51nod 1122 機器人走方格 V4[矩陣快速冪]

題面
計算到達從某個點到某個點的方案數,考慮構造矩陣:給四個格子編號,構造可達矩陣
然後快速冪計算完把合法的答案累計一下就好了

#include<cstdio>
#include<cstring>
#define N 4
#define P 1000000007
#define LL long long
using namespace std;

const int p[N][N]={
    {0,1,1,1},
    {1,0,1,1},
    {1,1,0,1},
    {1,1,1,0}
};

struct Matrix
{
    int v[5][5];
    Matrix(){memset(v,0,sizeof(v));}
    friend void print(Matrix a)
    {
        for(int i=0;i<N;++i)
            for(int j=0;j<N;++j) printf("%d%s",a.v[i][j],j==N?"\n":" ");
    }
    friend Matrix operator * (Matrix a,Matrix b)
    {
        Matrix res;
        for(int i=0;i<N;++i)
            for(int j=0;j<N;++j)
            {
                res.v[i][j]=0;
                for(int k=0;k<N;++k)
                    (res.v[i][j]+=1ll*a.v[i][k]*b.v[k][j]%P)%=P;
            }
        return res;
    }
    friend Matrix operator ^ (Matrix a,int b)
    {
        Matrix ans;
        for(int i=0;i<N;++i)  ans.v[i][i]=1;
        while(b)
        {
            if(b&1) ans=ans*a;
            b>>=1;a=a*a;
        }
        return ans;
    }
}a,b;

int n;

int main(){
    for(int i=0;i<N;++i)
        for(int j=0;j<N;++j)
            a.v[i][j]=p[i][j];
    scanf("%d",&n);
    Matrix ans=a^n;
    LL sum=0;
    for(int i=0;i<N;++i)
        for(int j=0;j<N;++j)
            for(int k=0;k<N;++k)
                for(int l=0;l<N;++l)
                    if(i!=j&&i!=k&&i!=l&&j!=k&&j!=l&&k!=l)
                        (sum+=1ll*ans.v[0][i]*ans.v[1][j]%P*ans.v[2][k]%P*ans.v[3][l]%P)%=P;
    printf("%lld\n",sum);
}