[i] = getw(M2, 1ll * (i - M) * (i - M) % M2, -op);
		mult(A, B, C, M2 + M - 1);
		rop(i, 0, M) a[i] = C[i + M] * getw(M2, 1ll * i * i % M2, op);
		if(op == -1) rop(i, 0, M) a[i].x = a[i].x / M, a[i].y = a[i].y / M;
	}
}PLE;

\(\color{red}{\text{多项式全集之三 多项式求逆与除法:}}\)

多项式求逆：

\(1.\)问题描述：

已知\(F(x)\)，且\(F(x)G(x)\equiv 1 (mod~x^n)\)，求\(G(x)\)

\(2.\)推导过程：

\begin{align}
B(x)&\equiv F(x){-1}&(mod~x)\
由于
G(x)&\equiv F(x)^{-1}& (mod~x^n)\
所以
G(x)&\equiv F(x)^{-1}& (mod~x^{\lceil \frac n 2 \rceil})\
G(x)& \equiv B(x)&(mod~x^{\lceil \frac n 2 \rceil})\
G(x)-B(x)& \equiv 0&(mod~x^{\lceil \frac n 2 \rceil})\
\end{align}
两边平方，得：

由于\([G(x)-B(x)]^2\)的第\(k<n\)项为

\[\sum_{i=0}^k[g_ix^i-b_ix^i][g_{k-i}x^{k-i}-b_{k-i}x^{k-i}] \]

\(i,k-i\)一定有一项\(<\frac n 2\)，所以
\begin{align}
[G(x)-B(x)]^2&\equiv 0 &(mod~x^n)\
G2(x)+B2(x)-2G(x)B(x)&\equiv 0&(mod~x^n)
\end{align}
两边同乘\(A(x)\)，得：
\begin{align}
A(x)G2(x)+A(x)B2(x)-2A(x)G(x)B(x)&\equiv 0& (mod~x^n)\
G(x)+A(x)B^2(x)-2B(x)&\equiv 0& (mod~x^n)\
G(x)&\equiv 2B(x)-A(x)B2(x)&(mod~xn)\
G(x)&\equiv B(x)[2-A(x)B(x)]&(mod~x^n)\
\end{align}

多项式除法：

\(1.\)问题描述：

已知一个\(n\)次多项式\(A(x)\)，一个\(m\)次多项式\(B(x)\)，且\(A(x)=B(x)C(x)+D(x)\)，求\(n-m\)次多项式\(C(x)\)，\(<m\)次多项式\(D(x)\)。

\(2.\)推导过程：

由\(A(x) = B(x)C(x)+D(x)\)得：
\begin{align}
A(\frac 1 x)&=B(\frac 1 x)C(\frac 1 x)+D(\frac 1 x)\
x^nA(\frac 1 x)&=x^nB(\frac 1 x)C(\frac 1 x)+x^nD(\frac 1 x)\
x^nA(\frac 1 x)&=x^mB(\frac 1 x)x^{n-m}C(\frac 1 x)+x{m-1}xD(\frac 1 x)\
A_r(x)&=B_r(x)C_r(x)+x^{n-m+1}D(x)\
A_r(x)&=B_r(x)C_r(x)&(mod ;x^{n-m+1})\
B_r(x)&=A_r^{-1}(x)C_r(x)&(mod ;x^{n-m+1})\
\end{align}
求逆可得\(B_r(x)\)，再反转得\(B(x)\)，然后乘\(C(x)\)去减\(A(x)\)得\(D(x)\).

喜闻乐见的模板：
namespace INV{
	typedef long long LL;
	int n, a[300005], b[300005];
	const int mod = 998244353;
	const int g = 3189;
	int qpow(int a, int b){
		int ans = 1;
		while(b){
			if(b & 1)	ans = 1ll * ans * a % mod;
			a = 1ll * a * a % mod;
			b >>= 1;
		}
		return ans;
	}
	struct p_l_e{
		int wz[300005], i_c[300005];
		void NTT(int *a, int N, int op){
			for(int i = 0; i < N; i++)
				if(i < wz[i]) swap(a[i], a[wz[i]]);
			for(int le = 2; le <= N; le <<= 1){
				int mid = le >> 1, wn = qpow(g, (mod - 1) / le);
				if(op == -1) wn = qpow(wn, mod - 2);
				for(int i = 0; i < N; i += le){
					LL w = 1; int x, y;
					for(int j = 0; j < mid; j++){
						x = a[i + j];
						y = w * a[i + j + mid] % mod;
						a[i + j] = (x + y) % mod;
						a[i + j + mid] = (x - y + mod) % mod;
						w = w * wn % mod;
					}
				}
			}
		}
		int init(int M){
			int N = 1, len = 0;
			while(N < M) N <<= 1, len++;	
			for(int i = 0; i < N; i++)
				wz[i] = (wz[i >> 1] >> 1) | ((i & 1) << (len - 1));
			return N;
		}
		void INV(int *a, int *b, int deg){
			if(deg == 1){b[0] = qpow(a[0], mod - 2); return;}
			INV(a, b, (deg + 1) >> 1);
			int N = init(deg + deg - 1);
			for(int i = 0; i < deg; i++) i_c[i] = a[i];
			for(int i = deg; i < N; i++) i_c[i] = 0;
			NTT(b, N, 1);NTT(i_c, N, 1);
			for(int i = 0; i < N; i++) b[i] = 1ll * b[i] * (2 - 1ll * b[i] * i_c[i] % mod + mod) % mod;
			NTT(b, N, -1);
			int t = qpow(N, mod - 2);
			for(int i = 0; i < N; i++) b[i] = 1ll * b[i] * t % mod;
			for(int i = deg; i < N; i++) b[i] = 0;
		}
	}PLE;
	
	void main(){
		scanf("%d", &n);
		for(int i = 0; i < n; i++)	scanf("%d", &a[i]);
		PLE.INV(a, b, n);
		for(int i = 0; i < n; i++)	printf("%d ",b[i]);
	}
}

\(\color{red}{\text{多项式全集之四 多项式ln与exp:}}\)

多项式ln：

\(1.\)做法：

设

\[G(x) =\ln F(x) \]

两边求导得

\[G'(x)=\frac {F'(x)} {F(x)} \]

积分回去即可。

\(2.\)应用：

\[e^{F(x)}=\sum_{k \ge 0}\frac{F^k(x)}{k!}=G(x) \]

\[F(x) = \ln G(x) \]

这个的组合意义是：无序组合。

设\(F(x)\)，\(f_i\)表示一些东西，那么这些东西有序组合的方案数为

\[F^0(x) + F^1(x)+F^2(x)+\cdots=\frac 1 {1 - F(x)} \]

而无序组成的方案数为：

\[\frac{F^0(x)}{0!} + \frac {F^1(x)}{1!}+\frac{F^2(x)}{2!}+\cdots=e^{F(x)} \]

如果无序组合方案数好求，那么求\(\ln\)就能得到\(F(x)\)。

例题

多项式\(\exp\)：
喜闻乐见的代码：

多项式\(\ln\):

namespace PLE_ln{
  struct polyme {
      int li[SZ], wz[SZ];
      void NTT(int *a, int N, int op) {
          rop(i, 0, N) if(i < wz[i]) swap(a[i], a[wz[i]]);
          for(int l = 2; l <= N; l <<= 1) {
              int mid = l >> 1;
              int x, y, w, wn = qpow(g, (mod - 1) / l);
              if(op) wn = qpow(wn, mod - 2);
              for(int i = 0; i < N; i += l) {
                  w = 1;
                  for(int j = 0; j < mid; ++j) {
                      x = a[i + j]; y = 1ll * w * a[i + j + mid] % mod;
                      a[i + j] = (x + y) % mod;
                      a[i + j + mid] = (x - y + mod) % mod;
                      w = 1ll * w * wn % mod;
                  }
              }
          }
      }
      void qd(int *a, int *b, int n) {
          rop(i, 0, n) b[i] = 1ll * a[i + 1] * (i + 1) % mod;
      }
      void jf(int *a, int *b, int n) {
          rop(i, 1, n) b[i] = 1ll * a[i - 1] * qpow(i, mod - 2) % mod;
      }
      void mult(int *a, int *b, int *c, int M) {
          int N = 1, len = 0;
          while(N < M) N <<= 1, len ++;
          rop(i, 0, N) wz[i] = (wz[i >> 1] >> 1) | ((i & 1) << (len - 1));
          NTT(a, N, 0); NTT(b, N, 0);
          rop(i, 0, N) c[i] = 1ll * a[i] * b[i] % mod;
          NTT(c, N, 1);
          int t = qpow(N, mod - 2);
          rop(i, 0, N) c[i] = 1ll * c[i] * t % mod;
      }
      void inv(int *a, int *b, int deg) {
          if(deg == 1) {b[0] = qpow(a[0], mod - 2) % mod; return;}
          inv(a, b, (deg + 1) >> 1);
          rop(i, 0, deg) li[i] = a[i];
          int N = 1, len = 0;
          while(N < deg + deg - 1) N <<= 1, len ++;
          rop(i, 0, N) wz[i] = (wz[i >> 1] >> 1) | ((i & 1) << (len - 1));
          rop(i, deg, N) li[i] = 0;
          NTT(li, N, 0); NTT(b, N, 0);
          rop(i, 0, N) b[i] = 1ll * b[i] * (2 - 1ll * li[i] * b[i] % mod + mod) % mod;
          NTT(b, N, 1);
          int t = qpow(N, mod - 2);
          for(int i = 0; i < N; i++) b[i] = 1ll * b[i] * t % mod;
          rop(i, deg, N) b[i] = 0;
      }
  }PLE;
  int a[SZ], da[SZ], ia[SZ], dla[SZ], la[SZ], n;

  void main() {

      scanf("%d", &n);
      rop(i, 0, n) scanf("%d", &a[i]);
      PLE.qd(a, da, n);
      PLE.inv(a, ia, n);
      PLE.mult(ia, da, dla, n + n - 1);
      PLE.jf(dla, la, n);
      rop(i, 0, n) printf("%d ", la[i]);
  }
}

\(\color{red}{\text{多项式全集之五 多项式快速幂与开根:}}\)

多项式快速幂方法一：

\(\color{red}{\text{多项式全集之六 多项式快速插值与多点求值:}}\)

\(\color{red}{\text{多项式全集之七 拉格朗日插值：}}\)

posted @ 2019-04-18 15:02  Smeow  阅读(381)  评论(0)  编辑  收藏  举报