每种父亲编号小于儿子编号的有标号二叉树的出现概率是相同的,问题相当于求所有n个点的此种树的所有结点两两距离之和。
设f[n]为答案,g[n]为所有此种树所有结点的深度之和,h[n]为此种树的个数。
枚举左右子树大小,则有f[n]=Σ{[f[i]+(g[i]+h[i]*i)·(n-i)]·h[n-i-1]+[f[n-i-1]+(g[n-i-1]+h[n-i-1]*(n-i-1))·(i+1)]·h[i]}·C(n-1,i),即对两棵子树分别统计贡献,C(n-1,i)即给左右子树分配编号。g[n]=Σ[(g[i]+h[i]*i)·h[n-i-1]+(g[n-i-1]+h[n-i-1]*(n-i-1))·h[i]]·C(n-1,i),h[n]=Σh[i]·h[n-i-1]·C(n-1,i),比较显然。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 2010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,P,C[N][N],f[N],g[N],h[N]; void inc(int &x,int y){x+=y;if (x>=P) x-=P;} int main() { #ifndef ONLINE_JUDGE freopen("bzoj5305.in","r",stdin); freopen("bzoj5305.out","w",stdout); const char LL[]="%I64d "; #else const char LL[]="%lld "; #endif n=read(),P=read(); C[0][0]=1; for (int i=1;i<=n;i++) { C[i][0]=C[i][i]=1; for (int j=1;j<i;j++) C[i][j]=(C[i-1][j-1]+C[i-1][j])%P; } f[0]=g[0]=0;h[0]=1; for (int i=1;i<=n;i++) { for (int j=0;j<i;j++) inc(h[i],1ll*h[j]*h[i-j-1]%P*C[i-1][j]%P); for (int j=0;j<i;j++) inc(g[i],((g[j]+1ll*j*h[j])%P*h[i-j-1]+(g[i-j-1]+1ll*(i-j-1)*h[i-j-1])%P*h[j])%P*C[i-1][j]%P); for (int j=0;j<i;j++) inc(f[i],((f[j]+(g[j]+1ll*j*h[j])%P*(i-j))%P*h[i-j-1]+(f[i-j-1]+(g[i-j-1]+1ll*(i-j-1)*h[i-j-1])%P*(j+1))%P*h[j])%P*C[i-1][j]%P); } cout<<f[n]; return 0; }