Team Work(CF 932 E)[bzoj5093][Lydsy1711月赛]图的价值

0 前言

这是一道小清新题

1 题目相关

1.1传送门

传送门

1.2 题目大意

给出n,kn,kn,k，求∑i=0n(ni)ik\sum_{i=0}^n\binom{n}{i}i^ki=0∑n(in)ik
结果模109+710^9+7109+7

1.3 数据范围

n≤109,k≤5000n\le10^9,k\le5000n≤109,k≤5000

2 算法

2.1 暴力

由于是cf题，所以不知道有什么好的部分分，大概暴力就是直接Θ(n)\Theta(n)Θ(n)处理组合数，递推iii的次幂

2.2 正解

这道题直接算似乎难以优化复杂度，毕竟nnn太大了，我们考虑推式子
我们有个经典的和第二类斯特林数相关的结论
xn=∑i=0n{ni}xi↓x^n=\sum_{i=0}^n\begin{Bmatrix}n\\i\end{Bmatrix}x^{i\downarrow}xn=i=0∑n{ni}xi↓证明
我们将其代入
∑i=0n(ni)ik=∑i=0n(ni)∑j=0k{kj}ij↓=∑i=0n∑j=0k(ni){kj}ij↓=∑i=0n∑j=0k{kj}(ni)ij↓=∑i=0n∑j=0k{kj}n!i!(n−i)!⋅i!(i−j)!=∑i=0n∑j=0k{kj}(n−j)!(n−i)!(i−j)!⋅n!(n−j)!=∑i=0n∑j=0k{kj}(n−jn−i)nj↓=∑j=0k{kj}nj↓∑i=0n(n−jn−i)=∑j=0k{kj}nj↓2n−j\begin{aligned} \sum_{i=0}^n\binom{n}{i}i^k&=\sum_{i=0}^n\binom{n}{i}\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}i^{j\downarrow}\\&=\sum_{i=0}^n\sum_{j=0}^k\binom{n}{i}\begin{Bmatrix}k\\j\end{Bmatrix}i^{j\downarrow}\\&=\sum_{i=0}^n\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}\binom{n}{i}i^{j\downarrow}\\ &=\sum_{i=0}^n\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}\frac{n!}{i!(n-i)!}·\frac{i!}{(i-j)!}\\ &=\sum_{i=0}^n\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}\frac{(n-j)!}{(n-i)!(i-j)!}·\frac{n!}{(n-j)!}\\ &=\sum_{i=0}^n\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}\binom{n-j}{n-i}n^{j\downarrow}\\ &=\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}n^{j\downarrow}\sum_{i=0}^n\binom{n-j}{n-i}\\ &=\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}n^{j\downarrow}2^{n-j}\\ \end{aligned}i=0∑n(in)ik=i=0∑n(in)j=0∑k{kj}ij↓=i=0∑nj=0∑k(in){kj}ij↓=i=0∑nj=0∑k{kj}(in)ij↓=i=0∑nj=0∑k{kj}i!(n−i)!n!⋅(i−j)!i!=i=0∑nj=0∑k{kj}(n−i)!(i−j)!(n−j)!⋅(n−j)!n!=i=0∑nj=0∑k{kj}(n−in−j)nj↓=j=0∑k{kj}nj↓i=0∑n(n−in−j)=j=0∑k{kj}nj↓2n−j
前面的推导中我们用到了二项式定理
我们发现，化到最后，下降幂和2的幂次都是可以Θ(k)\Theta(k)Θ(k)求的，我们现在要求的就是斯特林数了，我们显然会Θ(k2)\Theta(k^2)Θ(k2)的递推求斯特林数，那么就做完啦

代码

到这里这个算法已经可以通过此题了，贴出AC代码

#include<cstdio>
#include<cctype>
#define rg register
typedef long long LL;
template <typename T> inline T max(const T a,const T b){return a>b?a:b;}
template <typename T> inline T min(const T a,const T b){return a<b?a:b;}
template <typename T> inline void mind(T&a,const T b){a=a<b?a:b;}
template <typename T> inline void maxd(T&a,const T b){a=a>b?a:b;}
template <typename T> inline T abs(const T a){return a>0?a:-a;}
template <typename T> inline void swap(T&a,T&b){T c=a;a=b;b=c;}
template <typename T> inline T gcd(const T a,const T b){if(!b)return a;return gcd(b,a%b);}
template <typename T> inline T lcm(const T a,const T b){return a/gcd(a,b)*b;}
template <typename T> inline T square(const T x){return x*x;};
template <typename T> inline void read(T&x)
{char cu=getchar();x=0;bool fla=0;while(!isdigit(cu)){if(cu=='-')fla=1;cu=getchar();}while(isdigit(cu))x=x*10+cu-'0',cu=getchar();if(fla)x=-x;
}
template <typename T> inline void printe(const T x)
{if(x>=10)printe(x/10);putchar(x%10+'0');
}
template <typename T> inline void print(const T x)
{if(x<0)putchar('-'),printe(-x);else printe(x);
}
const LL mod=1000000007;
int n,k,s[5001][5001],bit,ny=500000004,xj=1,ans;
inline LL pow(LL x,LL y)
{LL res=1;for(;y;y>>=1,x=x*x%mod)if(y&1)res=res*x%mod;return res;
}
int main()
{read(n),read(k);s[0][0]=1;for(rg int i=1;i<=k;i++)for(rg int j=1;j<=i;j++)s[i][j]=((LL)j*s[i-1][j]+s[i-1][j-1])%mod;bit=pow(2ll,n);for(rg int i=1;i<=k;i++)xj=(LL)xj*(n-i+1)%mod,bit=(LL)bit*ny%mod,ans=((LL)s[k][i]*xj%mod*bit+ans)%mod;print(ans);return 0;
}

3 优化

3.1 加强版数据范围

显而易见，这道题可以优化
瓶颈在求第二类斯特林数上，我们可以用NTTNTTNTT优化复杂度，使得算法复杂度为O(klogk)\mathcal O(klogk)O(klogk)
然而本题的优化需要非NTTNTTNTT模数NTTNTTNTT，所以会比较麻烦
update by 2019.3.5:
加推一题kkk的数据范围增强版
[bzoj5093][Lydsy1711月赛]图的价值
我们发现这题的每个点都是独立的，那么答案就是
n∗2(n−12)∗(∑i=0n−1(n−1i)∗ik)n*2^{\binom{n-1}{2}}*\left(\sum_{i=0}^{n-1}\binom{n-1}{i}*i^k \right)n∗2(2n−1)∗(i=0∑n−1(in−1)∗ik)
kkk的范围比较大，并且模数为998244353998244353998244353，所以直接NTT预处理固定nnn的第二类斯特林数即可
代码（多项式板子较长勿喷）

#include<cstdio>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
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))
typedef long long ll;
#define rg register
template <typename T> inline T max(const T a,const T b){return a>b?a:b;}
template <typename T> inline T min(const T a,const T b){return a<b?a:b;}
template <typename T> inline T mind(T&a,const T b){a=a<b?a:b;}
template <typename T> inline T maxd(T&a,const T b){a=a>b?a:b;}
template <typename T> inline T abs(const T a){return a>0?a:-a;}
template <typename T> inline void Swap(T&a,T&b){T c=a;a=b;b=c;}
template <typename T> inline void swap(T*a,T*b){T c=a;a=b;b=c;}
template <typename T> inline T gcd(const T a,const T b){if(!b)return a;return gcd(b,a%b);}
template <typename T> inline T square(const T x){return x*x;};
template <typename T> inline void read(T&x)
{char cu=getchar();x=0;bool fla=0;while(!isdigit(cu)){if(cu=='-')fla=1;cu=getchar();}while(isdigit(cu))x=x*10+cu-'0',cu=getchar();if(fla)x=-x;
}
template <typename T> void printe(const T x)
{if(x>=10)printe(x/10);putchar(x%10+'0');
}
template <typename T> inline void print(const T x)
{if(x<0)putchar('-'),printe(-x);else printe(x);
}
const int maxn=2097152,mod=998244353;
inline int Md(const int x){return x>=mod?x-mod:x;}
template<typename T>
inline int pow(int x,T y)
{rg int res=1;x%=mod;for(;y;y>>=1,x=(ll)x*x%mod)if(y&1)res=(ll)res*x%mod;return res;
}
namespace Poly///namespace of Poly
{int W_[maxn],ha[maxn],hb[maxn],Inv[maxn];
inline void init(const int x)
{rg int tim=0,lenth=1;while(lenth<x)lenth<<=1,tim++;for(rg int i=1;i<lenth;i<<=1){const int WW=pow(3,(mod-1)/(i*2));W_[i]=1;for(rg int j=i+1,k=i<<1;j<k;j++)W_[j]=(ll)W_[j-1]*WW%mod;}Inv[0]=Inv[1]=1;for(rg int i=2;i<x;i++)Inv[i]=(ll)(mod-mod/i)*Inv[mod%i]%mod;
}
int L;
inline void DFT(int*A)//prepare:init L
{for(rg int i=0,j=0;i<L;i++){if(i>j)Swap(A[i],A[j]);for(rg int k=L>>1;(j^=k)<k;k>>=1);}for(rg int i=1;i<L;i<<=1)for(rg int j=0,J=i<<1;j<L;j+=J)for(rg int k=0;k<i;k++){const int x=A[j+k],y=(ll)A[j+k+i]*W_[i+k]%mod;A[j+k]=Md(x+y),A[j+k+i]=Md(mod+x-y);}
}
void IDFT(int*A)
{for(rg int i=1;i<L-i;i++)Swap(A[i],A[L-i]);DFT(A);
}
inline int Quadratic_residue(const int a)
{if(a==0)return 0;int b=(rand()<<14^rand())%mod;while(pow(b,(mod-1)>>1)!=mod-1)b=(rand()<<14^rand())%mod;int s=mod-1,t=0,x,inv=pow(a,mod-2),f=1;while(!(s&1))s>>=1,t++,f<<=1;t--,x=pow(a,(s+1)>>1),f>>=1;while(t){f>>=1;if(pow((int)((ll)inv*x%mod*x%mod),f)!=1)x=(ll)x*pow(b,s)%mod;t--,s<<=1;}return min(x,mod-x);
}
struct poly
{std::vector<int>A;poly(){A.resize(0);}poly(const int x){A.resize(1),A[0]=x;}inline int&operator[](const int x){return A[x];}inline int operator[](const int x)const{return A[x];}inline void clear(){A.clear();}inline unsigned int size()const{return A.size();}inline void resize(const unsigned int x){A.resize(x);}void RE(const int x){A.resize(x);for(rg int i=0;i<x;i++)A[i]=0; }void readin(const int MAX){A.resize(MAX);for(rg int i=0;i<MAX;i++)read(A[i]);}void putout()const{for(rg unsigned int i=0;i<A.size();i++)print(A[i]),putchar(' ');}inline poly operator +(const poly b)const{poly RES;RES.resize(max(size(),b.size()));for(rg unsigned int i=0;i<RES.size();i++)RES[i]=Md((i<size()?A[i]:0)+(i<b.size()?b[i]:0));return RES;}inline poly operator -(const poly b)const{poly RES;RES.resize(max(size(),b.size()));for(rg unsigned int i=0;i<RES.size();i++)RES[i]=Md((i<size()?A[i]:0)+mod-(i<b.size()?b[i]:0));return RES;}inline poly operator *(const int b)const{poly RES=*this;for(rg unsigned int i=0;i<RES.size();i++)RES[i]=(ll)RES[i]*b%mod;return RES;}inline poly operator /(const int b)const{poly RES=(*this)*pow(b,mod-2);return RES;}inline poly operator *(const poly b)const{const int RES=A.size()+b.size()-1;if(A.size()<=20||b.size()<=20){poly c;c.RE(RES);for(rg unsigned int i=0;i<size();i++)for(rg unsigned int j=0;j<b.size();j++)c[i+j]=((ll)A[i]*b[j]+c[i+j])%mod;return c;}L=1;while(L<RES)L<<=1;poly c;c.A.resize(RES);memset(ha,0,L<<2);memset(hb,0,L<<2);for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];for(rg unsigned int i=0;i<b.A.size();i++)hb[i]=b.A[i];DFT(ha),DFT(hb);for(rg int i=0;i<L;i++)ha[i]=(ll)ha[i]*hb[i]%mod;IDFT(ha);const int inv=pow(L,mod-2);for(rg int i=0;i<RES;i++)c.A[i]=(ll)ha[i]*inv%mod;return c;}inline poly inv()const{poly C;if(A.size()==1){C=*this;C[0]=pow(C[0],mod-2);return C;}C.resize((A.size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.inv();L=1;while(L<(int)size()*2-1)L<<=1;for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];for(rg unsigned int i=0;i<C.size();i++)hb[i]=C[i];memset(ha+A.size(),0,(L-A.size())<<2);memset(hb+C.size(),0,(L-C.size())<<2);DFT(ha),DFT(hb);for(rg int i=0;i<L;i++)ha[i]=(ll)(2-(ll)hb[i]*ha[i]%mod+mod)*hb[i]%mod;IDFT(ha);const int inv=pow(L,mod-2);C.resize(size());for(rg unsigned int i=0;i<size();i++)C[i]=(ll)ha[i]*inv%mod;return C;}
/*    inline poly inv()const{poly C;if(A.size()==1){C=*this;C[0]=pow(C[0],mod-2);return C;}C.resize((A.size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.inv();poly D=(poly)2-*this*C;D.resize(size());D=D*C;D.resize(size());return D;}*///大常数版本 inline void Reverse(const int n){A.resize(n);for(rg int i=0,j=n-1;i<j;i++,j--)Swap(A[i],A[j]);}inline poly operator /(const poly B)const{if(size()<B.size())return 0;poly a=*this,b=B;a.Reverse(size()),b.Reverse(B.size());b.resize(size()-B.size()+1);b=b.inv();b=b*a;b.Reverse(size()-B.size()+1);return b;}inline poly operator %(const poly B)const{poly C=(*this)-(*this)/B*B;C.resize(B.size()-1);return C;}inline poly diff()const{poly C;C.resize(size()-1);for(rg unsigned int i=1;i<size();i++)C[i-1]=(ll)A[i]*i%mod;return C;}inline poly inte()const{poly C;C.resize(size()+1);for(rg unsigned int i=0;i<size();i++)C[i+1]=(ll)A[i]*Inv[i+1]%mod;C[0]=0;return C;}inline poly ln()const{poly C=(diff()*inv()).inte();C.resize(size());return C;}inline poly exp()const{poly C;if(size()==1){C=*this;C[0]=1;return C;}C.resize((size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.exp();C.resize(size());poly D=(poly)1-C.ln()+*this;D=D*C;D.resize(size());return D;}inline poly sqrt()const{poly C;if(size()==1){C=*this;C[0]=Quadratic_residue(C[0]);return C;}C.resize((size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.sqrt();C.resize(size());C=(C+*this*C.inv())*(int)499122177;C.resize(size());return C;}inline poly operator >>(const unsigned int x)const{poly C;if(size()<x){C.resize(0);return C;}C.resize(size()-x);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i+x];return C;}inline poly operator <<(const unsigned int x)const{poly C;C.RE(size()+x);for(rg unsigned int i=0;i<size();i++)C[i+x]=A[i];return C;}inline poly Pow(const unsigned int x)const{for(rg unsigned int i=0;i<size();i++)if(A[i]){poly C=((((*this/A[i])>>i).ln()*x).exp()*pow(A[i],x))<<(min(i*x,size()));C.resize(size());return C;}return *this;}inline void cheng(const poly&B){for(rg unsigned int i=0;i<size();i++)A[i]=(ll)A[i]*B[i]%mod; }inline void jia(const poly&B){for(rg unsigned int i=0;i<size();i++)A[i]=Md(A[i]+B[i]); }inline void dft(){memset(ha,0,L<<2);for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];DFT(ha);resize(L);for(rg int i=0;i<L;i++)A[i]=ha[i];}inline void idft(){memset(ha,0,L<<2);for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];IDFT(ha);const int inv=pow(L,mod-2);for(rg int i=0;i<L;i++)A[i]=(ll)ha[i]*inv%mod;while(size()&&!A[size()-1])A.pop_back();}ll dai(int x)const{x=Md(x%mod+mod);ll res=0;for(rg unsigned int i=0,j=1;i<size();i++,j=(ll)j*x%mod)res=((ll)A[i]*j+res)%mod;return res;}
};
void fz(const int root,const int l,const int r,std::vector<int>&v,std::vector<poly>&A)
{if(l==r){A[root].resize(2);A[root][0]=(mod-v[l])%mod;A[root][1]=1;return;}const int mid=(l+r)>>1;fz(root<<1,l,mid,v,A),fz(root<<1|1,mid+1,r,v,A);A[root]=A[root<<1]*A[root<<1|1];
}
void calc(const int root,const int l,const int r,std::vector<int>&v,std::vector<poly>&A,std::vector<poly>&B)
{if(l==r){v[l]=B[root][0];return;}const int mid=(l+r)>>1;B[root<<1]=B[root]%A[root<<1];B[root<<1|1]=B[root]%A[root<<1|1];calc(root<<1,l,mid,v,A,B),calc(root<<1|1,mid+1,r,v,A,B);
}
void multi_point_evaluation(const poly a,std::vector<int>&v)
{std::vector<poly>A,B;A.resize(maxn),B.resize(maxn);fz(1,0,v.size()-1,v,A);B[1]=a%A[1];calc(1,0,v.size()-1,v,A,B);
}
void fz2(const int root,const int l,const int r,std::vector<int>&y,std::vector<poly>&A,std::vector<poly>&B)
{if(l==r){B[root].resize(1),B[root][0]=y[l];return;}const int mid=(l+r)>>1;fz2(root<<1,l,mid,y,A,B),fz2(root<<1|1,mid+1,r,y,A,B);B[root]=B[root<<1]*A[root<<1|1]+B[root<<1|1]*A[root<<1];
}
poly interpolation(std::vector<int>&x,std::vector<int>&y)
{std::vector<poly>A,B;A.resize(maxn),B.resize(maxn);fz(1,0,x.size()-1,x,A);multi_point_evaluation(A[1].diff(),x);for(rg unsigned int i=0;i<x.size();i++)y[i]=(ll)y[i]*pow(x[i],mod-2)%mod;fz2(1,0,x.size()-1,y,A,B);return B[1];
}
}///namespace of Poly
int n,k,bit,ny=499122177,xj=1,ans;
int fac[500001],inv[500001];
Poly::poly a,b;
int main()
{Poly::init(maxn);///namespace of Polyread(n),read(k),n--;fac[0]=1;for(rg int i=1;i<=k;i++)fac[i]=(ll)fac[i-1]*i%mod;inv[k]=pow(fac[k],mod-2);for(rg int i=k;i>=1;i--)inv[i-1]=(ll)inv[i]*i%mod;a.resize(k+1),b.resize(k+1);for(rg int i=0;i<=k;i++){a[i]=inv[i];if(i&1)a[i]=Md(mod-inv[i]);b[i]=(ll)pow(i,k)*inv[i]%mod;}a=a*b;bit=pow(2ll,n);for(rg int i=1;i<=k;i++)xj=(ll)xj*(n-i+1)%mod,bit=(ll)bit*ny%mod,ans=((ll)a[i]*xj%mod*bit+ans)%mod;print((ll)ans*(n+1)%mod*pow(2,(ll)n*(n-1)/2)%mod);return flush(),0;
}

4 总结

不错的裸题，直接带入就好了，貌似很多大佬都可以直接秒，算是一个斯特林数的应用吧