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多项式全家桶

多项式全家桶(更新至开根)

多项式常用复杂度分析 (T(n) = T(n/2) + O(nlogn) = O(nlogn))

FFT快速傅里叶变换

流程: 将多项式(Theta (nlog_n))转成点值表示形式进行卷积, 再(Theta (nlog_n)) 转回来

离散傅里叶变换:

朴素转为点值, 需要将一个一个x带入, 而这里傅里叶搞到了几个可以优化的复数根

利用复数, 在复平面上画出一个单位圆, 将单位圆n等分, 每一个等分点为$omega_n^i $,

接下来将(omega^1_n omega^2_ncdotsomega^n_n) n个根带入, 其中(omega_n^i = cos(frac{k}{n}2pi) + i * sin(frac{k}{n}2pi))

稍等, 还要几个小性质才可以:

性质一: (omega^{2k}_{2n} = omega_n^k)

性质二: (omega_n^{k+frac{n}{2}} = - omega_n^k)

接下来开始快速傅里叶变换:

设A(x) = (a_0 + a_1x + a_2x^2 + cdots + a_{n-1}x^{n-1})

利用分治, 将A按x的指数分为奇偶两部分

(A(x)=(a_0+a_2x^2+cdots + a_{n-2}x^{n-2})+(a_1x+a_3x^3+cdots) $ a_{n-1}x^{n-1})$

设偶部分 (A_1(x) = a_0 + a_2x + a_4x^2 + cdots + a_{n-2}x^{frac{n}{2}-1})

奇部分(A_2(x) = a_1 + a_3x + a_5x^2 + cdots + a_{n-1}x^{frac{n}{2}-1})

则有: (A(x) = A_1(x^2) + x A_2(x^2))

设k < n/2 带入(x = omega^k_n)

(A(omega^k_n) = A_1(omega^{2k}_n) + omega_n^kA_2(omega^{2k}_n))

再带入 (omega^{k+frac{n}{2}}_n)

得(A(omega^{k+frac{n}{2}}_n) = A_1(omega^{2k}_n) -omega_n^kA_2(omega^{2k}_n))

你fa♂现了吗, 分治时, 两个求值一次解决

IDFT: 离散傅里叶逆变换

设((y_0,y_1,cdots,y_{n-1})) 为多项式A(x) = (a_0 + a_1x + a_2x^2 + cdots + a_{n-1}x^{n-1})的离散傅里叶变换

设多项式 B(x) = A(x) = (y_0 + y_1x + y_2x^2 + cdots + y_{n-1}x^{n-1})

将n个单位根的倒数带入得新的离散傅里叶变换

此处略去推导, 得(a_i = frac{z_i}{n})

非递归版fft

每个数递归到底层时的二进制表示恰是原来的表示翻转得到的

如 6(011) 到 3(110)

再加上蝴蝶变换就可以啦

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int N = 3000600;


template <typename T>
void read(T &x) {
	x = 0; bool f = 0;
	char c = getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
	for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
	if (f) x = -x;
}

struct Complex {
	double x, y;
	Complex (double xx = 0,double yy = 0) {x=xx,y=yy;}
	Complex operator + (Complex &i) const {
		return Complex(x + i.x, y + i.y);
	}
	Complex operator - (Complex &i) const {
		return Complex(x - i.x, y - i.y);
	}
	Complex operator * (Complex &i) const {
		return Complex(x * i.x - y * i.y, x * i.y + y * i.x);
	}
}A[N], B[N];

const double Pi = acos(-1.0);

int n, m;

int lim = 1, L;
int r[N];

void FFT(Complex *a,int type) {
	for (int i = 0;i < lim; i++)
	if (r[i] > i) swap(a[r[i]], a[i]);
	for (int mid = 1;mid < lim; mid <<= 1) {
		Complex T(cos(Pi/mid), type * sin(Pi/mid));
		for (int j = 0;j < lim; j += (mid << 1)) {
			Complex t(1, 0);
			for (int k = 0;k < mid; k++, t = t * T) {
				Complex x = a[j + k], y = t * a[mid + j + k];
				a[k + j] = x + y;
				a[mid + j + k] = x - y;
			}
		}
	}
}

int main() {
	read(n), read(m);
	for (int i = 0;i <= n; i++) {
		int x; read(x);
		A[i].x = x;
	}
	for (int i = 0;i <= m; i++) {
		int x; read(x);
		B[i].x = x;
	}
	while (lim <= (n + m)) lim <<= 1, L++;
	for (int i = 0;i < lim; i++)
	r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
	
	FFT(A, 1); FFT(B, 1);
	for (int i = 0;i <= lim; i++) A[i] = A[i] * B[i];
	FFT(A, -1);
	for (int i = 0;i <= n + m; i++)
	printf ("%d ", (int)(A[i].x / lim + 0.5));
	return 0;
}

NTT快速数论变换

直接上代码

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int N = 3e6+6;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1)/G;
inline int read(void) {
	int x = 0, f = 1;
	char c = getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
	for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+c-'0';
	return x * f;
}
int n, m, r[N], L;
ll A[N], B[N];
int lim = 1;
ll pow(ll di,ll mi) {
	ll ans = 1, a = di;
	while (mi) {
		if (mi & 1) ans = ans * a % P;
		a = a * a % P;
		mi >>= 1;
	}
	return ans;
}
void NTT(ll *a, int tag) {
	for (int i = 0;i < lim; i++) 
	if (i < r[i]) swap(a[i], a[r[i]]);
	for (int mid = 1;mid < lim ;mid <<= 1) {
		ll wn = pow(tag == 1 ? G : Gi, (P-1) / (mid << 1));
		for (int j = 0;j < lim; j += (mid << 1)) {
			ll w = 1;
			for (int k = 0;k < mid ; k++, w = (w * wn) % P) {
				ll x = a[j+k], y = w * a[j+k+mid] % P;
				a[j+k] = (x + y) % P;
				a[j+k+mid] = (x - y + P) % P;
			}
		}
	}
}
int main() {
	n = read(), m = read();
	for (int i = 0;i <= n; i++) A[i] = read();
	for (int j = 0;j <= m; j++) B[j] = read();
	while (lim <= n + m) L++, lim <<= 1;
	for (int i = 0;i < lim; i++)
	r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
	NTT(A, 1), NTT(B, 1);
	for (int i = 0;i <= lim; i++) A[i] = (A[i] * B[i]) % P;
	NTT(A, -1);
	ll inv = pow(lim, P-2);
	for (int i = 0;i <= n + m; i++) 
	printf ("%d ", A[i] * inv % P);
	return 0;
}

多项式求逆

[ ext{求 }g equiv f^{-1} mod x^n \ ext{设 }h*f equiv 1 mod x^{lceil frac n2 ceil}\ (g - h)*f equiv 0 mod x^{lceil frac n2 ceil}\ (g-h)^2 equiv 0 mod x^n\ f*(g-h)^2 equiv 0 mod x^n\ g-2*h+h^2*f equiv 0 mod x^n\ gequiv h*(2-h*f) mod x^n\ ]

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1)/G;
const int N = 3e5+5;
template <typename T>
void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
	for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1) + c-'0';
	x *= f;
}
ll A[N], n, lim = 1, L;
int r[N];
ll pow(ll di,ll mi) {
	ll ans = 1;
	while (mi) {
		if (mi & 1) ans = (ans * di) % P;
		di = (di * di) % P;
		mi >>= 1;
	}
	return ans;
}
void NTT(ll *a, int type) {
	for (int i = 0;i < lim; i++)
	if (i < r[i]) swap(a[i], a[r[i]]);
	for (int mid = 1;mid < lim; mid <<= 1) {
		ll wn = pow(type == 1 ? G : Gi, (P-1)/(mid<<1));
		for (int j = 0;j < lim;j += (mid << 1)) {
			ll w = 1;
			for (int k = 0;k < mid; k++, w = (w * wn) % P) {
				ll x = a[j+k], y = w * a[j+k+mid] % P;
				a[j+k] = (x + y) % P, a[j+k+mid] = (x - y + P) % P;
			}
		}
	}
	if (type == 1) return;
	ll inv = pow(lim, P-2);
	for (int i = 0;i < lim; i++)
	a[i] = a[i] * inv % P;
}
	
ll B[N], C[N], D[N];

void Inv(int deg, ll *a, ll *b) {
	b[0] = pow(a[0], P - 2);
	int len = 1;
	for ( ; len <= (n << 1); len <<= 1) {
		lim = len << 1;
		for (int i = 1;i < lim; i++)
			r[i] = (r[i>>1]>>1) | ((i & 1) ? len : 0);
		for (int i = 0;i < len; i++) C[i] = a[i], D[i] = b[i];
		NTT(C, 1), NTT(D, 1);
		for (int i = 0;i < lim; i++) b[i] = (2 - C[i] * D[i] % P + P) * D[i] % P;
		NTT(b, -1);
		for (int i = len;i < lim; i++) b[i] = 0;
	}
}

int main() {
	read(n);
	for (int i = 0;i < n; i++) read(A[i]);
	Inv(n, A, B);
	for (int i = 0;i < n; i++) printf ("%lld ", B[i]);
	return 0;
}

分治FFT

应用CDQ分治思想即可

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3, Gi =  (P+1) / G;
const int N = 405000;
ll fpw(ll di, ll mi) {
	ll res = 1;
	while (mi) {
		if (mi & 1) res = res * di % P;
		di = di * di % P;
		mi >>= 1;
	}
	return res;
}


int lim, L;
ll r[N];
void NTT(ll *a,int type) {
	for (int i = 0;i < lim; i++) 
		if (i < r[i]) swap(a[i], a[r[i]]);
	for (int mid = 1;mid < lim; mid <<= 1) {
		ll wn = fpw(type == 1 ? G : Gi, (P-1) / (mid<<1));
		for (int i = 0;i < lim; i += mid << 1) {
			ll w = 1;
			for (int j = 0;j < mid; j++, w = w * wn % P) {
				ll x = a[j + i], y = a[mid + i + j] * w % P;
				a[i + j] = (x + y) % P;
				a[i + j + mid] = (x - y + P) % P;
			}
		}
	}
	if (type == 1) return;
	ll inv = fpw(lim, P-2);
	for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}

ll a[N], f[N], b[N], g[N];

void solve(int l,int R) {
	if (l == R) return;
	int mid = (l + R) >> 1;
	solve(l, mid); lim = 1, L = 0;
	int len = R - l + 1;
	for (int i = l;i <= mid; i++) a[i-l] = f[i];
	for (int i = 0;i < len; i++) b[i] = g[i];
	len += mid - l + 1;
	while (lim <= len) lim <<= 1, L++;
	for (int i = 0;i < lim; i++) 
		r[i] = (r[i>>1]>>1) | ((i&1)<<(L-1));
	NTT(a, 1), NTT(b, 1);
	for (int i = 0;i < lim; i++) a[i] = a[i] * b[i] % P;
	NTT(a, -1);
	for (int i = mid + 1;i <= R; i++) f[i] = (f[i] + a[i-l]) % P;
	for (int i = 0;i < lim; i++) a[i] = b[i] = 0;
	solve(mid + 1, R);
}

int read(void) {
	int x = 0; char c = getchar();
	while (!isdigit(c)) c=getchar();
	for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
	return x;
}

int n;
int main() {
	n = read();
	for (int i = 1;i < n; i++) g[i] = read();
	f[0] = 1;
	solve(0, n - 1);
	for (int i = 0;i < n; i++) printf ("%lld ", f[i]);
	return 0;
}

多项式ln：

[G equiv ln F\ G' equiv frac {F'}{F} ]

求逆再积分即可

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
using namespace std;
const int N = 505000;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1) / G;
int read(void) {
	int x = 0; bool f = 0;
	char c = getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
	for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
	return f ? -x : x;
}

ll fpw(ll di,ll mi) {
	ll ans = 1;
	while (mi) {
		if (mi & 1) ans = ans * di % P;
		di = di * di % P;
		mi >>= 1;
	}
	return ans;
}

ll A[N], B[N], C[N];
int r[N], n, lim, L;

void NTT(ll *a, int type) {
	for (int i = 0;i < lim; i++) 
		if (i < r[i]) swap(a[i], a[r[i]]);
	for (int j = 1;j < lim; j <<= 1) {
		ll wn = fpw(type == 1 ? G : Gi, (P-1) / (j<<1));
		for (int k = 0;k < lim; k += j << 1) {
			ll w = 1;
			for (int p = 0;p < j; p++, w = w * wn % P) {
				ll x = a[p+k], y = a[p+k+j] * w % P;
				a[p+k] = (x + y) % P;
				a[p+k+j] = (x - y + P) % P;
			}
		}
	}
	if (~type) return;
	ll inv = fpw(lim, P-2);
	for (int i = 0;i < lim; i++) 
		a[i] = a[i] * inv % P;
}

void work(int deg, ll *a, ll *b) {
	if (deg == 1) {
		b[0] = fpw(a[0], P-2);
		return;
	}
	work((deg+1)>>1, a, b);
	lim = 1, L = 0;
	while (lim < (deg<<1)) lim <<= 1, L++;
	for (int i = 0;i < lim; i++) 
		r[i] = (r[i>>1]>>1) | ((i&1)<<(L-1));
	for (int i = 0;i < deg; i++) C[i] = a[i];
	for (int i = deg;i < lim; i++) C[i] = 0;
	NTT(C, 1); NTT(b, 1);
	for (int i = 0;i < lim; i++) 
	b[i] = ((ll)2 - C[i] * b[i] % P + P) % P * b[i] % P;
	NTT(b, -1);
	for (int i = deg;i < lim; i++) b[i] = 0;
}

int m;
void qiudoor(void) {
	for (int i = 1;i < n; i++) A[i-1] = A[i] * i % P;
	A[n-1] = 0;
}

void jinitaimei(void) {
	for (int i = n-1;i >= 1; i--) A[i] = A[i-1] * fpw(i, P-2) % P;
	A[0] = 0;
}

int main() {
	n  = read();
	for (int i = 0;i < n; i++) A[i] = read();
	work(n, A, B); qiudoor();
	lim = 1, L = 0; m = n;
	while (lim <= n + m) lim <<= 1, L++;
	NTT(A, 1); NTT(B, 1);
	for (int i = 0;i < lim; i++) A[i] = A[i] * B[i] % P;
	NTT(A, -1); 
	jinitaimei();
	for (int i = 0;i < n; i++) printf ("%lld ", A[i] % P);
	return 0;
}

多项式exp

[牛顿迭代\ F(G_0) = 0 mod {x^{lceil frac n2 ceil}}\ F(G)=F(G_0) + frac{F'(G_0)}{1!}(G - G_0) + frac{F''(G_0)}{2!}(G-G_0)^2+cdots mod {x^n}\ 易知G - G_0 = 0mod {x^{lceil frac n2 ceil}}\ 因为后面二次三次方的系数已经大于x^n, 所以可以忽略\ 又 ecause F(G) = 0 mod x^n\ herefore G = G_0 - frac{F(G_0)}{F'(G_0)} ]

下面回到本题

[e^F = H\ F = lnH\ ln H-F = 0\ 设G = ln H-F, 即求G的零点\ 以H为变量,F为常量求导\ G'=frac {1}H\ H = H_0-frac{lnH_0-F}{frac {1}H_0}=H_0(1-lnH_0+F) ]

#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;

const int P = 998244353;
const int G = 3, Gi = (P+1) / G;
const int N = 400005;

template <typename T>
void read(T &x) {
    x = 0; int f = 0; char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    if (f) x=-x;
}

ll fpw(ll di, ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return res;
}

int lim, r[N], n, L;
ll A[N], B[N], C[N], D[N], E[N];


void NTT(ll *a, int type = 1) {
    for (int i = 0;i < lim; i++) 
        if (r[i] > i) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll p = mid << 1, wn = fpw(type == 1 ? G : Gi, (P-1) / (mid << 1));
        for (int j = 0;j < lim; j += p) {
            ll w = 1;
            for (int k = 0;k < mid; k++, w = (w * wn) % P) {
                ll x = a[j+k], y = a[j+k+mid] * w % P;
                a[j+k] = (x + y) % P, a[j+k+mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P - 2);
    for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}

void Inv(int deg, ll *a, ll *b) {
    b[0] = fpw(a[0], P-2); int len;
    for (len = 1;len < (deg << 1); len <<= 1) {
        lim = len << 1;
        for (int i = 0;i < len; i++) A[i] = a[i], B[i] = b[i];
        for (int i = 0;i < lim; i++) 
            r[i] = (r[i>>1]>>1) | ((i&1)?len:0);
        NTT(A, 1); NTT(B, 1);
        for (int i = 0;i < lim; i++) 
            b[i] = (2 - A[i] * B[i] % P + P) * B[i] % P;
        NTT(b, -1);
        for (int i = len;i < lim; i++) b[i] = 0;
    }
    for (int i = 0;i < len; i++) A[i] = B[i] = 0;
}

ll inv[N];

void init(void) {
    inv[1] = 1;
    for (int i = 2;i <= 200080; i++) 
        inv[i] = P - P / i * inv[P%i] % P;
}

void jinitaimei(ll *a, int deg) {
    for (int i = deg - 1;i >= 1; i--)
        a[i] = a[i-1] * inv[i] % P;
    a[0] = 0;
}

void qiudoor(ll *a, int deg) {
    for (int i = 1;i < deg; i++) a[i-1] = a[i] * i % P;
    a[deg-1] = 0;
}

void get_ln(ll *a, ll deg) {
    Inv(deg, a, C); qiudoor(a, deg);
    NTT(a, 1); NTT(C, 1);
    for (int i = 0;i < lim; i++) a[i] = a[i] * C[i] % P;
    NTT(a, -1); jinitaimei(a, deg);
}

void get_exp(int deg, ll *a, ll *b) {
    b[0] = 1; int len;
    for (len = 2;len < (deg << 1); len <<= 1) {
        lim = len << 1;
        for (int i = 0;i < len; i++) E[i] = b[i];
        get_ln(E, len);
        E[0] = (a[0] + 1 - E[0] + P) % P;
        for (int i = 1;i < len; i++) 
            E[i] = (a[i] - E[i] + P) % P;
        NTT(E, 1); NTT(b, 1);
        for (int i = 0;i < lim; i++) b[i] = b[i] * E[i] % P;
        NTT(b, -1);
        for (int i = len;i < lim; i++) b[i] = 0;
    }
}

ll f[N], g[N];
int main() {
    read(n); init();
    for (int i = 0;i < n; i++) read(f[i]);
    get_exp(n, f, g);
    for (int i = 0;i < n; i++) printf ("%lld ", g[i]);
    return 0;
}

多项式开根

方法一(大常数) :

[G^2 = F\ 2*lnG=lnF\ G = e^{frac {lnF}2} ]

方法二(牛顿迭代):

[G^2-F=0\ G=G_0-frac{(G_0^2-F)}{2G_0}=frac{G_0^2+F}{2G_0} ]

方法三(与牛顿迭代形式相同):

[G_0^2=F mod {x^{lceil frac n2 ceil}}\ G^2=F mod x^n\ G^2-G_0^2=0 mod {x^{lceil frac n2 ceil}}\ ecause G + G_0!= 0\ herefore G - G_0= 0\ (G-G_0)^2=0 mod x^n\ G^2-2*G*G_0+G_0^2=0\ F-2*G*G_0+G_0^2=0\ G=frac{F+G_0^2}{2*G_0} ]

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3, Gi = (P+1) / G;
const int INV2 = (P+1) / 2;
const int N = 300500;

int read(void) {
	int x = 0, f = 0;
	char c = getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
	for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
	return f ? -x : x;
}


ll fpw(ll di, ll mi) {
	ll res = 1;
	while (mi) {
		if (mi & 1) res = res * di % P;
		di = di * di % P;
		mi >>= 1;
	}
	return res;
}

int lim, r[N], n, L;
ll f[N], g[N], A[N], C[N], D[N];

void NTT(ll *a, int type) {
    for (int i = 0;i < lim; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = fpw(type == 1 ? G : Gi, (P-1)/(mid<<1));
        for (int j = 0;j < lim;j += (mid << 1)) {
            ll w = 1;
            for (int k = 0;k < mid; k++, w = (w * wn) % P) {
                ll x = a[j+k], y = w * a[j+k+mid] % P;
                a[j+k] = (x + y) % P, a[j+k+mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P-2);
    for (int i = 0;i < lim; i++)
    a[i] = a[i] * inv % P;
}
ll B[N];
void Inv(int deg, ll *a, ll *b) {
	b[0] = fpw(a[0], P-2);
	int len;
	for (len = 1;len < (deg<<1);len <<= 1) {
		lim = len << 1;
		for (int i = 0;i < len; i++) A[i] = a[i], B[i] = b[i];
		for (int i = 0;i < lim; i++) 
			r[i] = (r[i>>1]>>1) | ((i&1)?len:0);
		NTT(A, 1), NTT(B, 1);
		for (int i = 0;i < lim; i++)
			b[i] = (2 - A[i] * B[i] % P + P) * B[i] % P;
		NTT(b, -1);
		for (int i = len;i < lim; i++) b[i] = 0;
	}
	for (int i = 0;i < len; i++) A[i] = B[i] = 0;
	for (int i = deg;i < len; i++) b[i] = 0;
}
		


void Sqrt(int deg, ll *a, ll *b) {
	b[0] = 1;
	ll len;
	for (len = 1;len < (deg << 1);len <<= 1) {
		lim = len << 1;
		for (int i = 0;i < len; i++) C[i] = a[i];
		Inv(len, b, D);
		for (int i = 0;i < lim; i++) 
			r[i] = (r[i>>1]>>1) | ((i&1) ? len : 0);
		NTT(C, 1), NTT(D, 1);
		for (int i = 0;i < lim; i++) C[i] = C[i] * D[i] % P;
		NTT(C, -1);
		for (int i = 0;i < len; i++) b[i] = (b[i] + C[i]) * INV2 % P;
		for (int i = len;i < lim; i++) b[i] = 0;
	}
}

		

int main() {
	n = read();
	for (int i = 0;i < n; i++) f[i] = read();
	Sqrt(n, f, g);
	for (int i = 0;i < n; i++) printf ("%lld ", g[i]);
	cout << endl;
	return 0;
}

多项式除法

[F=Q*G+R (已知G, F)\ reverseA(x) = A(frac 1x)*x^n = rA\ F(frac 1x)*x^n=Q(frac 1x)*G(frac 1x)*x^n+R(frac 1x)*x^n\ rF=rQ*rG+rR*x^{n-m+1} mod x^{n-m+1}\ rQ=rF*rG^{-1}\ R = F-Q*G ]

注意因为mod的是(x^{n-m+1})大于Q的次数, 所以Q唯一

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1)/G;
const int N = 3e5+5;
template <typename T>
void read(T &x) {
    x = 0; int f = 1;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1) + c-'0';
    x *= f;
}
ll A[N], n, lim = 1, L;
int r[N];
ll pow(ll di,ll mi) {
    ll ans = 1;
    while (mi) {
        if (mi & 1) ans = (ans * di) % P;
        di = (di * di) % P;
        mi >>= 1;
    }
    return ans;
}
void NTT(ll *a, int type) {
    for (int i = 0;i < lim; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = pow(type == 1 ? G : Gi, (P-1)/(mid<<1));
        for (int j = 0;j < lim;j += (mid << 1)) {
            ll w = 1;
            for (int k = 0;k < mid; k++, w = (w * wn) % P) {
                ll x = a[j+k], y = w * a[j+k+mid] % P;
                a[j+k] = (x + y) % P, a[j+k+mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = pow(lim, P-2);
    for (int i = 0;i < lim; i++)
    a[i] = a[i] * inv % P;
}

ll B[N], C[N];
void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = pow(a[0], P-2);
        return;
    }
    work((deg+1)>>1, a, b);
    lim = 1, L = 0;
    while (lim < (deg<<1)) lim <<= 1, L++;
    for (int i = 0;i < lim; i++) 
    r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    for (int i = 0;i < deg; i++) C[i] = a[i];
    for (int i = deg;i < lim; i++) C[i] = 0;
    NTT(C, 1), NTT(b, 1);
    for (int i = 0;i < lim; i++)
    b[i] = ((ll)2 - C[i] * b[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}
ll f[N], g[N], q[N], ff[N], gg[N];
int m;
int main() {
	freopen("hs.in","r",stdin);
	freopen("hs.out","w",stdout);
    read(n), read(m);
    for (int i = n;i >= 0; i--) read(f[i]), ff[n-i] = f[i];
    for (int j = m;j >= 0; j--) read(g[j]), gg[m-j] = g[j];
    for (int i = n - m + 1;i <= n; i++) f[i] = g[i] = 0;
    work(n-m+1, g, B);
	NTT(B, 1); NTT(f, 1);
	for (int i = 0;i < lim; i++) q[i] = f[i] * B[i] % P;
	NTT(q, -1);
	reverse(q, q + n - m + 1);
	for (int i = 0;i <= n - m; i++) printf ("%lld ", q[i]);
	lim = 1, L = 0;
	while (lim <= n) lim <<= 1, L++;
	for (int i = n - m + 1;i < lim; i++) q[i] = 0;
	for (int i = 0;i < lim; i++) r[i] = (r[i>>1]>>1) | ((i&1)<<(L-1));
	NTT(gg, 1); NTT(q, 1);
	for (int i = 0;i < lim; i++) gg[i] = gg[i] * q[i] % P;
	NTT(gg, -1);
	putchar('
');
	for (int i = 0;i < m; i++) printf ("%lld ", (ff[i] - gg[i] + P) % P);
    return 0;
}

拉格朗日插值

ll fpw(ll x,ll mi) {
	ll res = 1;
	while (mi) {
		if (mi & 1) res = res * x % P;
		x = x * x % P;
		mi >>= 1;
	}
	return res;
}

int x[N], y[N], n, k;
int main() {
	read(n), read(k);
	for (int i = 1;i <= n; i++) read(x[i]), read(y[i]);
	ll res = 0;
	for (int i = 1;i <= n; i++) {
		ll ans1 = 1, ans2 = 1;
		for (int j = 1;j <= n; j++) {
			if (j == i) continue;
			ans1 = ans1 * (k - x[j]) % P;
			ans2 = ans2 * (x[i] - x[j]) % P;
		}
		res += y[i] * ans1 % P * fpw(ans2, P-2) % P;
	}
	res %= P;
	if (res < 0) res += P;
	cout << res << endl;
	return 0;
}

多项式快速幂

[F^k=G\ klnF=lnG\ e^{klnF}=G ]

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3, Gi = (P+1) / G;
const int N = 1005000;

int read(void) {
    int x = 0; char c = getchar();
    while (!isdigit(c)) c=getchar();
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    return x;
}

ll fpw(ll di, ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return res;
}

int lim, L;
int r[N];
void NTT(ll *a,int type) {
    for (int i = 0;i < lim; i++)
        if (r[i] > i) swap(a[r[i]], a[i]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = fpw(type == 1 ? G : Gi, (P-1) / (mid << 1));
        for (int i = 0;i < lim; i += mid << 1) {
            ll w = 1;
            for (int j = 0;j < mid; j++, w = w * wn % P) {
                int x = a[i + j], y = a[i + j + mid] * w % P;
                a[i + j] = (x + y) % P;
                a[i + j + mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P-2);
    for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}

ll a[N], b[N], c[N];
void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = fpw(a[0], P-2);
        return;
    }
    work((deg+1) >> 1, a, b);
    lim = 1, L = 0;
    while (lim < (deg << 1)) lim <<= 1, L++;
    for (int i = 0;i < lim ;i++) 
        r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    for (int i = 0;i < deg; i++) c[i] = a[i];
    for (int i = deg; i < lim; i++) c[i] = 0;
    NTT(b, 1); NTT(c, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = (2 - b[i] * c[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}
int n;
void qiudoor(ll *a, int deg) {
    for (int i = 1;i < n; i++) a[i-1] = a[i] * i % P;
    a[n-1] = 0;
}

ll inv[N];

void init(void) {
    inv[1] = 1;
    for (int i = 2;i <= n * 2; i++)
        inv[i] = (P - P / i * inv[P%i] % P) % P;
}

void jinitaimei(ll *a, int deg) {
    for (int i = n-1;i >= 1; i--) a[i] = a[i-1] * inv[i] % P;
    a[0] = 0;
}

ll A[N], B[N];

void get_ln(ll *a, ll n) {
    for (int i = 0;i < n*2; i++) B[i] = 0;
//  put(A, n), put(B, n*2);
    work(n, a, B); 
    qiudoor(a, n);
    NTT(a, 1); NTT(B, 1);
    for (int i = 0;i < lim; i++) a[i] = a[i] * B[i] % P;
    NTT(a, -1); jinitaimei(a, n);
}

void solve(int deg, ll *a, ll *b) {
    if (deg == 1) return (void) (b[0] = 1);
    solve(deg>>1, a,  b);
    for (int i = 0;i < deg; i++) A[i] = b[i];
    get_ln(A, deg); 
    A[0] = (a[0] + 1 - A[0] + P) % P;
    for (int i = 1;i < deg; i++)
        A[i] = (a[i] - A[i] + P) % P;
    NTT(A, 1); NTT(b, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = b[i] * A[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}

ll f[N], g[N];

ll get_k(void) {
    ll x = 0; char c = getchar();
    while (!isdigit(c)) c=getchar();
    for (;isdigit(c);c=getchar()) {
		x=(x<<3)+(x<<1)+(c^48);
		if (x >= P) x %= P;
	}
    return x;
}
ll k;
int main() {
    n = read(); init(); k = get_k();
    for (int i = 0;i < n; i++) f[i] = read();
    get_ln(f, n);
    for (int i = 0;i < n; i++) f[i] = f[i] * k % P;
    lim = 1;
    while (lim <= n) lim <<= 1;
    solve(lim, f, g);
    for (int i = 0;i < n; i++) 
        printf ("%d ", g[i]);
    return 0;
}

多项式三角函数: 直接欧拉公式即可

[e^{i heta}=sin heta+cosx heta ]

#include <iostream>
#include <cstring>
#include <cstdio>
#define ll long long
using namespace std;

const int P = 998244353;
const int I = 86583718;
const int G = 3, Gi = (P + 1) / G;
const int N = 300500;

template <typename T>
void read(T &x) {
    x = 0; int f = 0; char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    if (f) x=-x;
}

ll fpw(ll di, ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return res;
}

ll lim, r[N], n, L;
ll A[N], B[N], C[N], D[N], E[N];

void NTT(ll *a, int type = 1) {
    for (int i = 0;i < lim; i++) 
        if (r[i] > i) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll p = mid << 1, wn = fpw(type == 1 ? G : Gi, (P-1) / (mid << 1));
        for (int j = 0;j < lim; j += p) {
            ll w = 1;
            for (int k = 0;k < mid; k++, w = (w * wn) % P) {
                ll x = a[j+k], y = a[j+k+mid] * w % P;
                a[j+k] = (x + y) % P, a[j+k+mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P - 2);
    for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}


void Inv(int deg, ll *a, ll *b) {
    b[0] = fpw(a[0], P-2); int len;
    for (len = 1;len < (deg << 1); len <<= 1) {
        lim = len << 1;
        for (int i = 0;i < len; i++) A[i] = a[i], B[i] = b[i];
        for (int i = 0;i < lim; i++) 
            r[i] = (r[i>>1]>>1) | ((i&1)?len:0);
        NTT(A, 1); NTT(B, 1);
        for (int i = 0;i < lim; i++) 
            b[i] = (2 - A[i] * B[i] % P + P) * B[i] % P;
        NTT(b, -1);
        for (int i = len;i < lim; i++) b[i] = 0;
    }
    for (int i = 0;i < len; i++) A[i] = B[i] = 0;
}

ll inv[N];
void init(void) {
	inv[1] = 1;
	for (int i = 2;i <= 100500; i++)
		inv[i] = (P - P / i) * inv[P % i] % P;
}


void jinitaimei(ll *a, int deg) {
    for (int i = deg - 1;i >= 1; i--)
        a[i] = a[i-1] * inv[i] % P;
    a[0] = 0;
}

void qiudoor(ll *a, int deg) {
    for (int i = 1;i < deg; i++) a[i-1] = a[i] * i % P;
    a[deg-1] = 0;
}

void get_ln(ll *a, ll deg) {
    Inv(deg, a, C); qiudoor(a, deg);
    NTT(a, 1); NTT(C, 1);
    for (int i = 0;i < lim; i++) a[i] = a[i] * C[i] % P;
    NTT(a, -1); jinitaimei(a, deg);
}

void get_exp(int deg, ll *a, ll *b) {
    b[0] = 1; int len;
    for (len = 2;len < (deg << 1); len <<= 1) {
        lim = len << 1;
        for (int i = 0;i < len; i++) E[i] = b[i];
        get_ln(E, len);
        E[0] = (a[0] + 1 - E[0] + P) % P;
        for (int i = 1;i < len; i++) 
            E[i] = (a[i] - E[i] + P) % P;
        NTT(E, 1); NTT(b, 1);
        for (int i = 0;i < lim; i++) b[i] = b[i] * E[i] % P;
        NTT(b, -1);
        for (int i = len;i < lim; i++) b[i] = 0;
    }
}

ll f[N], t1[N], t2[N], type;
int main() {
	read(n), read(type); init();
	for (int i = 0;i < n; i++) read(f[i]), f[i] = f[i] * I % P;
	get_exp(n, f, t1); Inv(n, t1, t2);
	ll INV = fpw(2 * I % P, P - 2);
	for (int i = 0;i < n; i++) 
	if (type) printf ("%lld ", (t1[i] + t2[i]) * inv[2] % P);
	else printf ("%lld ", (t1[i] - t2[i] + P) * INV % P);
	return 0;
}

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原文地址：https://www.cnblogs.com/Hs-black/p/12004617.html