Codeforces - 185A 简单矩阵快速幂

Posted 2020-10-24 The Azure Arbitrator

题意:求第n个三角形内部的上三角形个数
对每个三角形分别维护上下三角形个数,记为\(dp[1][i],dp[2][i]\)
规律很明显是
\(dp[1][i+1]=3*dp[1][i]+dp[2][i]\)
\(dp[2][i+1]=3*dp[2][i]+dp[1][i]\)
别忘了快速幂里也要long long,白送了个TLE

/*H E A D*/
inline ll mod(ll a){return a%MOD;}
struct Matrix{
    ll mt[5][5],r,c;
    void init(int rr,int cc,bool flag=0){
        r=rr;c=cc;
        memset(mt,0,sizeof mt);
        if(flag) rep(i,1,r) mt[i][i]=1;
    }
    Matrix operator * (const Matrix &rhs)const{
        Matrix ans; ans.init(r,rhs.c);
        rep(i,1,r){
            rep(j,1,rhs.c){
                int t=max(r,rhs.c);
                rep(k,1,t){
                    ans.mt[i][j]+=mod(mt[i][k]*rhs.mt[k][j]);
                    ans.mt[i][j]=mod(ans.mt[i][j]);
                }
            }
        }
        return ans;
    }
};
Matrix fpw(Matrix A,ll n){
    Matrix ans;ans.init(A.r,A.c,1);
    while(n){
        if(n&1) ans=ans*A;
        n>>=1;
        A=A*A;
    }
    return ans;
}

int bas[3][3]={
    {0,0,0},
    {0,3,1},
    {0,1,3},
};
int bas2[3]={0,1,0}; 
ll n;
int main(){
    Matrix A;A.init(2,2);
    rep(i,1,2) rep(j,1,2) A.mt[i][j]=bas[i][j];
    Matrix b; b.init(2,1);
    rep(i,1,2) b.mt[i][1]=bas2[i];
    while(cin>>n){
        Matrix res=fpw(A,n); res=res*b;
        ll ans=mod(res.mt[1][1]);
        println(ans);
    }
    return 0;
}