BZOJ 3456 城市规划 ——NTT

搞出递推式。

发现可以变成三个函数的乘积。

移项之后就可以求逆+NTT做了。

miskoo博客中有讲

#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define ll long long
#define mp make_pair
#define md 1004535809
#define g 3
#define maxn 500005
 
int rev[maxn],n;
 
int ksm(int a,int b)
{
    int ret=1;
    for (;b;b>>=1,a=(ll)a*a%md) if (b&1) ret=(ll)ret*a%md;
    return ret;
}
 
void NTT(int *x,int n,int flag)
{
    F(i,0,n-1) if (rev[i]>i) swap(x[rev[i]],x[i]);
    for (int m=2;m<=n;m<<=1)
    {
        int wn=ksm(g,((md-1)/m*flag+md-1)%(md-1));
        for (int i=0;i<n;i+=m)
        {
            int w=1;
            for (int j=0;j<(m>>1);++j)
            {
                int u=x[i+j],v=(ll)x[i+j+(m>>1)]*w%md;
                x[i+j]=(u+v)%md; x[i+j+(m>>1)]=(u-v+md)%md;
                w=(ll)w*wn%md;
            }
        }
    }
    if (flag==-1)
    {
        int inv=ksm(n,md-2);
        F(i,0,n-1) x[i]=(ll)x[i]*inv%md;
    }
}
 
int fac[maxn],fac_inv[maxn],C[maxn],G[maxn],F[maxn],N,Inv_G[maxn];
 
void Get_Inv(int *a,int *b,int n)
{
    static int tmp[maxn];if (n==1){b[0]=ksm(a[0],md-2);return;}
    Get_Inv(a,b,n>>1);F(i,0,n-1)tmp[i]=a[i],tmp[i+n]=0;
    int L=0;while(!(n>>L&1))L++;
    F(i,0,(n<<1)-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<L);
    NTT(tmp,n<<1,1);NTT(b,n<<1,1);
    F(i,0,(n<<1)-1) tmp[i]=(ll)b[i]*(2LL-(ll)tmp[i]*b[i]%md+md)%md;
    NTT(tmp,n<<1,-1);F(i,0,n-1) b[i]=tmp[i],b[n+i]=0;
}
 
int main()
{
    scanf("%d",&n);
    fac[0]=1;F(i,1,maxn-1) fac[i]=(ll)fac[i-1]*i%md;
    fac_inv[maxn-1]=ksm(fac[maxn-1],md-2);
    D(i,maxn-2,0) fac_inv[i]=(ll)fac_inv[i+1]*(i+1)%md;
    for (N=1;N<=n;N<<=1);
    F(i,0,n) C[i]=(ll)ksm(2,(ll)i*(i-1)/2%(md-1))*fac_inv[i-1]%md;
    F(i,0,n) G[i]=(ll)ksm(2,(ll)i*(i-1)/2%(md-1))*fac_inv[i]%md;
    Get_Inv(G,Inv_G,N);
    NTT(C,N<<1,1);NTT(Inv_G,N<<1,1);
    F(i,0,(N<<1)-1) F[i]=(ll)C[i]*Inv_G[i]%md;
    NTT(F,N<<1,-1);
    printf("%d\n",(ll)F[n]*fac[n-1]%md);
}

　　

Po姐讲了另外一种方法。

哈哈哈，完全不会，抄抄抄

#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define ll long long
#define mp make_pair
#define maxn 500005
#define md 1004535809
#define g 3
 
 
int rev[maxn];
 
int ksm(int a,int b)
{
    int ret=1;
    for (;b;b>>=1,a=(ll)a*a%md) if (b&1) ret=(ll)ret*a%md;
    return ret;
}
 
void NTT(int *x,int n,int flag)
{
    F(i,0,n-1) if (rev[i]>i) swap(x[rev[i]],x[i]);
    for (int m=2;m<=n;m<<=1)
    {
        int wn=ksm(g,((md-1)/m*flag+md-1)%(md-1));
        for (int i=0;i<n;i+=m)
        {
            int w=1;
            for (int j=0;j<(m>>1);++j)
            {
                int u=x[i+j],v=(ll)x[i+j+(m>>1)]*w%md;
                x[i+j]=(u+v)%md; x[i+j+(m>>1)]=(u-v+md)%md;
                w=(ll)w*wn%md;
            }
        }
    }
    if (flag==-1)
    {
        int inv=ksm(n,md-2);
        F(i,0,n-1) x[i]=(ll)x[i]*inv%md;
    }
}
 
 
 
int n,G[maxn],F[maxn],Inv_G[maxn],N,fac[maxn],fac_inv[maxn],Der_G[maxn];
 
void Get_Inv(int *a,int *b,int n)
{
    static int tmp[maxn];if (n==1){b[0]=ksm(a[0],md-2);return;}
    Get_Inv(a,b,n>>1);F(i,0,n-1)tmp[i]=a[i],tmp[n+i]=0;
    int L=0;while(!(n>>L&1)) L++;
    F(i,0,(n<<1)-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<L);
    NTT(tmp,n<<1,1);NTT(b,n<<1,1);
    F(i,0,(n<<1)-1) tmp[i]=(ll)b[i]*(2-(ll)tmp[i]*b[i]%md+md)%md;
    NTT(tmp,n<<1,-1); F(i,0,n-1) b[i]=tmp[i],b[i+n]=0;
}
 
 
int main()
{
    scanf("%d",&n);
    fac[0]=1;F(i,1,maxn-1) fac[i]=(ll)fac[i-1]*i%md;
    fac_inv[maxn-1]=ksm(fac[maxn-1],md-2);
    for (N=1;N<=n;N<<=1);
    D(i,maxn-2,0)fac_inv[i]=(ll)fac_inv[i+1]*(i+1)%md;
    F(i,0,n)G[i]=(ll)ksm(2,(ll)i*(i-1)/2%(md-1))*fac_inv[i]%md;
    Get_Inv(G,Inv_G,N);
    F(i,1,N-1) Der_G[i-1]=(ll)G[i]*i%md;
    int L=0;while(!(N>>L&1)) L++;
    F(i,0,(N<<1)-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<L);
    NTT(Der_G,N<<1,1);NTT(Inv_G,N<<1,1);
    F(i,0,(N<<1)-1)F[i]=(ll)Der_G[i]*Inv_G[i]%md;
    NTT(F,N<<1,-1);
    printf("%d\n",(ll)F[n-1]*ksm(n,md-2)%md*fac[n]%md);
}