标签：namespace 多点 struct exp sqrt tin check type 函数
#include<bits/stdc++.h>
using namespace std;
const int N=(1<<21)+10;
const int mod=998244353;
const int gg=3;
const int giv=(mod+1)/3;
typedef long long ll;
namespace Math{
int inv[N],tp=2;
inline void init(int n){
if(tp==2) inv[0]=inv[1]=1;
for(tp;tp<=n;++tp)
inv[tp]=1ll*(mod-mod/tp)*inv[mod%tp]%mod;
}
inline int ksm(int a,ll b){
int ret=1;
for(;b;a=1ll*a*a%mod,b>>=1) (b&1)&&(ret=1ll*ret*a%mod);
return ret;
}
inline int add(int x,int y){return (x+y>=mod)?x+y-mod:x+y;}
inline int dec(int x,int y){return (x-y<0)?x-y+mod:x-y;}
}
using namespace Math;
namespace Container{
struct poly{
vector<int>v;
inline int& operator[](int x){while(x>=v.size())v.push_back(0);return v[x];}
inline poly(int x=0):v(1){v[0]=x;}
inline int size(){return v.size();}
inline void resize(int x){v.resize(x);}
inline void mem(int l,int r,int x){fill(v.begin()+l,v.begin()+r+1,x);}
};
inline poly operator +(poly x,poly y){
int mx=max(x.size(),y.size());
for(int i=0;i<mx;++i) x[i]=add(x[i],y[i]);
return x;
}
inline poly operator -(poly x,poly y){
int mx=max(x.size(),y.size());
for(int i=0;i<mx;++i) x[i]=dec(x[i],y[i]);
return x;
}
inline poly operator *(poly x,poly y){
int mx=max(x.size(),y.size());
for(int i=0;i<mx;++i) x[i]=1ll*x[i]*y[i]%mod;
return x;
}
inline poly operator *(poly x,int y){
for(int i=0;i<x.size();++i) x[i]=1ll*x[i]*y%mod;
return x;
}
}
using namespace Container;
namespace basic{
int r[N],Wn[N];
inline void NTT(int lim,poly& f,int tp){
for(int i=0;i<lim;++i) if(i<r[i]) swap(f[i],f[r[i]]);
for(int mid=1;mid<lim;mid<<=1){
int len=mid<<1,wn=ksm(tp==1?gg:giv,(mod-1)/len);
Wn[0]=1;for(int i=1;i<mid;++i) Wn[i]=1ll*Wn[i-1]*wn%mod;
for(int l=0;l+len-1<lim;l+=len){
for(int k=l;k<=l+mid-1;++k){
int w1=f[k],w2=1ll*Wn[k-l]*f[k+mid]%mod;
f[k]=add(w1,w2);f[k+mid]=dec(w1,w2);
}
}
}
}
inline poly poly_mul(int n,int m,poly f,poly g){
int lim=1,len=0;
while(lim<(n+m)) lim<<=1,len++;
for(int i=0;i<lim;++i) r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
if(f.size()>n) f.mem(n,min(lim,f.size())-1,0);
if(g.size()>m) g.mem(m,min(lim,g.size())-1,0);
NTT(lim,f,1);NTT(lim,g,1);
for(int i=0;i<lim;++i) f[i]=1ll*f[i]*g[i]%mod;
NTT(lim,f,-1);
int iv=ksm(lim,mod-2);
for(int i=0;i<lim;++i) f[i]=1ll*f[i]*iv%mod;
return f;
}
inline void getdao(int n,poly& f){
for(int i=1;i<n;++i){
int t=f[i];
f[i-1]=1ll*i*t%mod;
}
f[n-1]=0;
}
inline void jifen(int n,poly& f){
init(n);
for(int i=n-1;i;--i) f[i]=1ll*inv[i]*f[i-1]%mod;
f[0]=0;
}
inline void get(int x,poly& f){
for(int i=0;i<x;++i) scanf("%d",&f[i]);
}
inline void print(int x,poly f){
for(int i=0;i<x;++i) printf("%d ",f[i]);puts("");
}
}
using namespace basic;
namespace Cipolla{
int I,fl=0;
struct pt{
int a,b;
pt(int _a=0,int _b=0){a=_a;b=_b;}
};
inline pt operator *(pt x,pt y){
pt ret;
ret.a=add(1ll*x.a*y.a%mod,1ll*x.b*y.b%mod*I%mod);
ret.b=add(1ll*x.a*y.b%mod,1ll*x.b*y.a%mod);
return ret;
}
inline bool check(int x){
return ksm(x,(mod-1)/2)==1;
}
inline int random(){
return 1ll*rand()*rand()%mod;
}
inline pt qpow(pt a,int b){
pt ret=pt(1,0);
for(;b;a=a*a,b>>=1) if(b&1) ret=ret*a;
return ret;
}
inline int cipolla(int n){
if(!fl) srand(time(0)),fl=1;
if(!check(n)) return 0;
int a=random();
while(!a||check(dec(1ll*a*a%mod,n))) a=random();
I=dec(1ll*a*a%mod,n);
int ans=qpow(pt(a,1),(mod+1)/2).a;
return min(ans,mod-ans);
}
}
using namespace Cipolla;
namespace Poly{
inline poly getinv(int n,poly f){
if(n==1){poly g;g[0]=ksm(f[0],mod-2);return g;}
poly g=getinv(n+1>>1,f);poly p=g;
int lim=1,len=0;
while(lim<(n<<1)) lim<<=1,len++;
for(int i=0;i<lim;++i) r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
if(f.size()>n) f.mem(n,min(lim,f.size())-1,0);
if(p.size()>n) p.mem(n,min(lim,g.size())-1,0);
NTT(lim,f,1);NTT(lim,p,1);
for(int i=0;i<lim;++i) f[i]=1ll*f[i]*p[i]%mod*p[i]%mod;
NTT(lim,f,-1);
int iv=ksm(lim,mod-2);
poly h;
for(int i=0;i<n;++i) h[i]=dec(2ll*g[i]%mod,1ll*f[i]*iv%mod);
return h;
}
inline poly getln(int n,poly f){
poly g=getinv(n,f);getdao(n,f);
poly h=poly_mul(n,n,f,g);
jifen(n,h);
return h;
}
inline poly getexp(int n,poly f){
if(n==1){poly g;g[0]=1;return g;}
poly g=getexp(n+1>>1,f),B=getln(n,g);
for(int i=0;i<n;++i) B[i]=dec((i==0),dec(B[i],f[i]));
return poly_mul(n,n,g,B);
}
inline poly Pow(int n,int k,poly f){
poly g=getln(n,f);
for(int i=0;i<n;++i) g[i]=1ll*g[i]*k%mod;
return getexp(n,g);
}
inline poly Sqrt(int n,poly f){
if(n==1){poly g;g[0]=cipolla(f[0]);return g;}
poly g=Sqrt((n+1)>>1,f);
poly p=getinv(n,g);
poly h=poly_mul(n,n,p,f);
for(int i=0;i<n;++i) h[i]=1ll*((mod+1)/2)*add(h[i],g[i])%mod;
return h;
}
inline poly rev(int n,poly f){
for(int i=0;i<(n>>1);++i) swap(f[i],f[n-i-1]);
return f;
}
inline poly divide(int n,int m,poly A,poly B){
A=rev(n,A);B=rev(m,B);
B=getinv(n-m+1,B);
poly C=poly_mul(n-m+1,n-m+1,A,B);
return rev(n-m+1,C);
}
inline poly Mod(int n,int m,poly A,poly B,poly C){
B=poly_mul(m,n-m+1,B,C);
for(int i=0;i<m-1;++i) A[i]=dec(A[i],B[i]);
return A;
}
inline poly Mod(int n,int m,poly A,poly B){
poly C=divide(n,m,A,B);
return Mod(n,m,A,B,C);
}
inline poly Sin(int n,poly f){
int II=ksm(gg,(mod-1)/4);
poly g1=getexp(n,f*II),g2=getexp(n,f*(mod-II));
return (g1-g2)*ksm(add(II,II),mod-2);
}
inline poly Cos(int n,poly f){
int II=ksm(gg,(mod-1)/4);
poly g1=getexp(n,f*II),g2=getexp(n,f*(mod-II));
return (g1+g2)*((mod+1)/2);
}
}
using namespace Poly;
int n,m,a[N],ans[N];poly f,g;
namespace fastpow{
int flag,k,kk,ne;
poly f,ret;
char s[N];
inline int readpow1(){
long long x=0,f=1;int len=strlen(s+1);
for(int i=1;i<=len;++i){char ch=s[i];x=((x<<3)+(x<<1)+(ch^48));if(x>=n) flag=1;while(x>=mod) x-=mod;}
return f==-1?mod-x:x;
}
inline int readpow2(){
long long x=0,f=1;int m=mod-1,len=strlen(s+1);
for(int i=1;i<=len;++i){char ch=s[i];x=((x<<3)+(x<<1)+(ch^48));while(x>=m) x-=m;}
return f==-1?m-x:x;
}
inline int init(int &n,poly& f){
int now=0;
while(f[now]==0) ++now;
int iv=ksm(f[now],mod-2),ans=ksm(f[now],kk);
for(int i=now;i<n;++i) f[i-now]=1ll*f[i]*iv%mod;
long long x=1ll*now*k;
if(x>=n||(flag&&now!=0)){
for(int i=0;i<n;++i) ret[i]=0;
return -1;
}
for(int i=0;i<x;++i) ret[i]=0;
ne=x;
n-=now;
return ans;
}//init:将f的最低非零系数变为1
inline poly notone_pow(int n,int k,poly f){
ne=n;kk=k;int t=n;int p=init(n,f);
if(p==-1) return ret;
f=Poly::Pow(n,k,f);
for(int i=ne;i<t;++i) ret[i]=f[i-ne];
return ret;
}
}
namespace DCFFT{
inline void solve(int l,int r){
if(l>=r) return ;
int mid=(l+r)>>1;
solve(l,mid);
poly f;for(int i=l;i<=mid;++i) f[i-l]=ans[i];
poly h=poly_mul(mid-l+1,r-l+1,f,g);
for(int i=mid+1;i<=r;++i) ans[i]=add(ans[i],h[i-l]);
solve(mid+1,r);
}
}
//Tip:上面所有的n都表示n-1次多项式
int main(){
return 0;
}
标签：namespace 多点 struct exp sqrt tin check type 函数
原文地址：https://www.cnblogs.com/tqxboomzero/p/14170694.html
