20200104 A pattern simulation game problems

topic

 

 

 

 

analysis:

The old rules, expected to be ready to meet at any time to surrender. . .

Generally thought of by-bit processing, and then were down search, then the number of combinations of simple calculations it. . .

But the communication between the two sides of the block count how even expect it?

Well, to surrender. . .

Under term problem solution. . .

Sure enough, the search is credited. .

First we let F (n, m) denotes the n points takes a value [0, 2 ^ m) is the sum of all minimum spanning tree Consideration

Then Ans = F (n, m) / 2 ^ (n * m)

Then set G (S, T, m) represents a portion of the point set of size S, size another portion is T, the value of the point right after between [0, 2 ^ m), where all the sum of the weights minimum edge

So F (n, m) can be found in mind:

F(n,m)=

Sigma (I =. 1 ... n-) C (n-, I) (which point is selected) * (

F. (I,. 1-m) * 2 ^ ((Ni) * (. 1-m)) (part of the search program down then multiplier) +

F. (Ni,. 1-m) * 2 ^ (I * (. 1-m)) (the other part) +

G (I, Ni,. 1-m) (intermediate communication side) +

^ 2 (. 1-m) * 2 ^ (n-* (. 1-m)) (Consideration must spend multiplied by the number of programs) )

 

Then we come to find G:

Cheese we provided a number of function P (S, T, m, K) represents a set of points S, T, value [0, when 2 ^ m), the right edge is greater than a minimum value of equal to K

G (S, T, m) can be subtly into sigma (i = 1 ... (2 ^ m-1)) P (S, T, m, i)

Heavy secret, the brain can fill it, and about the same feeling with the prefix

Then P is very good demand, and violently S and T will continue to go down by 01 divided, into the statistics a bit when all the answers are not divided

 

Search triple in mind, but also so difficult equation

Immortal title Orz

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>

#define maxn 55
#define maxm 9
#define MOD 258280327

using namespace std;

inline long long getint()
{
    long long num=0,flag=1;char c;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
    return num*flag;
}

int n,m;
long long C[maxn][maxn],F[maxn][maxm],G[maxn][maxn][maxm],P[maxn][maxn][maxm][1<<maxm],pw[maxn*maxm];

inline long long ksm(long long num,long long k)
{
    long long ret=1;
    for(;k;k>>=1,num=num*num%MOD)if(k&1)ret=ret*num%MOD;
    return ret;
}

inline long long getP(int S,int T,int M,int K)
{
    if(S>T)swap(S,T);
    if(!S||K<=0)return pw[(S+T)*M];
    if(K>=(1<<M))return 0;
    if(~P[S][T][M][K])return P[S][T][M][K];
    long long tmp=0;
    for(int i=0;i<=S;i++)for(int j=0;j<=T;j++)
        if((i==0&&j==T)||(i==S&&j==0))tmp=(tmp+getP(S,T,M-1,K-(1<<(M-1))))%MOD;
        else tmp=(tmp+getP(i,j,M-1,K)*getP(S-i,T-j,M-1,K)%MOD*C[S][i]%MOD*C[T][j]%MOD)%MOD;
    return P[S][T][M][K]=tmp;
}

inline long long getG(int S,int T,int M)
{
    if(!M)return 0;
    if(S>T)swap(S,T);
    if(~G[S][T][M])return G[S][T][M];
    long long tmp=0;
    for(int i=1;i<(1<<M);i++)
        tmp=(tmp+getP(S,T,M,i))%MOD;
    return G[S][T][M]=tmp;
}

inline long long getF(int N,int M)
{
    if(!M||N<2)return 0;
    if(~F[N][M])return F[N][M];
    long long tmp=2*getF(N,M-1)%MOD;
    for(int i=1;i<N;i++)
        tmp=(tmp+C[N][i]*(getF(i,M-1)*pw[(N-i)*(M-1)]%MOD+getF(N-i,M-1)*pw[i*(M-1)]%MOD+getG(i,N-i,M-1)+(1<<(M-1))*pw[N*(M-1)]%MOD))%MOD;
    return F[N][M]=tmp;
}

int main()
{
    n=getint(),m=getint();
    pw[0]=1;for(int i=1;i<=n*m;i++)pw[i]=pw[i-1]*2%MOD;
    for(int i=0;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])%MOD;
    }
    memset(F,-1,sizeof F),memset(G,-1,sizeof G),memset(P,-1,sizeof P);
    printf("%lld\n",getF(n,m)*ksm(pw[n*m],MOD-2)%MOD);
}

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