【Codeforces 623E】Transforming Sequence（MTT）

传送门

设$f[i][j]$表示$i$个数，$j$位有$1$时的方案数
那么有
f a , i ∗ f b , j ∗ ( i + j i ) 2 b i − > f a + b , i + j f_{a,i}*f_{b,j}*{i+j\choose i}2^{bi}->f_{a+b,i+j} fa,i​∗fb,j​∗(ii+j​)2bi−>fa+b,i+j​
于是倍增$dp,fft$优化
要写$mtt$

#include
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){static char ibuf[RLEN],*ib,*ob;(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));return (ib==ob)?EOF:*ib++;
}
inline int read(){char ch=gc();int res=0;bool f=1;while(!isdigit(ch))f^=ch=='-',ch=gc();while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
struct plx{double x,y;plx(double _x=0,double _y=0):x(_x),y(_y){}friend inline plx operator +(cs plx &a,cs plx &b){return plx(a.x+b.x,a.y+b.y);}friend inline plx operator -(cs plx &a, cs plx &b){return plx(a.x-b.x,a.y-b.y);}friend inline plx operator *(cs plx &a,cs plx &b){return plx(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}friend inline plx operator /(cs plx &a,cs int &b){return plx(a.x/b,a.y/b);}inline plx conj(){return plx(x,-y);}inline void operator /=(cs int &b){*this=*this/b;}
};
cs int N=30005,C=16,M=(1<<15)-1,L=(1<<C)+1;
int fac[N],ifac[N];
inline void init_inv(cs int len=N-5){fac[0]=ifac[0]=1;for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);ifac[len]=Inv(fac[len]);for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
}
inline int c(int n,int m){return n<m?0:mul(fac[n],mul(ifac[m],ifac[n-m]));}
cs double pi=acos(-1);
int rev[L],lim;
ll n;int k;
inline void init_rev(int lim){for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
plx *w[C+1];
inline void init_w(){for(int i=1;i<=C;i++)w[i]=new plx[1<<(i-1)];for(int i=0,l=(1<<(C-1));i<l;i++)w[C][i]=plx(cos(pi*i/l),sin(pi*i/l));for(int i=C-1;i;i--)for(int j=0,l=1<<(i-1);j<l;j++)w[i][j]=w[i+1][j<<1];
}
inline void fft(plx *f,int lim,int kd){for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);plx a0,a1;for(int mid=1,l=1;mid<lim;mid<<=1,l++)for(int i=0;i<lim;i+=mid<<1)for(int j=0;j<mid;j++)a0=f[i+j],a1=f[i+j+mid]*w[l][j],f[i+j]=a0+a1,f[i+j+mid]=a0-a1;if(kd==-1){reverse(f+1,f+lim);for(int i=0;i<lim;i++)f[i]/=lim;}
}
inline void mul(int *A,int *B,int *ret){static plx a[L],b[L],c[L],d[L],da,db,dc,dd;for(int i=0;i<=k;i++)a[i]=plx(A[i]&M,A[i]>>15),b[i]=plx(B[i]&M,B[i]>>15);for(int i=k+1;i<=lim;i++)a[i]=b[i]=plx();fft(a,lim,1),fft(b,lim,1);for(int i=0;i<lim;i++){int j=(lim-i)&(lim-1);da=(a[i]+a[j].conj())*plx(0.5,0);db=(a[j].conj()-a[i])*plx(0,0.5);dc=(b[i]+b[j].conj())*plx(0.5,0);dd=(b[j].conj()-b[i])*plx(0,0.5);c[i]=(da*dc)+((da*dd)*plx(0,1));d[i]=(db*dd)+((db*dc)*plx(0,1));}fft(c,lim,-1),fft(d,lim,-1);for(int i=0;i<=k;i++){ll da=(ll)(d[i].x+0.5)%mod,db=(ll)(d[i].y+0.5)%mod,dc=(ll)(c[i].y+0.5)%mod,dd=(ll)(c[i].x+0.5)%mod;ret[i]=((da<<30)+((db+dc)<<15)+dd)%mod;}
}
int tmp[N];
inline void init(int *a,int ii){for(int i=0;i<=k;i++)tmp[i]=mul(a[i],ksm(2,1ll*i*ii%(mod-1)));
}
int a[N],res[N];
int main(){#ifdef Stargazerfreopen("lx.in","r",stdin);#endifcin>>n>>k;if(n>k)return puts("0"),0;init_w(),init_inv();for(int i=1;i<=k;i++)res[i]=a[i]=ifac[i];lim=1,n--;while(lim<(k+1)*2)lim<<=1;init_rev(lim);for(int i=1;n;init(a,i),mul(tmp,a,a),i<<=1,n>>=1)if(n&1){init(res,i),mul(tmp,a,res);}for(int i=0;i<=k;i++)Mul(res[i],fac[i]);int ret=0;for(int i=0;i<=k;i++)Add(ret,mul(c(k,i),res[i]));cout<<ret;
}

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