[luoguP4705]玩游戏
前言
好的思路题
题目相关
传送门
题意简介
给定长度为 n n n的序列 a a a和长度为 m m m的序列 b b b
对所有 k ∈ [ 1 , t ] k\in[1,t] k∈[1,t]，求在两个序列里分别随机取一个数的和的 k k k次方的期望，模 998244353 998244353 998244353
数据范围
1 ≤ n , m ≤ 1 0 5 , 1 ≤ t ≤ 1 0 5 1\le n,m\le10^5,1\le t\le 10^5 1≤n,m≤105,1≤t≤105
题解
化一下式子，求期望相当于求所有情况的和取平均值
∑ ( a + b ) k = ∑ ∑ i = 0 k ( k i ) a i b k − i = ∑ i = 0 k ( k i ) ( ∑ a i ) ( ∑ b k − i ) = ∑ i = 0 k ( k i ) ( ∑ a i ) ( ∑ b k − i ) = ∑ i = 0 k k ! i ! ( k − i ) ! ( ∑ a i ) ( ∑ b k − i ) = ∑ i = 0 k k ! ∑ a i i ! ∑ b k − i ( k − i ) ! = k ! ∑ i = 0 k ∑ a i i ! ∑ b k − i ( k − i ) ! \begin{aligned} \sum(a+b)^k&=\sum\sum_{i=0}^k\binom{k}{i}a^ib^{k-i}\\ &=\sum_{i=0}^k\binom{k}{i}(\sum a^i)(\sum b^{k-i})\\ &=\sum_{i=0}^k\binom{k}{i}(\sum a^i)(\sum b^{k-i})\\ &=\sum_{i=0}^k\frac{k!}{i!(k-i)!}(\sum a^i)(\sum b^{k-i})\\ &=\sum_{i=0}^kk!\frac{\sum a^i}{i!}\frac{\sum b^{k-i}}{(k-i)!}\\ &=k!\sum_{i=0}^k\frac{\sum a^i}{i!}\frac{\sum b^{k-i}}{(k-i)!}\\ \end{aligned} ∑(a+b)k=∑i=0∑k(ik)aibk−i=i=0∑k(ik)(∑ai)(∑bk−i)=i=0∑k(ik)(∑ai)(∑bk−i)=i=0∑ki!(k−i)!k!(∑ai)(∑bk−i)=i=0∑kk!i!∑ai(k−i)!∑bk−i=k!i=0∑ki!∑ai(k−i)!∑bk−i
我们如果求出所有 ∑ a i \sum a^i ∑ai和 ∑ b i \sum b^i ∑bi
考虑生成函数
F i ( x ) = ∑ j = 0 ∞ a i j x j F_i(x)=\sum_{j=0}^{\infty}a_i^jx^j Fi(x)=j=0∑∞aijxj
根据等比数列求和得到
F i ( x ) = 1 1 − a i x F_i(x)=\frac{1}{1-a_ix} Fi(x)=1−aix1
我们对所有生成函数求和
∑ i = 1 n F i ( x ) = ∑ i = 1 n 1 1 − a i x \sum_{i=1}^nF_i(x)=\sum_{i=1}^n\frac{1}{1-a_ix} i=1∑nFi(x)=i=1∑n1−aix1
然后分治求分子、分母，用多项式求逆将分母乘进分子里
上面的式子算完后别忘了由于是期望所以要除以n*m
算法总复杂度 O ( n l o g 2 n ) \mathcal O(nlog^2n) O(nlog2n)
代码
这份代码优化了DFT与IDFT的次数，速度大大提高
#include
}
int L;
inline void NTT(int*A,const int fla)//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)
{
const int w=fla==-1?FW_[i]:W_[i];
for(rg int j=0,J=i<<1;j<L;j+=J)
{
int K=1;
for(rg int k=0;k<i;k++,K=(ll)K*w%mod)
{
const int x=A[j+k],y=(ll)A[j+k+i]*K%mod;
A[j+k]=Md(x+y),A[j+k+i]=Md(mod+x-y);
}
}
}
}
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;
L=1;while(L<RES)L<<=1;
poly c;c.A.resize(RES);
memset(ha,0,sizeof(int)*L);
memset(hb,0,sizeof(int)*L);
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];
NTT(ha,1),NTT(hb,1);
for(rg int i=0;i<L;i++)ha[i]=(ll)ha[i]*hb[i]%mod;
NTT(ha,-1);
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,sizeof(int)*(L-A.size()));
memset(hb+C.size(),0,sizeof(int)*(L-C.size()));
NTT(ha,1),NTT(hb,1);
for(rg int i=0;i<L;i++)ha[i]=(ll)(2-(ll)hb[i]*ha[i]%mod+mod)*hb[i]%mod;
NTT(ha,-1);
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//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
{
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,sizeof(int)*L);
for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];
NTT(ha,1);
resize(L);
for(rg int i=0;i<L;i++)A[i]=ha[i];
}
inline void IDFT()
{
memset(ha,0,sizeof(int)*L);
for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];
NTT(ha,-1);
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();
}
};
}///////namespace of Poly
Poly::poly a[maxn],za[maxn],b[maxn],zb[maxn],A,B;
int n,m,t;
void fz(Poly::poly*mu,Poly::poly*zi,const int l,const int r)
{
if(l==r)return;
const int mid=(l+r)>>1;
fz(mu,zi,l,mid),fz(mu,zi,mid+1,r);
const int RES=mu[l].size()<<1|1;
Poly::L=1;while(Poly::L<RES)Poly::L<<=1;
const int ll=l,rr=mid+1;
zi[ll].DFT(),mu[ll].DFT(),zi[rr].DFT(),mu[rr].DFT();
zi[ll].cheng(mu[rr]);
zi[rr].cheng(mu[ll]);
zi[ll].jia(zi[rr]);
zi[ll].IDFT();
mu[ll].cheng(mu[rr]);
mu[ll].IDFT();
}
int fac[maxn+1],inv[maxn+1];
int main()
{
fac[0]=1;for(rg int i=1;i<=maxn;i++)fac[i]=(ll)fac[i-1]*i%mod;
inv[maxn]=pow(fac[maxn],mod-2);for(rg int i=maxn;i>=1;i--)inv[i-1]=(ll)inv[i]*i%mod;
Poly::init(maxn);///////namespace of Poly
read(n),read(m);
for(rg int i=1;i<=n;i++)
a[i].resize(2),a[i][0]=1,read(a[i][1]),a[i][1]=Md(mod-a[i][1]),
za[i].resize(1),za[i][0]=1;
for(rg int i=1;i<=m;i++)
b[i].resize(2),b[i][0]=1,read(b[i][1]),b[i][1]=Md(mod-b[i][1]),
zb[i].resize(1),zb[i][0]=1;
for(rg int i=1;i<=m;i++);
read(t);
fz(a,za,1,n);
fz(b,zb,1,m);
a[1].resize(100001);
b[1].resize(100001);
A=za[1]*a[1].inv();
B=zb[1]*b[1].inv();
A.resize(100001);
B.resize(100001);
for(rg int i=1;i<=100000;i++)A[i]=(ll)A[i]*inv[i]%mod,B[i]=(ll)B[i]*inv[i]%mod;
A=A*B;
const int INV=pow((ll)n*m%mod,mod-2);
for(rg int i=1;i<=t;i++)print((ll)A[i]*fac[i]%mod*INV%mod),putchar('\n');
return flush(),0;
}
题解2
这题有更清真的做法
求所有 ∑ a i \sum a^i ∑ai和 ∑ b i \sum b^i ∑bi的时候能用ln优化
考虑生成函数
F i ( x ) = ∑ j = 0 ∞ a i j x j F_i(x)=\sum_{j=0}^{\infty}a_i^jx^j Fi(x)=j=0∑∞aijxj
根据等比数列求和得到
F i ( x ) = 1 1 − a i x F_i(x)=\frac{1}{1-a_ix} Fi(x)=1−aix1
我们对所有生成函数求和，设
f ( x ) = ∑ i = 1 n F i ( x ) = ∑ i = 1 n 1 1 − a i x f(x)=\sum_{i=1}^nF_i(x)=\sum_{i=1}^n\frac{1}{1-a_ix} f(x)=i=1∑nFi(x)=i=1∑n1−aix1
直接求常数很大
我们考虑求导
众所周知 ( l n ( x ) ) ′ = 1 x , ( a x ) ′ = a (ln(x))'=\frac1x,(ax)'=a (ln(x))′=x1,(ax)′=a
根据复合函数求导法则
( l n ( 1 − a i x ) ) ′ = − a i 1 − a i x (ln(1-a_ix))'=\frac{-a_i}{1-a_ix} (ln(1−aix))′=1−aix−ai
设 g ( x ) = ∑ i = 1 n − a i 1 − a i x g(x)=\sum_{i=1}^n\frac{-a_i}{1-a_ix} g(x)=i=1∑n1−aix−ai
我们发现 f ( x ) + g ( x ) x = n f(x)+g(x)x=n f(x)+g(x)x=n
移项得 f ( x ) = n − g ( x ) x f(x)=n-g(x)x f(x)=n−g(x)x
那么我们只要求出 g ( x ) g(x) g(x)就能求出 f ( x ) f(x) f(x)
g ( x ) = ∑ i = 1 n − a i 1 − a i x = ∑ i = 1 n ( l n ( 1 − a i x ) ) ′ = ( ∑ i = 1 n l n ( 1 − a i x ) ) ′ = ( l n ( ∏ i = 1 n ( 1 − a i x ) ) ) ′ \begin{aligned} g(x)&=\sum_{i=1}^n\frac{-a_i}{1-a_ix}\\ &=\sum_{i=1}^n(ln(1-a_ix))'\\ &=(\sum_{i=1}^nln(1-a_ix))'\\ &=(ln(\prod_{i=1}^n(1-a_ix)))'\\ \end{aligned} g(x)=i=1∑n1−aix−ai=i=1∑n(ln(1−aix))′=(i=1∑nln(1−aix))′=(ln(i=1∏n(1−aix)))′
这个算法总复杂度也是 O ( n l o g 2 n ) \mathcal O(nlog^2n) O(nlog2n)
不贴代码了，我写的ln常数太大了，过不了
懒得卡常了
总结
一道经典模型好题