清真多項式題

BZOJ #3625

codeforces #438E

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題意

每個點的權值可以在集合$?S$中任取

求點權和恰好為$[1..n]$的不同的二叉樹數量

數據范圍全是$ 10^5$

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$ Solution $?

設集合$?S$的生成函數為$ C$

考慮暴力$ DP$,枚舉當前根節點的權值以及左右子樹的權值

有

$f[0]=1$

$ f[x]=\sum\limits_{i=0}^x \sum\limits_{j=0}^{x-i}f[i]f[j]·[C[x-i-j]=1]$

化成生成函數形式即為

$ f=C·f^2+1$

其中$+1$是因為$ C$中不含常數項而多項式$ f$中含有值為$ 1$的常數項

二次方程求解得

$f=\frac{1 \pm \sqrt{1-4C}}{2C}$

因為分母沒有常數項，分子顯然也不應有常數項

即

$f=\frac{1 - \sqrt{1-4C}}{2C}$

分母沒有常數項不能求逆，則分子分母同乘$ 1+ \sqrt{1-4c}$得

$ f=\frac{2}{1+\sqrt{1-4C}}$

直接上多項式開根，求逆的板子即可

時間復雜度$ O(n \ log \ n)$

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$ my \ code $

這是一份自帶巨大常數的板子

單$ log 10w$需要跑$ 1s+$

#include<ctime> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<vector> #define rt register int #define ll long long using namespace std; namespace fast_IO{const int IN_LEN=10000000,OUT_LEN=10000000;char ibuf[IN_LEN],obuf[OUT_LEN],*ih=ibuf+IN_LEN,*oh=obuf,*lastin=ibuf+IN_LEN,*lastout=obuf+OUT_LEN-1;inline char getchar_(){return (ih==lastin)&&(lastin=(ih=ibuf)+fread(ibuf,1,IN_LEN,stdin),ih==lastin)?EOF:*ih++;}inline void putchar_(const char x){if(oh==lastout)fwrite(obuf,1,oh-obuf,stdout),oh=obuf;*oh++=x;}inline void flush(){fwrite(obuf,1,oh-obuf,stdout);} } using namespace fast_IO; //#define getchar() getchar_() //#define putchar(x) putchar_((x)) inline ll read(){ll x=0;char zf=1;char ch=getchar();while(ch!='-'&&!isdigit(ch))ch=getchar();if(ch=='-')zf=-1,ch=getchar();while(isdigit(ch))x=x*10+ch-'0',ch=getchar();return x*zf; } void write(ll y){if(y<0)putchar('-'),y=-y;if(y>9)write(y/10);putchar(y%10+48);} void writeln(const ll y){write(y);putchar('\n');} int k,m,n,x,y,z,cnt,ans;namespace poly{#define p 998244353vector<int>R;vector<int>get(int n){vector<int>ret(n);for(rt i=0;i<n;i++)ret[i]=read();return ret;}void print(const vector<int>A){for(rt i=0;i<A.size();i++)write((A[i]+p)%p),putchar(' ');}int ksm(int x,int y=p-2){int ans=1;for(rt i=y;i;i>>=1,x=1ll*x*x%p)if(i&1)ans=1ll*ans*x%p;return ans;}void NTT(int n,vector<int>&A,int fla){A.resize(n);for(rt i=0;i<n;i++)if(i>R[i])swap(A[i],A[R[i]]);for(rt i=1;i<n;i<<=1){int w=ksm(3,(p-1)/2/i);for(rt j=0;j<n;j+=i<<1){int K=1;for(rt k=0;k<i;k++,K=1ll*K*w%p){int x=A[j+k],y=1ll*K*A[i+j+k]%p;A[j+k]=(x+y)%p,A[i+j+k]=(x-y)%p;}}}if(fla==-1){reverse(A.begin()+1,A.end());int invn=ksm(n);for(rt i=0;i<n;i++)A[i]=1ll*A[i]*invn%p;}}vector<int>Resize(int n,vector<int>A){A.resize(n);return A;}vector<int>Mul(vector<int>x,vector<int>y){int lim=1,sz=x.size()+y.size()-1;while(lim<=sz)lim<<=1;R.resize(lim);for(rt i=0;i<lim;i++)R[i]=(R[i>>1]>>1)|(i&1)*(lim>>1);NTT(lim,x,1);NTT(lim,y,1);for(rt i=0;i<lim;i++)x[i]=1ll*x[i]*y[i]%p;NTT(lim,x,-1);x.resize(sz);return x;}vector<int>Inv(vector<int>A,int n=-1){if(n==-1)n=A.size();if(n==1)return vector<int>(1,ksm(A[0]));vector<int>b=Inv(A,(n+1)/2);int lim=1;while(lim<=n+n)lim<<=1;R.resize(lim);for(rt i=0;i<lim;i++)R[i]=(R[i>>1]>>1)|(i&1)*(lim>>1);A.resize(n);NTT(lim,A,1);NTT(lim,b,1);for(rt i=0;i<lim;i++)A[i]=1ll*b[i]*(2ll-1ll*A[i]*b[i]%p)%p;NTT(lim,A,-1);A.resize(n);return A;}vector<int>Div(vector<int>A,vector<int>B){int n=A.size(),m=B.size();reverse(A.begin(),A.end());reverse(B.begin(),B.end());A.resize(n-m+1),B.resize(n-m+1);int lim=1;while(lim<=2*(n-m+1))lim<<=1;R.resize(lim);for(rt i=0;i<lim;i++)R[i]=(R[i>>1]>>1)|(i&1)*(lim>>1);vector<int>ans=Resize(n-m+1,Mul(A,Inv(B)));reverse(ans.begin(),ans.end());return ans;}vector<int>Add(vector<int>A,vector<int>B){int len=max(A.size(),B.size());A.resize(len);for(rt i=0;i<len;i++)(A[i]+=B[i])%=p;return A;}vector<int>Sub(vector<int>A,vector<int>B){int len=max(A.size(),B.size());A.resize(len);for(rt i=0;i<len;i++)(A[i]-=B[i])%=p;return A;}vector<int>Mul(int x,vector<int>A){for(rt i=0;i<A.size();i++)A[i]=1ll*A[i]*x%p;return A;}vector<int>deriv(vector<int>A){//求導 for(rt i=1;i<A.size();i++)(A[i-1]=1ll*A[i]*i%p);A.pop_back();return A;}vector<int>integ(vector<int>A){//積分 A.push_back(0);for(rt i=A.size()-2;i>=0;i--)A[i+1]=1ll*A[i]*ksm(i+1)%p;A[0]=0;return A;}vector<int>Ln(const vector<int>A){return integ(Resize(A.size()-1,Mul(deriv(A),Inv(A))));}vector<int>Exp(vector<int>A,int n=-1){if(n==-1)n=A.size();if(n==1)return vector<int>(1,1);vector<int>A0=Resize(n,Exp(A,(n+1)>>1));vector<int>now=Resize(n,Ln(A0));for(rt i=0;i<n;i++)now[i]=(A[i]-now[i])%p;now[0]++;return Resize(n,Mul(A0,now));}struct cp{ll a,b,z;//a+bsqrt(z)cp operator *(const cp s)const{return {(1ll*a*s.a%p+1ll*b*s.b%p*z%p)%p,(1ll*a*s.b%p+1ll*b*s.a)%p,z};}};cp ksm(cp x,int y){cp ans={1,0,x.z};for(rt i=y;i;i>>=1,x=x*x)if(i&1){ans=x*ans;}return ans;}int Sqrt(int n){//求二次剩馀 if(ksm(n,(p-1)/2)!=1)return -1;while(1){x=rand()%p;if(ksm((1ll*x*x%p-n%p+p)%p,(p-1)/2)==1)continue;cp ret=ksm({x,1,(1ll*x*x%p+p-n)%p},(p+1)/2); return min(ret.a,p-ret.a);}}vector<int>GetSqrt(vector<int>A,int n=-1){if(n==-1)n=A.size();if(n==1)return vector<int>(1,Sqrt(A[0]));vector<int>ans=Resize(n,GetSqrt(A,n+1>>1)),C(A.begin(),A.begin()+n);return Resize(n,Mul(ksm(2),Add(ans,Mul(Inv(ans),C))));}vector<int>Pow(vector<int>A,int k){A[0]=1;return Exp(Mul(k,Ln(A)));}//#undef p }; using namespace poly; int inv[100010],v[100010]; int main(){m=read();n=read();vector<int>c(n+1);for(rt i=1;i<=m;i++){x=read();if(x<=n)c[x]++;}vector<int>ans=Mul(-4,c);ans[0]++;ans=GetSqrt(ans);ans[0]++;ans=Mul(2,Inv(ans));for(rt i=1;i<=n;i++)writeln((ans[i]+p)%p);return flush(),0; }

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轉載于:https://www.cnblogs.com/DreamlessDreams/p/10177306.html

超強干貨來襲 云風專訪：近40年碼齡，通宵達旦的技術人生

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