BZOJ5305 HAOI2018 apple tree

Good question qwq

Not very comfortable to write 1k

Own expectations and the ability to count or almost

First we look at this problem we found a * n!

This is how it goes

Because each node would add a small but a position location so 2 more optional position number is added every time i

And then we look at this question has some good properties, such as binary tree

First, based on common routines we can calculate the contribution of an edge is szi * (n-szi)

So in the end we only need to discuss how many edges connected to the node is szi

We observed that only 2,000 n n ^ 2 can prompt us

So we can enumerate a re-enumeration sz x of x

That must be back again x sz-1 point in selected sub-tree long x's and because of this sub-tree which must be smaller than the father son so generated with the way the subject is the same as described

This contribution is C (nx, sz-1) * sz!

We continue to consider out a total of n-sz sub-tree nodes consider the previous node x total of x! Kinds of ways to generate

Consider other nodes after a total of x (nx-sz + 1) th to the inside can not be long x is (x + 1-2) * (x + 2-2) * (n-sz + 1-2) kinds of ways to generate

The above two bit is combined (n-sz-1)! I (i-1)

So we finished it

The final contribution formula is to multiply the above and then adding

$\sum_{x=2}^n\sum_{sz=1}^{n-x+1}(n-sz-1)!i(i-1)sz!C(n-x,sz-1)sz(n-sz)$

Step by step we can push the launch da

Due to the modulo a prime number so we did not seek to ensure the C and Pascal's Triangle factorial normal requirements in accordance with it qwq

//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 2001
using namespace std;

int C[N][N],fac[N],n,m;
void pre()
{
    C[0][0]=C[1][0]=C[1][1]=1; fac[1]=fac[0]=1;
    for(int i=2;i<=n;i++)
    {
        fac[i]=1ll*fac[i-1]*i%m; C[i][0]=C[i][i]=1;
        for(int j=1;j<n;j++)
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%m;
    }
}
int main()
{
    scanf("%d%d",&n,&m); pre(); int ans=0;
    for(int x=2;x<=n;x++)    for(int sz=1;sz<=n-x+1;sz++)
    {
        int tmp=1ll*fac[sz]*C[n-x][sz-1]%m*x%m*(x-1)%m*fac[n-sz-1]%m*sz%m*(n-sz)%m;
        ans=(ans+tmp)%m;
    } 
    printf("%d\n",ans);
    return 0;
}

View Code