題解 P4491 [HAOI2018]染色
阿新 • • 發佈:2020-12-30
題意
長度為\(N\)的序列, 每個位置都可以被染成\(M\)種顏色中的某一種.
如果恰 好出現了\(S\)次的顏色有\(K\)種, 會產生\(W_k\)的貢獻.
對於所有可能的染色方案, 他能獲得的愉悅度的和對 \(1004535809\)取模的結果是多少.
題解
最多有\(lim=\min(m,\frac{n}{s})\)種顏色。
記最少選\(i\)種的答案為\(F[i]\),有:
\[F[i]=\tbinom{m}{i}\frac{n!}{(s!)^i(n-is)!}(m-i)^{n-is} \]容斥一下得到答案為:
\[ans[i]=\sum_{j=i}^{lim}(-1)^{j-i}\tbinom{j}{i}F[j] \]拆開來:
\[ans[i]=\sum_{j=i}^{lim}(-1)^{j-i}\frac{j!}{i!(j-i)!}F[j] \]\[ans[i]\times i!=\sum_{j=i}^{lim}\frac{(-1)^{j-i}}{(j-i)!}\times F[j]j! \]記\(g[i]=\frac{(-1)^i}{i!},f[i]=i!\times F[i]\)
\[ans[i]\times i!=\sum_{j=i}^{lim}g[j-i]\times f[j] \]翻轉\(f\),得到:
\[ans[i]\times i!=\sum_{j=i}^{lim}g[j-i]\times f[lim-j] \]卷積顯而易見。
程式碼
#include<bits/stdc++.h> namespace in{ char buf[1<<21],*p1=buf,*p2=buf; inline int getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;} template <typename T>inline void read(T& t){ t=0;int f=0;char ch=getc();while (!isdigit(ch)){if(ch=='-')f = 1;ch=getc();} while(isdigit(ch)){t=t*10+(ch-48);ch = getc();}if(f)t=-t; } template <typename T,typename... Args> inline void read(T& t, Args&... args){read(t);read(args...);} } namespace out{ char buffer[1<<21];int p1=-1;const int p2 = (1<<21)-1; inline void flush(){fwrite(buffer,1,p1+1,stdout),p1=-1;} inline void putc(const char &x) {if(p1==p2)flush();buffer[++p1]=x;} template <typename T>void write(T x) { static char buf[15];static int len=-1;if(x>=0){do{buf[++len]=x%10+48,x/=10;}while (x);}else{putc('-');do {buf[++len]=-(x%10)+48,x/=10;}while(x);} while (len>=0)putc(buf[len]),--len; } } using namespace std; template<const int mod> struct modint{ int x; modint<mod>(int o=0){x=o;} modint<mod> &operator = (int o){return x=o,*this;} modint<mod> &operator +=(modint<mod> o){return x=x+o.x>=mod?x+o.x-mod:x+o.x,*this;} modint<mod> &operator -=(modint<mod> o){return x=x-o.x<0?x-o.x+mod:x-o.x,*this;} modint<mod> &operator *=(modint<mod> o){return x=1ll*x*o.x%mod,*this;} modint<mod> &operator ^=(int b){ modint<mod> a=*this,c=1; for(;b;b>>=1,a*=a)if(b&1)c*=a; return x=c.x,*this; } modint<mod> &operator /=(modint<mod> o){return *this *=o^=mod-2;} modint<mod> &operator +=(int o){return x=x+o>=mod?x+o-mod:x+o,*this;} modint<mod> &operator -=(int o){return x=x-o<0?x-o+mod:x-o,*this;} modint<mod> &operator *=(int o){return x=1ll*x*o%mod,*this;} modint<mod> &operator /=(int o){return *this *= ((modint<mod>(o))^=mod-2);} template<class I>friend modint<mod> operator +(modint<mod> a,I b){return a+=b;} template<class I>friend modint<mod> operator -(modint<mod> a,I b){return a-=b;} template<class I>friend modint<mod> operator *(modint<mod> a,I b){return a*=b;} template<class I>friend modint<mod> operator /(modint<mod> a,I b){return a/=b;} friend modint<mod> operator ^(modint<mod> a,int b){return a^=b;} friend bool operator ==(modint<mod> a,int b){return a.x==b;} friend bool operator !=(modint<mod> a,int b){return a.x!=b;} bool operator ! () {return !x;} modint<mod> operator - () {return x?mod-x:0;} modint<mod> &operator++(int){return *this+=1;} }; const int N=1e7+5; const int mod=1004535809; const modint<mod> GG=3,Ginv=modint<mod>(1)/3; struct poly{ vector<modint<mod>>a; modint<mod>&operator[](int i){return a[i];} int size(){return a.size();} void resize(int n){a.resize(n);} void reverse(){std::reverse(a.begin(),a.end());} }; int rev[N]; inline int ext(int n){int k=0;while((1<<k)<n)k++;return k;} inline void init(int k){int n=1<<k;for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(k-1));} inline void ntt(poly&a,int k,int typ){ int n=1<<k; for(int i=0;i<n;i++)if(i<rev[i])swap(a[i],a[rev[i]]); for(int mid=1;mid<n;mid<<=1){ modint<mod> wn=(typ>0?GG:Ginv)^((mod-1)/(mid<<1)); for(int r=mid<<1,j=0;j<n;j+=r){ modint<mod> w=1; for(int k=0;k<mid;k++,w=w*wn){ modint<mod> x=a[j+k],y=w*a[j+k+mid]; a[j+k]=x+y,a[j+k+mid]=x-y; } } } if(typ<0){ modint<mod> inv=modint<mod>(1)/n; for(int i=0;i<n-1;i++)a[i]*=inv; } } inline poly one(){poly a;a.a.push_back(1);return a;} poly operator +(poly a,poly b){ int n=max(a.size(),b.size());a.resize(n),b.resize(n); for(int i=0;i<n;i++)a[i]+=b[i];return a; } poly operator -(poly a,poly b){ int n=max(a.size(),b.size());a.resize(n),b.resize(n); for(int i=0;i<n;i++)a[i]-=b[i];return a; } inline poly operator*(poly a,poly b){ int n=a.size()+b.size()-1,k=ext(n); a.resize(1<<k),b.resize(1<<k),init(k); ntt(a,k,1);ntt(b,k,1);for(int i=0;i<(1<<k);i++)a[i]*=b[i]; ntt(a,k,-1),a.resize(n);return a; } int n,m,s;int w[N]; modint<mod>fac[N]; poly ans,f,g; modint<mod>C(int n,int m){if(n<m)return 0;return (fac[n]/fac[n-m])/fac[m];} modint<mod>F(int i){return C(m,i)*(modint<mod>(m-i)^(n-i*s))*fac[n]/(((fac[s]^i)*fac[n-i*s]));} signed main(){ freopen("color15.in","r",stdin); fac[0]=1;for(int i=1;i<N;i++)fac[i]=fac[i-1]*i; in::read(n,m,s);for(int i=0;i<=m;i++)in::read(w[i]); int lim=min(m,n/s);f.resize(lim+1);g.resize(lim+1); for(int i=0;i<=lim;i++)f[i]=fac[i]*F(i);f.reverse(); //for(int i=0;i<=f.size();i++)cout<<f[i].x<<" ";cout<<endl; for(int i=0;i<=lim;i++)if(i&1)g[i]=-modint<mod>(1)/fac[i];else g[i]=modint<mod>(1)/fac[i]; ans=f*g;ans.resize(lim+1);ans.reverse(); //for(int i=0;i<ans.size();i++)cout<<ans[i].x<<" "; modint<mod>Ans=0;for(int i=0;i<=lim;i++)Ans+=w[i]*ans[i]/fac[i]; out::write(Ans.x); out::flush(); return 0; }