多项式入门(FFT,NTT,MTT)

目录

前言

又到了愉快的数学时间了！

由于蒟蒻作者数学特别烂，贴个友链然后直接上代码吧

本博客由某谷搬运过来，目的只是存板子。

\(update\ 2021.2.2\) 更新了部分代码，已经基本学懂，懒得更新讲解了。u1s1,到了高中后有了一定数学基础，就是比初中傻白甜的时候学得快。

\(update\ 2021.2.3\) 学懂了蝴蝶变换之后又更新了一波板子，不得不说，迭代版本真的快得离谱。

\(update\ 2021.7.27\) 折叠了代码，增加文章可读性。

FFT

友链

Aaplloo

OneInDark

练习

板题(UOJ)

板题(洛谷)

力(洛谷)

代码

板题代码

$Mine\ \text{(递归)}$ 
//12252024832524
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define TT template<typename T>
using namespace std; 

typedef long long LL;
const int MAXN = 1 << 21 | 5;
const double PI = acos(-1);
int lena,lenb;

LL Read()
{
	LL x = 0,f = 1;char c = getchar();
	while(c > '9' || c < '0'){if(c == '-')f = -1;c = getchar();}
	while(c >= '0' && c <= '9'){x = (x*10) + (c^48);c = getchar();}
	return x * f;
}
TT void Put1(T x)
{
	if(x > 9) Put1(x/10);
	putchar(x%10^48);
}
TT void Put(T x,char c = -1)
{
	if(x < 0) putchar('-'),x = -x;
	Put1(x);
	if(c >= 0) putchar(c);
}
TT T Max(T x,T y){return x > y ? x : y;}
TT T Min(T x,T y){return x < y ? x : y;}
TT T Abs(T x){return x < 0 ? -x : x;}

struct cp//complex
{
	double x,y;
	cp(){}
	cp(double x1,double y1){
		x = x1;
		y = y1;
	}
	cp operator + (const cp &A) const {return cp(x+A.x,y+A.y);}
	cp operator - (const cp &A) const {return cp(x-A.x,y-A.y);}
	cp operator * (const cp &A) const {return cp(x*A.x-y*A.y,x*A.y+y*A.x);}
}a[MAXN],b[MAXN];

void FFT(int len,cp * a,int f)
{
	if(len == 1) return;
	cp a1[len >> 1],a2[len >> 1];
	for(int i = 0;i < len;i += 2) a1[i >> 1] = a[i],a2[i >> 1] = a[i+1];
	FFT(len>>1,a1,f);
	FFT(len>>1,a2,f);
	cp w = cp(cos(2*PI/len),f*sin(2*PI/len)),k = cp(1,0);
	len >>= 1;
	for(int i = 0;i < len;++ i,k = k * w)
	{
		a[i] = a1[i] + k * a2[i];
		a[i+len] = a1[i] - k * a2[i];
	}
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".in","w",stdout);
	lena = Read(); lenb = Read();
	for(int i = 0;i <= lena;++ i) a[i].x = Read();
	for(int i = 0;i <= lenb;++ i) b[i].x = Read();
	int len = 1;
	while(len <= lena + lenb) len <<= 1;
	FFT(len,a,1);
	FFT(len,b,1);
	for(int i = 0;i <= len;++ i) a[i] = a[i] * b[i];
	FFT(len,a,-1);
	for(int i = 0;i <= lena+lenb;++ i) Put((int)(a[i].x/len + 0.5),' ');
	return 0;
}
$Mine\ \text{(迭代)}$ 
//12252024832524
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define TT template<typename T>
using namespace std; 

typedef long long LL;
const int MAXN = 1 << 21 | 5;
const double PI = acos(-1);
int lena,lenb,len = 1,l = -1;
int rev[MAXN];

LL Read()
{
	LL x = 0,f = 1;char c = getchar();
	while(c > '9' || c < '0'){if(c == '-')f = -1;c = getchar();}
	while(c >= '0' && c <= '9'){x = (x*10) + (c^48);c = getchar();}
	return x * f;
}
TT void Put1(T x)
{
	if(x > 9) Put1(x/10);
	putchar(x%10^48);
}
TT void Put(T x,char c = -1)
{
	if(x < 0) putchar('-'),x = -x;
	Put1(x);
	if(c >= 0) putchar(c);
}
TT T Max(T x,T y){return x > y ? x : y;}
TT T Min(T x,T y){return x < y ? x : y;}
TT T Abs(T x){return x < 0 ? -x : x;}

struct cp
{
	double x,y;
	cp(){}
	cp(double x1,double y1){
		x = x1;
		y = y1;
	}
	cp operator + (const cp &A)const{return cp(x+A.x,y+A.y);}
	cp operator - (const cp &A)const{return cp(x-A.x,y-A.y);}
	cp operator * (const cp &A)const{return cp(x*A.x-y*A.y,x*A.y+y*A.x);}
}a[MAXN],b[MAXN];

void FFT(cp *a,int opt)
{
	for(int i = 0;i < len;++ i) if(i < rev[i]) swap(a[i],a[rev[i]]);
	for(int i = 1;i < len;i <<= 1)
	{
		cp w = cp(cos(PI/i),opt*sin(PI/i));
		for(int j = 0,p = i << 1;j < len;j += p)
		{
			cp s = cp(1,0);
			for(int k = 0;k < i;++ k,s = s * w)
			{
				cp X = a[j+k],Y = s * a[i+j+k];
				a[j+k] = X + Y;
				a[i+j+k] = X - Y;
			}
		}
	}
	if(opt == -1) for(int i = 0;i < len;++ i) a[i].x /= len;
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	lena = Read(); lenb = Read();
	for(int i = 0;i <= lena;++ i) a[i].x = Read();
	for(int i = 0;i <= lenb;++ i) b[i].x = Read();
	while(len <= lena + lenb) len <<= 1,l++;
	for(int i = 0;i < len;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << l);
	FFT(a,1);
	FFT(b,1);
	for(int i = 0;i < len;++ i) a[i] = a[i] * b[i];
	FFT(a,-1);
	for(int i = 0;i <= lena+lenb;++ i) Put((int)(a[i].x + 0.5),' ');
	return 0;
}
$French's\space code\ \text{(迭代)}$ 
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define maxn 10000005
#define x first
#define y second
const double pi=acos(-1.0);
int n,m;
int limit=1;
pair<double,double> a[maxn];
pair<double,double> b[maxn];
int l;
int r[maxn];
pair<double,double> operator + (pair<double,double> a,pair<double,double> b)
{
	return make_pair(a.x+b.x,a.y+b.y);
}
pair<double,double> operator - (pair<double,double> a,pair<double,double> b)
{
	return make_pair(a.x-b.x,a.y-b.y);
}
pair<double,double> operator * (pair<double,double> a,pair<double,double> b)
{
	return make_pair(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
}
#undef x
#undef y
void fft(pair<double,double> *a,int t)
{
	for(int i=0;i<limit;i++)
		if(i<r[i])
			swap(a[i],a[r[i]]);
	for(int mid=1;mid<limit;mid<<=1)
	{
		pair<double,double> wn=make_pair(cos(pi/mid),t*sin(pi/mid));
		for(int r=mid<<1,j=0;j<limit;j+=r)
		{
			pair<double,double> w=make_pair(1,0);
			for(int k=0;k<mid;k++,w=w*wn)
			{
				pair<double,double> x=a[j+k],y=w*a[j+mid+k];
				a[j+k]=x+y;
				a[j+mid+k]=x-y;
			}
		}
	}
}
void work()
{
	fft(a,1);
	fft(b,1);
	for(int i=0;i<=limit;i++)
		a[i]=a[i]*b[i];
	fft(a,-1);
	for(int i=0;i<=n+m;i++)
		cout<<(int)(a[i].first/limit+0.5)<<' ';
}
void perpare()
{
	while(limit<=n+m)
	{
		limit<<=1;
		l++;
	}
	for(int i=0;i<limit;i++)
		r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
}
int main()
{
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=0;i<=n;i++)
		cin>>a[i].first;
	for(int i=0;i<=m;i++)
		cin>>b[i].first;
	perpare();
	work();
	return 0;
}

NTT

讲解

难得一见的讲解：

由于 \(FFT\) 会有精度问题，而且不能取模，所以 \(NTT\) 就诞生了。

我们只需将 \(FFT\) 中的 \(\omega\) 换成 \(NTT\) 中的模数的原根 \(g\) 就好了。

如果我们不需要取模，只需要找一个很大的模数就好了，这样取模就相当于没有取模。

当然最后除 \(len\) 的时候改为乘逆元就好了。

练习

板题(UOJ)

板题(洛谷)

其实你可以用 \(NTT\) 过所有 \(FFT\) 的题。 好像并不是

代码

板题代码

$Mine\ \text{(递归)}$ 
//12252024832524
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define TT template<typename T>
using namespace std; 

typedef long long LL;
const int MAXN = 1 << 21 | 5;
const int MOD = 998244353;
const int PHI = 998244352;
const int GINV = 332748118;
const int G = 3;
int lena,lenb;
int a[MAXN],b[MAXN];

LL Read()
{
	LL x = 0,f = 1;char c = getchar();
	while(c > '9' || c < '0'){if(c == '-')f = -1;c = getchar();}
	while(c >= '0' && c <= '9'){x = (x*10) + (c^48);c = getchar();}
	return x * f;
}
TT void Put1(T x)
{
	if(x > 9) Put1(x/10);
	putchar(x%10^48);
}
TT void Put(T x,char c = -1)
{
	if(x < 0) putchar('-'),x = -x;
	Put1(x);
	if(c >= 0) putchar(c);
}
TT T Max(T x,T y){return x > y ? x : y;}
TT T Min(T x,T y){return x < y ? x : y;}
TT T Abs(T x){return x < 0 ? -x : x;}

int qpow(int x,int y)
{
	int ret = 1;
	while(y){if(y & 1) ret = 1ll * ret * x % MOD;x = 1ll * x * x % MOD;y >>= 1;}
	return ret;
}
void NTT(int len,int *a,int f)
{
	if(len == 1) return;
	int a1[len >> 1],a2[len >> 1];
	for(int i = 0;i < len;i += 2) a1[i >> 1] = a[i],a2[i >> 1] = a[i+1];
	NTT(len>>1,a1,f);
	NTT(len>>1,a2,f);
	int w = qpow(f == 1 ? G : GINV,PHI/len),k = 1;
	len >>= 1;
	for(int i = 0;i < len;++ i,k = 1ll * k * w % MOD)
	{
		a[i] = (a1[i] + 1ll * k * a2[i]) % MOD;
		a[i+len] = (a1[i] - 1ll * k * a2[i]) % MOD;
	}
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".in","w",stdout);
	lena = Read(); lenb = Read();
	for(int i = 0;i <= lena;++ i) a[i] = Read();
	for(int i = 0;i <= lenb;++ i) b[i] = Read();
	int len = 1;
	while(len <= lena + lenb) len <<= 1;
	NTT(len,a,1);
	NTT(len,b,1);
	for(int i = 0;i <= len;++ i) a[i] = 1ll * a[i] * b[i] % MOD;
	NTT(len,a,-1);
	const int invlen = qpow(len,MOD-2);
	for(int i = 0;i <= lena+lenb;++ i) Put((1ll * a[i] * invlen % MOD + MOD) % MOD,' ');
	return 0;
}
$Mine\ \text{(迭代)}$ 
//12252024832524
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define TT template<typename T>
using namespace std; 

typedef long long LL;
const int MAXN = 1 << 21 | 5;
const int MOD = 998244353;
const int PHI = 998244352;
const int GINV = 332748118;
const int G = 3;
int lena,lenb,len = 1,l = -1;
int a[MAXN],b[MAXN],rev[MAXN];

LL Read()
{
	LL x = 0,f = 1;char c = getchar();
	while(c > '9' || c < '0'){if(c == '-')f = -1;c = getchar();}
	while(c >= '0' && c <= '9'){x = (x*10) + (c^48);c = getchar();}
	return x * f;
}
TT void Put1(T x)
{
	if(x > 9) Put1(x/10);
	putchar(x%10^48);
}
TT void Put(T x,char c = -1)
{
	if(x < 0) putchar('-'),x = -x;
	Put1(x);
	if(c >= 0) putchar(c);
}
TT T Max(T x,T y){return x > y ? x : y;}
TT T Min(T x,T y){return x < y ? x : y;}
TT T Abs(T x){return x < 0 ? -x : x;}

int qpow(int x,int y)
{
	int ret = 1;
	while(y){if(y & 1) ret = 1ll * ret * x % MOD;x = 1ll * x * x % MOD;y >>= 1;}
	return ret;
}
void NTT(int *a,int opt)
{
	for(int i = 0;i < len;++ i) if(i < rev[i]) swap(a[i],a[rev[i]]);
	for(int i = 1;i < len;i <<= 1)
	{
		int w = qpow(opt == 1 ? G : GINV,PHI / (i << 1));
		for(int j = 0,p = i << 1;j < len;j += p)
		{
			int mi = 1;
			for(int k = 0;k < i;++ k,mi = 1ll * mi * w % MOD)
			{
				int X = a[j+k],Y = 1ll * mi * a[i+j+k] % MOD;
				a[j+k] = (X + Y) % MOD;
				a[i+j+k] = (X - Y + MOD) % MOD;
			}
		}
	}
	int invlen = qpow(len,MOD-2);
	if(opt == -1) for(int i = 0;i < len;++ i) a[i] = 1ll * a[i] * invlen % MOD;
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	lena = Read(); lenb = Read();
	for(int i = 0;i <= lena;++ i) a[i] = Read();
	for(int i = 0;i <= lenb;++ i) b[i] = Read();
	while(len <= lena + lenb) len <<= 1,l++;
	for(int i = 0;i < len;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << l);
	NTT(a,1);
	NTT(b,1);
	for(int i = 0;i < len;++ i) a[i] = 1ll * a[i] * b[i] % MOD;
	NTT(a,-1);
	for(int i = 0;i <= lena+lenb;++ i) Put(a[i],' ');
	return 0;
}

MTT

讲解

lvzelong2014巨佬

记得用long double

练习

板题(洛谷)

代码

板题(洛谷)代码 
//12252024832524
#include <bits/stdc++.h>
#define TT template<typename T>
using namespace std;

typedef long long LL;
const int MAXN = 1 << 18 | 5,M = 32767;
const long double PI = acos(-1); 
int lena,lenb,mod;
int A[MAXN],B[MAXN],ans[MAXN];

LL Read()
{
	LL x = 0,f = 1; char c = getchar();
	while(c > '9' || c < '0'){if(c == '-') f = -1;c = getchar();}
	while(c >= '0' && c <= '9'){x = (x*10) + (c^48);c = getchar();}
	return x * f;
}
TT void Put1(T x)
{
	if(x > 9) Put1(x/10);
	putchar(x%10^48);
}
TT void Put(T x,char c = -1)
{
	if(x < 0) putchar('-'),x = -x;
	Put1(x); if(c >= 0) putchar(c);
}
TT T Max(T x,T y){return x > y ? x : y;}
TT T Min(T x,T y){return x < y ? x : y;}
TT T Abs(T x){return x < 0 ? -x : x;}

struct cp
{
	long double x,y;
	cp operator + (const cp &C)const{return cp{x+C.x,y+C.y};}
	cp operator - (const cp &C)const{return cp{x-C.x,y-C.y};}
	cp operator * (const cp &C)const{return cp{x*C.x-y*C.y,x*C.y+y*C.x};}
}a[MAXN],b[MAXN],I,ab,ak,bb,bk,da[MAXN],db[MAXN],dc[MAXN],dd[MAXN];
cp cj(cp C){return cp{C.x,-C.y};}

int rev[MAXN],len;
void pre(int L)
{
	int l = -1; len = 1;
	while(len <= L) len <<= 1,++l;
	for(int i = 1;i < len;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << l);
}
void FFT(cp *a,int f)//val = kM + b
{
	for(int i = 0;i < len;++ i) if(i < rev[i]) swap(a[i],a[rev[i]]);
	for(int i = 1;i < len;i <<= 1)
	{
		cp w = cp{cos(PI/i),f*sin(PI/i)};
		for(int j = 0,p = i << 1;j < len;j += p)
		{
			cp mi = cp{1,0};
			for(int k = 0;k < i;++ k,mi = mi * w)
			{
				cp X = a[j+k],Y = mi * a[i+j+k];
				a[j+k] = X+Y; 
				a[i+j+k] = X-Y; 
			}
		}
	}
	if(f == -1) for(int i = 0;i < len;++ i) a[i].x /= len,a[i].y /= len;
}
void mul(int *ma,int *mb,int *mc,int L,const int MOD)
{
	pre(L);
//	for(int i = 0;i < len;++ i) a[i] = b[i] = I;
	for(int i = 0;i <= lena;++ i) a[i] = cp{ma[i] & M,ma[i] >> 15};
	for(int i = 0;i <= lenb;++ i) b[i] = cp{mb[i] & M,mb[i] >> 15};
	FFT(a,1); FFT(b,1);
	for(int i = 0,j;i < len;++ i)
	{
		j = (len-i) & (len-1);
		ab = (a[i] + cj(a[j])) * cp{0.5,0};
		ak = (a[i] - cj(a[j])) * cp{0,-0.5};
		bb = (b[i] + cj(b[j])) * cp{0.5,0};
		bk = (b[i] - cj(b[j])) * cp{0,-0.5};
		da[i] = ab * bb; db[i] = ab * bk; dc[i] = ak * bb; dd[i] = ak * bk;
	}
	for(int i = 0;i < len;++ i) a[i] = da[i] + db[i] * cp{0,1};
	for(int i = 0;i < len;++ i) b[i] = dc[i] + dd[i] * cp{0,1};
	FFT(a,-1); FFT(b,-1);
	for(int i = 0;i <= lena+lenb;++ i)
	{
		LL v1 = ((LL)(a[i].x+0.5)) % MOD,v2 = ((LL)(a[i].y+0.5)) % MOD,v3 = ((LL)(b[i].x+0.5)) % MOD,v4 = ((LL)(b[i].y+0.5)) % MOD;
		mc[i] = (v1 + ((v2 + v3) << 15) + (v4 << 30)) % MOD;
	}
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	lena = Read(); lenb = Read(); mod = Read();
	for(int i = 0;i <= lena;++ i) A[i] = Read();
	for(int i = 0;i <= lenb;++ i) B[i] = Read();
	mul(A,B,ans,lena+lenb,mod);
	for(int i = 0;i <= lena+lenb;++ i) Put(ans[i],i == lena+lenb ? '\n' : ' ');
	return 0;
}
posted @ 2021-02-02 15:32  皮皮刘  阅读(428)  评论(0)  编辑  收藏  举报