多项式工业基础与全家桶

多项式工业基础与全家桶

开坑待填，放个常数巨大的板子先

别忘了这道题！P3338 [ZJOI2014]力

#define Maxn 200005
#define mod 998244353
inline int ksm(int x,int y=mod-2)
{int ret=1;for(;y;y>>=1,x=1ll*x*x%mod) if(y&1) ret=1ll*ret*x%mod;return ret;
}
const int G=3,invG=ksm(3,mod-2);
int tr[Maxn<<1];
struct Poly
{int len,f[Maxn<<1];Poly(int L=0) { len=L,memset(f,0,sizeof(f)); }inline void ntt(int n,int opt){for(int i=0;i<n;i++) tr[i]=(tr[i>>1]>>1)|((i&1)?(n>>1):0);for(int i=0;i<n;i++) if(i<tr[i]) swap(f[i],f[tr[i]]);for(int p=2;p<=n;p<<=1){int tG=ksm((opt==1)?G:invG,(mod-1)/p),L=p>>1;for(int i=0;i<n;i+=p){int buf=1;for(int j=0;j<L;j++){int tmp=1ll*buf*f[i+j+L]%mod;f[i+j+L]=f[i+j]-tmp;if(f[i+j+L]<0) f[i+j+L]+=mod;f[i+j]=f[i+j]+tmp;if(f[i+j]>=mod) f[i+j]-=mod;buf=1ll*buf*tG%mod;}}}}Poly friend operator + (Poly x,Poly y){x.len=max(x.len,y.len);for(int i=0;i<=x.len;i++) x.f[i]+=y.f[i];for(int i=0;i<=x.len;i++) if(x.f[i]>=mod) x.f[i]-=mod;return x;}Poly friend operator - (Poly x,Poly y){x.len=max(x.len,y.len);for(int i=0;i<=x.len;i++) x.f[i]-=y.f[i];for(int i=0;i<=x.len;i++) if(x.f[i]<0) x.f[i]+=mod;return x;}Poly friend operator * (Poly x,int y){for(int i=0;i<=x.len;i++) x.f[i]=1ll*x.f[i]*y%mod;return x;}Poly friend operator * (Poly x,Poly y){int L=1;for(;L<=(x.len+y.len);L<<=1);int invl=ksm(L);x.ntt(L,1),y.ntt(L,1);for(int i=0;i<L;i++) x.f[i]=1ll*x.f[i]*y.f[i]%mod;x.ntt(L,-1);for(int i=0;i<L;i++) x.f[i]=1ll*x.f[i]*invl%mod;x.len=x.len+y.len;return x;}Poly friend operator % (Poly x,int y){for(int i=y+1;i<=x.len;i++) x.f[i]=0;x.len=y;return x;}Poly invf(){Poly ret(0);ret.f[0]=ksm(f[0]);int L=1; for(;L<len;L<<=1);for(int p=2;p<=L;p<<=1) ret=((ret*2)-(*this%(p-1))*ret*ret)%(p-1);return ret;}Poly friend operator / (Poly x,Poly y){y=y.invf();int L=1;for(;L<=(x.len+y.len);L<<=1);int invl=ksm(L);x.ntt(L,1),y.ntt(L,1);for(int i=0;i<L;i++) x.f[i]=1ll*x.f[i]*y.f[i]%mod;x.ntt(L,-1);for(int i=0;i<L;i++) x.f[i]=1ll*x.f[i]*invl%mod;x.len=x.len+y.len;return x;}
};