多项式板子

FFT

给定一个\(n\)次多项式\(F(x)\),和一个\(m\)次多项式\(G(x)\),请求出\(F(x)\)\(G(x)\)的卷积。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=3e6+1000;
const double pi=acos(-1);
int n,m,r[maxn],len;
struct complex{double x,y;}a[maxn],b[maxn];
complex operator-(complex a,complex b){return (complex){a.x-b.x,a.y-b.y};}
complex operator+(complex a,complex b){return (complex){a.x+b.x,a.y+b.y};}
complex operator*(complex a,complex b){return (complex){a.x*b.x-a.y*b.y,a.x*b.y+b.x*a.y};}
void fft(complex *a,int f)
{
    for(rg int i=0;i<n;i++)if(r[i]>i)swap(a[r[i]],a[i]);
    for(rg int i=1;i<n;i<<=1)
    {
        complex wn=(complex){cos(pi/i),f*sin(pi/i)};
        for(rg int j=0;j<n;j+=(i<<1))
        {
            complex w=(complex){1,0};
            for(rg int k=0;k<i;k++)
            {
                complex x=a[j+k],y=w*a[j+k+i];
                a[j+k]=x+y,a[j+k+i]=x-y,w=w*wn;
            }
        }
    }
    if(f==-1)for(rg int i=0;i<=m;i++)a[i].x=a[i].x/n+0.1;
}
int main()
{
    read(n),read(m);
    for(rg int i=0,x;i<=n;i++)read(x),a[i].x=x;
    for(rg int i=0,x;i<=m;i++)read(x),b[i].x=x;
    m+=n;for(n=1;n<=m;n<<=1)len++;
    for(rg int i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    fft(a,1),fft(b,1);
    for(rg int i=0;i<=n;i++)a[i]=a[i]*b[i];
    fft(a,-1);
    for(rg int i=0;i<=m;i++)printf("%d ",(int)a[i].x);
}

NTT

给定一个\(n\)次多项式\(F(x)\),和一个\(m\)次多项式\(G(x)\),请求出\(F(x)\)\(G(x)\)的卷积,并对\(998244353\)取模。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=3e6+10,mod=998244353,g=3,ginv=332748118;
int n,m,a[maxn],b[maxn],len,r[maxn];
int mi(int a,int b)
{
    int ans=1;
    while(b)
    {
        if(b&1)ans=1ll*ans*a%mod;
        b>>=1,a=1ll*a*a%mod;
    }
    return ans;
}
void ntt(int *a,int f)
{
    for(rg int i=0;i<n;i++)if(r[i]>i)swap(a[r[i]],a[i]);
    for(rg int i=1;i<n;i<<=1)
    {
        int wn=mi(f?g:ginv,(mod-1)/(i<<1));
        for(rg int j=0;j<n;j+=(i<<1))
        {
            int w=1;
            for(rg int k=0;k<i;k++)
            {
                int x=a[j+k],y=1ll*a[j+k+i]*w%mod;
                a[j+k]=(x+y)%mod,a[j+k+i]=(x-y+mod)%mod,w=1ll*w*wn%mod;
            }
        }
    }
}
int main()
{
    read(n),read(m);
    for(rg int i=0;i<=n;i++)read(a[i]);
    for(rg int i=0;i<=m;i++)read(b[i]);
    m+=n;for(n=1;n<=m;n<<=1)len++;
    for(rg int i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    ntt(a,1),ntt(b,1);
    for(rg int i=0;i<n;i++)a[i]=1ll*a[i]*b[i]%mod;
    ntt(a,0);int inv=mi(n,mod-2);
    for(rg int i=0;i<=m;i++)printf("%d ",1ll*inv*a[i]%mod);
}

多项式求逆

给定一个多项式\(F(x)\) ,请求出一个多项式\(G(x)\), 满足\(F(x)∗G(x)≡1(mod\ x^n)\)。系数对 \(998244353\)取模。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=3e6+10,mod=998244353,g=3,ginv=332748118;
int n,m,a[maxn],b[maxn],c[maxn],r[maxn];
int mi(int a,int b)
{
    int ans=1;
    while(b)
    {
        if(b&1)ans=1ll*ans*a%mod;
        b>>=1,a=1ll*a*a%mod;
    }
    return ans;
}
void ntt(int *a,int n,int f)
{
    for(rg int i=0;i<n;i++)if(r[i]>i)swap(a[r[i]],a[i]);
    for(rg int i=1;i<n;i<<=1)
    {
        int wn=mi(f?g:ginv,(mod-1)/(i<<1));
        for(rg int j=0;j<n;j+=(i<<1))
        {
            int w=1;
            for(rg int k=0;k<i;k++)
            {
                int x=a[j+k],y=1ll*a[j+k+i]*w%mod;
                a[j+k]=(x+y)%mod,a[j+k+i]=(x-y+mod)%mod,w=1ll*w*wn%mod;
            }
        }
    }
    if(f)return ;
    int inv=mi(n,mod-2);
    for(rg int i=0;i<n;i++)a[i]=1ll*a[i]*inv%mod;
}
void solve(int n,int *a,int *b)
{
    if(n==1){b[0]=mi(a[0],mod-2);return ;}
    solve((n+1)>>1,a,b);int m,len=0;
    for(m=1;m<=n*2;m<<=1)len++;
    for(rg int i=0;i<n;i++)c[i]=a[i];
    for(rg int i=n;i<m;i++)c[i]=0;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    ntt(b,m,1),ntt(c,m,1);
    for(rg int i=0;i<m;i++)b[i]=((2ll*b[i]-1ll*c[i]*b[i]%mod*b[i]%mod)+mod)%mod;
    ntt(b,m,0);
    for(rg int i=n;i<m;i++)b[i]=0;
}
int main()
{
    read(n);
    for(rg int i=0;i<n;i++)read(a[i]);
    solve(n,a,b);
    for(rg int i=0;i<n;i++)printf("%d ",b[i]);
}

多项式求导

其实就是幂函数的求导
\[ f(x)=\sum_{i=0}a_ix_i\\ f'(x)=\sum_{i=1}a_{i}x^{i-1} \]

多项式求ln

给出\(n-1\)次多项式\(A(x)\),求一个 \(\bmod{\:x^n}\)下的多项式\(B(x)\),满足\(B(x)\equiv \ln A(x)\).
\(\text{mod } 998244353\)下进行,且\(a_i\in [0, 998244353] \cap \mathbb{Z}\)
代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=8e5+10,mod=998244353,g=3,gi=332748118;
int n,m,a[maxn],b[maxn],c[maxn],len,r[maxn],d[maxn],e[maxn];
int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
int del(int x,int y){return x-y<0?x-y+mod:x-y;}
int mul(int x,int y){return 1ll*x*y-1ll*x*y/mod*mod;}
int mi(int a,int b)
{
    int ans=1;
    while(b)
    {
        if(b&1)ans=mul(a,ans);
        b>>=1,a=mul(a,a);
    }
    return ans;
}
void ntt(int *a,int n,int f)
{
    for(rg int i=0;i<n;i++)if(r[i]>i)swap(a[i],a[r[i]]);
    for(rg int i=1;i<n;i<<=1)
    {
        int wn=mi((f?g:gi),(mod-1)/(i<<1));
        for(rg int j=0;j<n;j+=(i<<1))
        {
            int w=1;
            for(rg int k=0;k<i;k++)
            {
                int x=a[j+k],y=mul(a[j+k+i],w);
                a[j+k]=add(x,y),a[j+k+i]=del(x,y),w=mul(w,wn);
            }
        }
    }
    if(f)return ;int inv=mi(n,mod-2);
    for(rg int i=0;i<n;i++)a[i]=mul(a[i],inv);
}
void solve(int n,int *a,int *b)
{
    if(n==1){b[0]=mi(a[0],mod-2);return ;}
    solve((n+1)>>1,a,b);len=0;
    for(m=1;m<=n*2;m<<=1)len++;
    for(rg int i=0;i<n;i++)c[i]=a[i];
    for(rg int i=n;i<m;i++)c[i]=0;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    ntt(c,m,1),ntt(b,m,1);
    for(rg int i=0;i<m;i++)b[i]=del(mul(2,b[i]),mul(mul(b[i],b[i]),c[i]));
    ntt(b,m,0);
    for(rg int i=n;i<m;i++)b[i]=0;
}
void getln(int *a,int n)
{
    solve(n,a,b);len=0;
    for(rg int i=1;i<n;i++)d[i-1]=mul(a[i],i);
    for(m=1;m<=n;m<<=1)len++;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    ntt(d,m,1),ntt(b,m,1);
    for(rg int i=0;i<m;i++)d[i]=mul(d[i],b[i]);
    ntt(d,m,0);e[0]=0;
    for(rg int i=1;i<m;i++)e[i]=mul(d[i-1],mi(i,mod-2));
}
int main()
{
    read(n);
    for(rg int i=0;i<n;i++)read(a[i]);
    for(m=1;m<=n;m<<=1);getln(a,m);
    for(rg int i=0;i<n;i++)printf("%d ",e[i]);
}

多项式求exp

给出\(n-1\)次多项式\(A(x)\),求一个\(\mod\ x^n\)下的多项式\(B(x)\),满足\(B(x)≡e^A(x)\).
代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=8e5+10,mod=998244353,g=3,gi=332748118;
int n,a[maxn],r[maxn],c[maxn],f[maxn],b[maxn],s[maxn],w[maxn],h[maxn];
int mul(int x,int y){return 1ll*x*y-1ll*x*y/mod*mod;}
int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
int del(int x,int y){return x-y<0?x-y+mod:x-y;}
int mi(int a,int b){
    int ans=1;
    while(b){
        if(b&1)ans=mul(ans,a);
        b>>=1,a=mul(a,a);
    }
    return ans;
}
void ntt(int *a,int n,int f){
    for(rg int i=0;i<n;i++)if(r[i]>i)swap(a[i],a[r[i]]);
    for(rg int i=1;i<n;i<<=1){
        int wn=mi(f?g:gi,(mod-1)/(i<<1));
        for(rg int j=0;j<n;j+=(i<<1)){
            int w=1;
            for(rg int k=0;k<i;k++){
                int x=a[j+k],y=mul(w,a[j+k+i]);
                a[j+k]=add(x,y),a[j+k+i]=del(x,y),w=mul(w,wn); 
            }
        }
    }
    if(f)return ;int inv=mi(n,mod-2);
    for(rg int i=0;i<n;i++)a[i]=mul(a[i],inv); 
}
void get_inv(int *a,int *b,int n){
    if(n==1)return b[0]=mi(a[0],mod-2),void();
    get_inv(a,b,(n+1)>>1);
    int m,len=0;
    for(m=1;m<=n<<1;m<<=1)len++;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    for(rg int i=0;i<n;i++)c[i]=a[i];
    for(rg int i=n;i<m;i++)c[i]=0;
    ntt(c,m,1),ntt(b,m,1);
    for(rg int i=0;i<m;i++)
        b[i]=del(mul(2,b[i]),mul(c[i],mul(b[i],b[i])));
    ntt(b,m,0);
    for(rg int i=n;i<m;i++)b[i]=0;
}
void get_ln(int *a,int *b,int n){
    for(rg int i=0;i<n;i++)w[i]=mul(a[i+1],i+1);
    get_inv(a,h,n);int m,len=0;
    for(m=1;m<=n<<1;m<<=1)len++;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    ntt(w,m,1),ntt(h,m,1);
    for(rg int i=0;i<m;i++)w[i]=mul(w[i],h[i]);
    ntt(w,m,0);b[0]=0;
    for(rg int i=0;i<m;i++)b[i+1]=mul(w[i],mi(i+1,mod-2)),h[i]=0;
    for(rg int i=n;i<m;i++)b[i]=0;
}
void get_exp(int *a,int *b,int n){
    if(n==1)return b[0]=1,void();
    get_exp(a,b,(n+1)>>1);
    get_ln(b,s,n);
    int m,len=0;
    for(m=1;m<=n<<1;m<<=1)len++;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    for(rg int i=n;i<m;i++)s[i]=0;
    for(rg int i=0;i<n;i++)s[i]=del(a[i],s[i]);s[0]=add(s[0],1);
    ntt(s,m,1),ntt(b,m,1);
    for(rg int i=0;i<m;i++)b[i]=mul(b[i],s[i]);
    ntt(b,m,0);
    for(rg int i=n;i<m;i++)b[i]=0;
}
int main()
{
    read(n);
    for(rg int i=0;i<n;i++)read(a[i]);
    get_exp(a,f,n);
    for(rg int i=0;i<n;i++)printf("%d ",f[i]);
}

多项式快速幂

给定一个\(n-1\)次多项式\(A(x)\),求一个在\(mod\ x^n\)意义下的多项式\(B(x)\),使\(B(x)≡A^k(x) (mod \ x^n)\)
代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=8e5+10,mod=998244353,g=3,gi=332748118;
char ch[maxn];
int n,m,tot,k,now=1,r[maxn],a[maxn],b[maxn],w[maxn],h[maxn],d[maxn],c[maxn],f[maxn],s[maxn];
int mul(int x,int y){return 1ll*x*y-1ll*x*y/mod*mod;}
int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
int del(int x,int y){return x-y<0?x-y+mod:x-y;}
int mi(int a,int b){
    int ans=1;
    while(b){
        if(b&1)ans=mul(ans,a);
        b>>=1,a=mul(a,a);
    }
    return ans;
}
void ntt(int *a,int n,int f){
    for(rg int i=0;i<n;i++)if(r[i]>i)swap(a[i],a[r[i]]);
    for(rg int i=1;i<n;i<<=1){
        int wn=mi(f?g:gi,(mod-1)/(i<<1));
        for(rg int j=0;j<n;j+=i<<1){
            int w=1;
            for(rg int k=0;k<i;k++){
                int x=a[j+k],y=mul(w,a[j+k+i]);
                a[j+k]=add(x,y),a[j+k+i]=del(x,y),w=mul(w,wn);
            }
        }
    }
    if(f)return ;int inv=mi(n,mod-2);
    for(rg int i=0;i<n;i++)a[i]=mul(a[i],inv);
}
void get_inv(int *a,int *b,int n){
    if(n==1)return b[0]=mi(a[0],mod-2),void(); 
    get_inv(a,b,(n+1)>>1);int m,len=0;for(m=1;m<=n<<1;m<<=1)len++;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    for(rg int i=0;i<n;i++)c[i]=a[i];
    for(rg int i=n;i<m;i++)c[i]=0;
    ntt(c,m,1),ntt(b,m,1);
    for(rg int i=0;i<m;i++)b[i]=del(mul(2,b[i]),mul(c[i],mul(b[i],b[i])));
    ntt(b,m,0);
    for(rg int i=n;i<m;i++)b[i]=0;
}
void get_ln(int *a,int *b,int n){
    for(rg int i=0;i<n;i++)w[i]=mul(a[i+1],i+1);
    get_inv(a,h,n);int m,len=0;for(m=1;m<=n<<1;m<<=1)len++;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    ntt(h,m,1),ntt(w,m,1);
    for(rg int i=0;i<m;i++)w[i]=mul(w[i],h[i]);
    ntt(w,m,0);b[0]=0;
    for(rg int i=0;i<m;i++)b[i+1]=mul(w[i],mi(i+1,mod-2)),h[i]=w[i]=0;
    for(rg int i=n;i<m;i++)b[i]=0; 
}
void get_exp(int *a,int *b,int n){
    if(n==1)return b[0]=1,void();
    get_exp(a,b,(n+1)>>1),get_ln(b,s,n);
    int m,len=0;for(m=1;m<=n<<1;m<<=1)len++;
    for(rg int i=0;i<m;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
    for(rg int i=0;i<n;i++)s[i]=del(a[i],s[i]);s[0]=add(s[0],1);
    ntt(s,m,1),ntt(b,m,1);
    for(rg int i=0;i<m;i++)b[i]=mul(b[i],s[i]);
    ntt(b,m,0);
    for(rg int i=n;i<m;i++)b[i]=0;
}
int main()
{
    read(n),read(k);
    for(rg int i=0;i<n;i++)read(a[i]);
    get_ln(a,d,n);for(rg int i=0;i<n;i++)d[i]=mul(k,d[i]);
    get_exp(d,f,n);for(rg int i=0;i<n;i++)printf("%d ",f[i]);
}
posted @ 2019-03-30 18:59 蒟蒻--lichenxi 阅读(...) 评论(...) 编辑 收藏