[多项式] COGS 有标号的二分图计数系列

我是抄的 orzz

QAQ_bipartite_one

∑k=0nCkn∗2(n−k)∗k

注意这里是2(n−k)∗k不是2(n−k)+k 怎么卷？

2(n−k)∗k=(2√)n2(2√)k2∗(2√)(n−k)2

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;const int P=998244353;const int G=3;const int SQRT=116195171;const int INV=(P+1)>>1;const int N=300005;inline ll Pow(ll a,int b){  ll ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}inline ll Inv(ll a){  return Pow(a,P-2);}ll fac[N];inline void Init(int n){  fac[0]=1; for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;}int num;int w[2][N];inline void Pre(int m){  num=m;  int g=Pow(G,(P-1)/m);  w[0][0]=w[1][0]=1;  for (int i=1;i<num;i++) w[1][i]=(ll)w[1][i-1]*g%P;  for (int i=1;i<num;i++) w[0][i]=w[1][num-i];}int R[N];inline void FFT(int n,int *a,int r){  for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);  for (int i=1;i<n;i<<=1)    for (int j=0;j<n;j+=(i<<1))      for (int k=0;k<i;k++){    int x=a[j+k],y=(ll)a[j+k+i]*w[r][num/(i<<1)*k]%P;    a[j+k]=(x+y)%P; a[j+i+k]=(x+P-y)%P;      }  if (!r) for (int i=0,inv=Pow(n,P-2);i<n;i++) a[i]=(ll)a[i]*inv%P;}int n,m;int a[N];int main(){  freopen("QAQ_bipartite_one.in","r",stdin);  freopen("QAQ_bipartite_one.out","w",stdout);  scanf("%d",&n); Init(n);  int L=0; m=1;  while (m<=(n<<1)) m<<=1,L++;  for (int i=1;i<m;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));  Pre(m);  for (int i=0;i<=n;i++)    a[i]=Inv(fac[i]*Pow(SQRT,(ll)i*i%(P-1))%P);  FFT(m,a,1);  for (int i=0;i<m;i++) a[i]=(ll)a[i]*a[i]%P;  FFT(m,a,0);  printf("%d\n",(ll)a[n]*fac[n]%P*Pow(SQRT,(ll)n*n%(P-1))%P);  return 0;}

QAQ_bipartite_two

设上一题的也就是有色的指数生成函数是F(x)，这题的无色的指数生成函数为G(x)，无色的强制必须连通的指数生成函数是H(x)，那么

G=eH

F=∑i=0∞2iHii!=e2H

因为k个联通块染色会有2k

所以F=G2 直接多项式开根

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const int P=998244353;const int G=3;const int SQRT=116195171;const int INV=(P+1)>>1;const int N=600005;inline ll Pow(ll a,int b){  ll ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}inline int Root(int a){  if (a==1) return 1;}inline ll Inv(ll a){  return Pow(a,P-2);}ll fac[N];inline void Init(int n){  fac[0]=1; for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;}int num;int w[2][N];inline void Pre(int m){  num=m;  int g=Pow(G,(P-1)/m);  w[0][0]=w[1][0]=1;  for (int i=1;i<num;i++) w[1][i]=(ll)w[1][i-1]*g%P;  for (int i=1;i<num;i++) w[0][i]=w[1][num-i];}int R[N];inline void FFT(int *a,int n,int r){  for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);  for (int i=1;i<n;i<<=1)    for (int j=0;j<n;j+=(i<<1))      for (int k=0;k<i;k++){    int x=a[j+k],y=(ll)a[j+k+i]*w[r][num/(i<<1)*k]%P;    a[j+k]=(x+y)%P; a[j+i+k]=(x+P-y)%P;      }  if (!r) for (int i=0,inv=Pow(n,P-2);i<n;i++) a[i]=(ll)a[i]*inv%P;}inline void GetInv(int *a,int *b,int n){  static int tmp[N];  if (n==1) return void(b[0]=Inv(a[0]));  GetInv(a,b,n>>1);  for (int i=0;i<n;i++) tmp[i]=a[i],tmp[n+i]=0;  int L=0; while (!(n>>L&1)) L++;  for (int i=1;i<(n<<1);i++) R[i]=(R[i>>1]>>1)|((i&1)<<L);  FFT(tmp,n<<1,1); FFT(b,n<<1,1);  for (int i=0;i<(n<<1);i++)    tmp[i]=(ll)b[i]*(2+P-(ll)tmp[i]*b[i]%P)%P;  FFT(tmp,n<<1,0);  for (int i=0;i<n;i++) b[i]=tmp[i],b[n+i]=0;}const int INV2=(P+1)/2;inline void GetRoot(int *a,int *b,int n){  static int tmp[N],invb[N];  if (n==1) return void(b[0]=Root(a[0]));  GetRoot(a,b,n>>1);  memset(invb,0,sizeof(int)*n); GetInv(b,invb,n);  int L=0; while (!(n>>L&1)) L++;  for (int i=1;i<(n<<1);i++) R[i]=(R[i>>1]>>1)|((i&1)<<L);  for (int i=0;i<n;i++) tmp[i]=a[i],tmp[n+i]=0;  FFT(tmp,n<<1,1); FFT(b,n<<1,1); FFT(invb,n<<1,1);  for (int i=0;i<(n<<1);i++)    tmp[i]=(ll)((ll)b[i]*b[i]+tmp[i])%P*invb[i]%P*INV2%P;  FFT(tmp,n<<1,0);  for (int i=0;i<n;i++) b[i]=tmp[i],b[n+i]=0;}int n,m;int a[N],b[N];ll pw[N];int main(){  freopen("QAQ_bipartite_two.in","r",stdin);  freopen("QAQ_bipartite_two.out","w",stdout);  scanf("%d",&n); Init(n);  int L=0; m=1;  while (m<=(n<<1)) m<<=1,L++;  for (int i=1;i<m;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));  Pre(m<<1);  for (int i=0;i<=n;i++)    a[i]=Inv(fac[i]*(pw[i]=Pow(SQRT,(ll)i*i%(P-1)))%P);  FFT(a,m,1);  for (int i=0;i<m;i++) a[i]=(ll)a[i]*a[i]%P;  FFT(a,m,0);  for (int i=0;i<=n;i++) a[i]=a[i]*pw[i]%P;  for (int i=n+1;i<m;i++) a[i]=0;  m=1; while (m<=n) m<<=1;  GetRoot(a,b,m);  printf("%d",(ll)b[n]*fac[n]%P);  return 0;}

QAQ_bipartite_thr

F=e2H 多项式ln

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const int P=998244353;const int G=3;const int SQRT=116195171;const int INV2=(P+1)>>1;const int N=600005;inline ll Pow(ll a,int b){  ll ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}inline int Root(int a){  if (a==1) return 1;}inline ll Inv(ll a){  return Pow(a,P-2);}ll fac[N],inv[N];inline void Init(int n){  fac[0]=1; for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;  inv[1]=1; for (int i=2;i<=n;i++) inv[i]=(ll)(P-P/i)*inv[P%i]%P;  inv[0]=1; for (int i=1;i<=n;i++) inv[i]=inv[i-1]*inv[i]%P;}int num;int w[2][N];inline void Pre(int m){  num=m;  int g=Pow(G,(P-1)/m);  w[0][0]=w[1][0]=1;  for (int i=1;i<num;i++) w[1][i]=(ll)w[1][i-1]*g%P;  for (int i=1;i<num;i++) w[0][i]=w[1][num-i];}int R[N];inline void FFT(int *a,int n,int r){  for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);  for (int i=1;i<n;i<<=1)    for (int j=0;j<n;j+=(i<<1))      for (int k=0;k<i;k++){    int x=a[j+k],y=(ll)a[j+k+i]*w[r][num/(i<<1)*k]%P;    a[j+k]=(x+y)%P; a[j+i+k]=(x+P-y)%P;      }  if (!r) for (int i=0,inv=Pow(n,P-2);i<n;i++) a[i]=(ll)a[i]*inv%P;}inline void GetInv(int *a,int *b,int n){  static int tmp[N];  if (n==1) return void(b[0]=Inv(a[0]));  GetInv(a,b,n>>1);  for (int i=0;i<n;i++) tmp[i]=a[i],tmp[n+i]=0;  int L=0; while (!(n>>L&1)) L++;  for (int i=1;i<(n<<1);i++) R[i]=(R[i>>1]>>1)|((i&1)<<L);  FFT(tmp,n<<1,1); FFT(b,n<<1,1);  for (int i=0;i<(n<<1);i++)    tmp[i]=(ll)b[i]*(2+P-(ll)tmp[i]*b[i]%P)%P;  FFT(tmp,n<<1,0);  for (int i=0;i<n;i++) b[i]=tmp[i],b[n+i]=0;}int n,m;int a[N],b[N],c[N],f[N];ll pw[N];int main(){  freopen("QAQ_bipartite_thr.in","r",stdin);  freopen("QAQ_bipartite_thr.out","w",stdout);  scanf("%d",&n); Init(n);  int L=0; m=1;  while (m<=(n<<1)) m<<=1,L++;  for (int i=1;i<m;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));  Pre(m<<1);  for (int i=0;i<=n;i++)    a[i]=Inv(fac[i]*(pw[i]=Pow(SQRT,(ll)i*i%(P-1)))%P);  FFT(a,m,1);  for (int i=0;i<m;i++) a[i]=(ll)a[i]*a[i]%P;  FFT(a,m,0);  for (int i=0;i<=n+1;i++) f[i]=(ll)a[i]*pw[i]%P*fac[i]%P,a[i]=(ll)a[i]*pw[i]%P;  for (int i=0;i<=n;i++) c[i]=f[i+1]*inv[i]%P;  for (int i=n+1;i<=m;i++) a[i]=0;  GetInv(a,b,m>>1);  FFT(b,m,1); FFT(c,m,1);  for (int i=0;i<m;i++) b[i]=(ll)c[i]*b[i]%P;  FFT(b,m,0);  printf("%d\n",(ll)b[n-1]*fac[n-1]%P*INV2%P);  return 0;}