Codeforces Round #250 (Div. 1)E. The Child and Binary Tree
阿新 • • 發佈:2018-08-23
tor base sig target pri inline class long long inf ,\((B(x)+B'(x))*(B(x)-B'(x))\equiv 0\),取\(B(x)=B'(x)\)
\(B(x)^2-2*B(x)*B'(x)+B'(x)^2\equiv0\),
\(B(x)\equiv \frac{A(x)+B'(x)^2}{2*B'(x)}\)
題意:有一個集合,求有多少形態不同的二叉樹滿足每個點的權值都屬於這個集合並且總點權等於i
題解:先用生成函數搞出來\(f(x)=f(x)^2*c(x)+1\)
然後轉化一下變成\(f(x)=\frac{2}{1+\sqrt{1-4*c(x)}}\)
然後多項式開根和多項式求逆即可(先對下面的項開根,然後再求逆)
多項式開根:
\(B(x)^2=A(x)(mod x^{\lfoor \frac{n}{2} \rfloor)\)
\(B'(x)^2=A(x)(mod x^{\lfoor \frac{n}{2} \rfloor)\)
\(B(x)^2-B'(x)^2\equiv 0\)
\(B(x)^2-2*B(x)*B'(x)+B'(x)^2\equiv0\),
\(B(x)\equiv \frac{A(x)+B'(x)^2}{2*B'(x)}\)
//#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#pragma GCC optimize("unroll-loops") //#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #include<bits/stdc++.h> #define fi first #define se second #define db double #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define vi vector<int> #define mod 998244353 #define ld long double #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pll pair<ll,ll> #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> //#define cd complex<double> #define ull unsigned long long #define base 1000000000000000000 #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define fin freopen("a.txt","r",stdin) #define fout freopen("a.txt","w",stdout) #define fio ios::sync_with_stdio(false);cin.tie(0) template<typename T> inline T const& MAX(T const &a,T const &b){return a>b?a:b;} template<typename T> inline T const& MIN(T const &a,T const &b){return a<b?a:b;} inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;} inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;} inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;} inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;} using namespace std; const double eps=1e-8; const ll INF=0x3f3f3f3f3f3f3f3f; const int N=100000+10,maxn=400000+10,inf=0x3f3f3f3f; ll a[N<<3],b[N<<3],c[N<<3],d[N<<3],tmp[N<<3],inv2=qp(2,mod-2); int rev[N<<3]; void getrev(int bit) { for(int i=0;i<(1<<bit);i++) rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1)); } void ntt(ll *a,int n,int dft) { for(int i=0;i<n;i++) if(i<rev[i]) swap(a[i],a[rev[i]]); for(int step=1;step<n;step<<=1) { ll wn=qp(3,(mod-1)/(step*2)); if(dft==-1)wn=qp(wn,mod-2); for(int j=0;j<n;j+=step<<1) { ll wnk=1; for(int k=j;k<j+step;k++) { ll x=a[k]; ll y=wnk*a[k+step]%mod; a[k]=(x+y)%mod;a[k+step]=(x-y+mod)%mod; wnk=wnk*wn%mod; } } } if(dft==-1) { ll inv=qp(n,mod-2); for(int i=0;i<n;i++)a[i]=a[i]*inv%mod; } } void pol_inv(int deg,ll *a,ll *b) { if(deg==1){b[0]=qp(a[0],mod-2);return ;} pol_inv((deg+1)>>1,a,b); int sz=0;while((1<<sz)<=deg)sz++; getrev(sz);int len=1<<sz; for(int i=0;i<deg;i++)tmp[i]=a[i]; for(int i=deg;i<len;i++)tmp[i]=0; ntt(tmp,len,1),ntt(b,len,1); for(int i=0;i<len;i++) b[i]=(2ll-tmp[i]*b[i]%mod+mod)%mod*b[i]%mod; ntt(b,len,-1); for(int i=deg;i<len;i++)b[i]=0; } void pol_sqrt(int deg,ll *a,ll *b) { if(deg==1){b[0]=1;return ;} pol_sqrt((deg+1)>>1,a,b); int sz=0;while((1<<sz)<=deg)sz++; getrev(sz);int len=1<<sz; for(int i=0;i<len;i++)d[i]=0; pol_inv(deg,b,d); for(int i=0;i<deg;i++)tmp[i]=a[i]; for(int i=deg;i<len;i++)tmp[i]=0; ntt(tmp,len,1),ntt(b,len,1),ntt(d,len,1); for(int i=0;i<len;i++) b[i]=(tmp[i]*d[i]%mod+b[i])%mod*inv2%mod; ntt(b,len,-1); for(int i=deg;i<len;i++)b[i]=0; } int main() { int n,m;scanf("%d%d",&n,&m); for(int i=0,x;i<n;i++) { scanf("%d",&x); a[x]=mod-4; } a[0]=1; int len=1;while(len<=m)len<<=1; pol_sqrt(len,a,b); ++b[0]; pol_inv(len,b,c); for(int i=1;i<=m;i++)printf("%lld\n",c[i]*2%mod); return 0; } /******************** ********************/
Codeforces Round #250 (Div. 1)E. The Child and Binary Tree