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.
bzoj3456: 城市规划
Description
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：
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #define maxn 262148 7 #define mod 1004535809 8 #define g 3 9 using namespace std; 10 typedef long long int64; 11 char ch; 12 int n; 13 int a[maxn],b[maxn],c[maxn],Wn[2][maxn],wn,w,t1,t2; 14 int f[maxn],pow2[maxn],fac[maxn],inv_fac[maxn],inv_n[maxn]; 15 int re[19][maxn]; 16 bool ok; 17 inline void read(int &x){ 18 for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1; 19 for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar()); 20 if (ok) x=-x; 21 } 22 inline int rev(int len,int v){ 23 int t=0; 24 for (int i=0;i<len;i++) t<<=1,t|=v&1,v>>=1; 25 return t; 26 } 27 inline int ksm(int a,int b){ 28 int64 t=1; 29 for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 30 return t; 31 } 32 inline int ksm(int a,int64 b){ 33 int64 t=1; 34 for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 35 return t; 36 } 37 inline void ntt(int *a,int n,int len,int op){ 38 for (int i=0,t=re[len][i];i<n;i++,t=re[len][i]) if (i<t) swap(a[i],a[t]); 39 for (int s=2;s<=n;s<<=1){ 40 wn=Wn[op][s]; 41 for (int i=0;i<n;i+=s){ 42 w=1; 43 for (int j=i;j<i+(s>>1);j++,w=1LL*w*wn%mod){ 44 t1=a[j],t2=1LL*w*a[j+(s>>1)]%mod; 45 a[j]=(t1+t2)%mod,a[j+(s>>1)]=(t1-t2+mod)%mod; 46 } 47 } 48 } 49 if (op==1){ 50 int x=inv_n[n]; 51 for (int i=0;i<n;i++) a[i]=1LL*a[i]*x%mod; 52 } 53 } 54 inline void solve(int l,int r){ 55 if (l==r){ 56 f[l]=(pow2[l]-(int)(1LL*fac[l-1]*f[l]%mod)+mod)%mod; 57 return; 58 } 59 int m=(l+r)>>1; 60 solve(l,m); 61 int n=1,len=0; 62 while (n<((r-l+1)<<1)) n<<=1,len++; 63 for (int i=0;i<n;i++) a[i]=0; 64 for (int i=0;i<n;i++) b[i]=0; 65 for (int i=l;i<=m;i++) a[i-l]=1LL*f[i]*inv_fac[i-1]%mod; 66 for (int i=1;i<r-l+1;i++) b[i]=1LL*pow2[i]*inv_fac[i]%mod; 67 ntt(a,n,len,0),ntt(b,n,len,0); 68 for (int i=0;i<n;i++) c[i]=1LL*a[i]*b[i]%mod; 69 ntt(c,n,len,1); 70 for (int i=m+1;i<=r;i++) f[i]=(f[i]+c[i-l])%mod; 71 solve(m+1,r); 72 } 73 void init(){ 74 read(n); 75 for (int i=1;i<=n;i++) pow2[i]=ksm(2,(1LL*i*(i-1))>>1); 76 fac[0]=1; 77 for (int i=1;i<=n;i++) fac[i]=1LL*i*fac[i-1]%mod; 78 for (int i=0;i<=n;i++) inv_fac[i]=ksm(fac[i],mod-2); 79 for (int i=2;i<(n<<2);i<<=1) Wn[0][i]=ksm(g,(mod-1)/i); 80 for (int i=2;i<(n<<2);i<<=1) Wn[1][i]=ksm(Wn[0][i],mod-2); 81 for (int i=2;i<(n<<2);i<<=1) inv_n[i]=ksm(i,mod-2); 82 for (int i=1;(1<<i)<(n<<2);i++){ 83 for (int j=0;j<(1<<i);j++) re[i][j]=rev(i,j); 84 } 85 } 86 int main(){ 87 init(),solve(1,n); 88 printf("%d\n",f[n]); 89 return 0; 90 }
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：
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #define maxn 262148 7 #define mod 1004535809 8 #define g 3 9 using namespace std; 10 typedef long long int64; 11 char ch; 12 int m,n,len,N; 13 int a[maxn],b[maxn],c[maxn],f[maxn],t[maxn],Wn[2][maxn],wn,w,t1,t2; 14 int pow2[maxn],fac[maxn],inv_fac[maxn],inv_n[maxn]; 15 int re[19][maxn]; 16 bool ok; 17 inline void read(int &x){ 18 for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1; 19 for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar()); 20 if (ok) x=-x; 21 } 22 inline int rev(int len,int v){ 23 int t=0; 24 for (int i=0;i<len;i++) t<<=1,t|=v&1,v>>=1; 25 return t; 26 } 27 inline int ksm(int a,int b){ 28 int64 t=1; 29 for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 30 return t; 31 } 32 inline int ksm(int a,int64 b){ 33 int64 t=1; 34 for (;b;b>>=1){if (b&1) t=1LL*t*a%mod; a=1LL*a*a%mod;} 35 return t; 36 } 37 inline void ntt(int *a,int n,int len,int op){ 38 for (int i=0,t=re[len][i];i<n;i++,t=re[len][i]) if (i<t) swap(a[i],a[t]); 39 for (int s=2;s<=n;s<<=1){ 40 wn=Wn[op][s]; 41 for (int i=0;i<n;i+=s){ 42 w=1; 43 for (int j=i;j<i+(s>>1);j++,w=1LL*w*wn%mod){ 44 t1=a[j],t2=1LL*w*a[j+(s>>1)]%mod; 45 a[j]=(t1+t2)%mod,a[j+(s>>1)]=(t1-t2+mod)%mod; 46 } 47 } 48 } 49 if (op==1){ 50 int x=inv_n[n]; 51 for (int i=0;i<n;i++) a[i]=1LL*a[i]*x%mod; 52 } 53 } 54 void init(){ 55 read(m),n=m+1,N=1; 56 while (N<(n<<1)) N<<=1,len++; 57 for (int i=0;i<=n;i++) pow2[i]=ksm(2,(1LL*i*(i-1))>>1); 58 fac[0]=1; 59 for (int i=1;i<=n;i++) fac[i]=1LL*i*fac[i-1]%mod; 60 for (int i=0;i<=n;i++) inv_fac[i]=ksm(fac[i],mod-2); 61 for (int i=2;i<(n<<2);i<<=1) Wn[0][i]=ksm(g,(mod-1)/i); 62 for (int i=2;i<(n<<2);i<<=1) Wn[1][i]=ksm(Wn[0][i],mod-2); 63 for (int i=2;i<(n<<2);i<<=1) inv_n[i]=ksm(i,mod-2); 64 for (int i=1;(1<<i)<(n<<2);i++) for (int j=0;j<(1<<i);j++) re[i][j]=rev(i,j); 65 for (int i=0;i<=m;i++) a[i]=1LL*pow2[i]*inv_fac[i]%mod; 66 for (int i=1;i<=m;i++) c[i]=1LL*pow2[i]*inv_fac[i-1]%mod; 67 } 68 void get_inv(int deg,int *a,int *b){ 69 if (deg==1){b[0]=ksm(a[0],mod-2);return;} 70 get_inv((deg+1)>>1,a,b); 71 int n=1,len=0; 72 while (n<(deg<<1)) n<<=1,len++; 73 for (int i=0;i<deg;i++) t[i]=a[i]; 74 for (int i=deg;i<n;i++) t[i]=0; 75 ntt(t,n,len,0),ntt(b,n,len,0); 76 for (int i=0;i<n;i++) b[i]=(2LL-1LL*t[i]*b[i]%mod+mod)*b[i]%mod; 77 ntt(b,n,len,1); 78 for (int i=deg;i<n;i++) b[i]=0; 79 } 80 int main(){ 81 init(); 82 get_inv(n,a,b); 83 ntt(b,N,len,0),ntt(c,N,len,0); 84 for (int i=0;i<N;i++) f[i]=1LL*b[i]*c[i]%mod; 85 ntt(f,N,len,1); 86 printf("%d\n",(int)(1LL*f[m]*fac[m-1]%mod)); 87 return 0; 88 }
