bzoj3456: 城市规划

Description

 刚刚解决完电力网络的问题, 阿狸又被领导的任务给难住了.
 刚才说过, 阿狸的国家有n个城市, 现在国家需要在某些城市对之间建立一些贸易路线, 使得整个国家的任意两个城市都直接或间接的连通. 为了省钱, 每两个城市之间最多只能有一条直接的贸易路径. 对于两个建立路线的方案, 如果存在一个城市对, 在两个方案中是否建立路线不一样, 那么这两个方案就是不同的, 否则就是相同的. 现在你需要求出一共有多少不同的方案.
 好了, 这就是困扰阿狸的问题. 换句话说, 你需要求出n个点的简单(无重边无自环)无向连通图数目.
 由于这个数字可能非常大, 你只需要输出方案数mod 1004535809(479 * 2 ^ 21 + 1)即可.

Input

 仅一行一个整数n(<=130000)
 

Output

 仅一行一个整数, 为方案数 mod 1004535809.
 

Sample Input

3

Sample Output

4

HINT

 对于 100%的数据, n <= 130000

Source

方法一：cdq+ntt

设f[n]为n个点的答案，则

这就可以做了

code：

#include<cstdio> 
#include<iostream> 
#include<cmath> 
#include<cstring> 
#include<algorithm> 
#define maxn 262148 
#define mod 1004535809 
#define g 3 
using namespace std; 
typedef long long int64; 
char ch; 
int n; 
int a[maxn],b[maxn],c[maxn],Wn[2][maxn],wn,w,t1,t2; 
int f[maxn],pow2[maxn],fac[maxn],inv_fac[maxn],inv_n[maxn]; 
int re[19][maxn]; 
bool ok; 
inline void read(int &x){ 
    for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1; 
    for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar()); 
    if (ok) x=-x; 
} 
inline int rev(int len,int v){ 
    int t=0; 
    for (int i=0;i<len;i++) t<<=1,t|=v&1,v>>=1; 
    return t; 
} 
inline int ksm(int a,int b){ 
    int64 t=1; 
    for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 
    return t; 
} 
inline int ksm(int a,int64 b){ 
    int64 t=1; 
    for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 
    return t; 
} 
inline void ntt(int *a,int n,int len,int op){
    for (int i=0,t=re[len][i];i<n;i++,t=re[len][i]) if (i<t) swap(a[i],a[t]); 
    for (int s=2;s<=n;s<<=1){ 
        wn=Wn[op][s]; 
        for (int i=0;i<n;i+=s){ 
            w=1; 
            for (int j=i;j<i+(s>>1);j++,w=1LL*w*wn%mod){ 
                t1=a[j],t2=1LL*w*a[j+(s>>1)]%mod; 
                a[j]=(t1+t2)%mod,a[j+(s>>1)]=(t1-t2+mod)%mod; 
            } 
        } 
    } 
    if (op==1){ 
        int x=inv_n[n]; 
        for (int i=0;i<n;i++) a[i]=1LL*a[i]*x%mod; 
    } 
} 
inline void solve(int l,int r){ 
    if (l==r){ 
        f[l]=(pow2[l]-(int)(1LL*fac[l-1]*f[l]%mod)+mod)%mod; 
        return; 
    } 
    int m=(l+r)>>1; 
    solve(l,m); 
    int n=1,len=0; 
    while (n<((r-l+1)<<1)) n<<=1,len++; 
    for (int i=0;i<n;i++) a[i]=0; 
    for (int i=0;i<n;i++) b[i]=0; 
    for (int i=l;i<=m;i++) a[i-l]=1LL*f[i]*inv_fac[i-1]%mod; 
    for (int i=1;i<r-l+1;i++) b[i]=1LL*pow2[i]*inv_fac[i]%mod; 
    ntt(a,n,len,0),ntt(b,n,len,0); 
    for (int i=0;i<n;i++) c[i]=1LL*a[i]*b[i]%mod; 
    ntt(c,n,len,1); 
    for (int i=m+1;i<=r;i++) f[i]=(f[i]+c[i-l])%mod; 
    solve(m+1,r); 
} 
void init(){ 
    read(n); 
    for (int i=1;i<=n;i++) pow2[i]=ksm(2,(1LL*i*(i-1))>>1); 
    fac[0]=1; 
    for (int i=1;i<=n;i++) fac[i]=1LL*i*fac[i-1]%mod; 
    for (int i=0;i<=n;i++) inv_fac[i]=ksm(fac[i],mod-2); 
    for (int i=2;i<(n<<2);i<<=1) Wn[0][i]=ksm(g,(mod-1)/i); 
    for (int i=2;i<(n<<2);i<<=1) Wn[1][i]=ksm(Wn[0][i],mod-2); 
    for (int i=2;i<(n<<2);i<<=1) inv_n[i]=ksm(i,mod-2); 
    for (int i=1;(1<<i)<(n<<2);i++){ 
        for (int j=0;j<(1<<i);j++) re[i][j]=rev(i,j);   
    } 
} 
int main(){ 
    init(),solve(1,n); 
    printf("%d
",f[n]); 
    return 0; 
} 

方法二：多项式的逆元

http://blog.miskcoo.com/2015/05/bzoj-3456

如何求多项式的逆元http://blog.miskcoo.com/2015/05/polynomial-inverse

code：

#include<cstdio> 
#include<iostream> 
#include<cmath> 
#include<cstring> 
#include<algorithm> 
#define maxn 262148 
#define mod 1004535809 
#define g 3 
using namespace std; 
typedef long long int64; 
char ch; 
int m,n,len,N; 
int a[maxn],b[maxn],c[maxn],f[maxn],t[maxn],Wn[2][maxn],wn,w,t1,t2; 
int pow2[maxn],fac[maxn],inv_fac[maxn],inv_n[maxn]; 
int re[19][maxn]; 
bool ok; 
inline void read(int &x){ 
    for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1; 
    for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar()); 
    if (ok) x=-x; 
} 
inline int rev(int len,int v){ 
    int t=0; 
    for (int i=0;i<len;i++) t<<=1,t|=v&1,v>>=1; 
    return t; 
} 
inline int ksm(int a,int b){ 
    int64 t=1; 
    for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 
    return t; 
} 
inline int ksm(int a,int64 b){ 
    int64 t=1; 
    for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 
    return t; 
} 
inline void ntt(int *a,int n,int len,int op){ 
    for (int i=0,t=re[len][i];i<n;i++,t=re[len][i]) if (i<t) swap(a[i],a[t]); 
    for (int s=2;s<=n;s<<=1){ 
        wn=Wn[op][s]; 
        for (int i=0;i<n;i+=s){ 
            w=1; 
            for (int j=i;j<i+(s>>1);j++,w=1LL*w*wn%mod){ 
                t1=a[j],t2=1LL*w*a[j+(s>>1)]%mod; 
                a[j]=(t1+t2)%mod,a[j+(s>>1)]=(t1-t2+mod)%mod; 
            } 
        } 
    } 
    if (op==1){ 
        int x=inv_n[n]; 
        for (int i=0;i<n;i++) a[i]=1LL*a[i]*x%mod; 
    } 
} 
void init(){ 
    read(m),n=m+1,N=1; 
    while (N<(n<<1)) N<<=1,len++; 
    for (int i=0;i<=n;i++) pow2[i]=ksm(2,(1LL*i*(i-1))>>1); 
    fac[0]=1; 
    for (int i=1;i<=n;i++) fac[i]=1LL*i*fac[i-1]%mod; 
    for (int i=0;i<=n;i++) inv_fac[i]=ksm(fac[i],mod-2); 
    for (int i=2;i<(n<<2);i<<=1) Wn[0][i]=ksm(g,(mod-1)/i); 
    for (int i=2;i<(n<<2);i<<=1) Wn[1][i]=ksm(Wn[0][i],mod-2); 
    for (int i=2;i<(n<<2);i<<=1) inv_n[i]=ksm(i,mod-2); 
    for (int i=1;(1<<i)<(n<<2);i++)  for (int j=0;j<(1<<i);j++) re[i][j]=rev(i,j); 
    for (int i=0;i<=m;i++) a[i]=1LL*pow2[i]*inv_fac[i]%mod; 
    for (int i=1;i<=m;i++) c[i]=1LL*pow2[i]*inv_fac[i-1]%mod; 
} 
void get_inv(int deg,int *a,int *b){ 
    if (deg==1){b[0]=ksm(a[0],mod-2);return;} 
    get_inv((deg+1)>>1,a,b); 
    int n=1,len=0; 
    while (n<(deg<<1)) n<<=1,len++; 
    for (int i=0;i<deg;i++) t[i]=a[i]; 
    for (int i=deg;i<n;i++) t[i]=0; 
    ntt(t,n,len,0),ntt(b,n,len,0); 
    for (int i=0;i<n;i++) b[i]=(2LL-1LL*t[i]*b[i]%mod+mod)*b[i]%mod; 
    ntt(b,n,len,1); 
    for (int i=deg;i<n;i++) b[i]=0; 
} 
int main(){ 
    init(); 
    get_inv(n,a,b); 
    ntt(b,N,len,0),ntt(c,N,len,0); 
    for (int i=0;i<N;i++) f[i]=1LL*b[i]*c[i]%mod; 
    ntt(f,N,len,1); 
    printf("%d
",(int)(1LL*f[m]*fac[m-1]%mod)); 
    return 0; 
}