1303. 斐波那契前 n 项和

\[f_n = f_{n-1}+f_{n-2} \\ S_n=S_{n-1}+f_n \\ [f_n,f_{n-1},S_n]= [f_{n-1},f_{n-2},S_{n-1}] \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 1\\ 0 & 0 & 1 \end{bmatrix} =[f_1,f_0,S_1] \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n-1} \]

const int N=3;
int n,mod;

void mul(int f[],int m[][N])
{
    int c[N]={0};
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++)
            c[i]=(c[i]+(LL)f[j]*m[j][i])%mod;
    memcpy(f,c,sizeof c);
}

void mulself(int m[][N])
{
    int c[N][N]={0};
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++)
            for(int k=0;k<N;k++)
                c[i][j]=(c[i][j]+(LL)m[i][k]*m[k][j])%mod;
    memcpy(m,c,sizeof c);
}

int main()
{
    int f[N]={1,0,1};
    int m[N][N]={{1,1,1},
                 {1,0,1},
                 {0,0,1}};

    cin>>n>>mod;

    if(n == 1)
        cout<<f[0]<<endl;
    else
    {
        n--;
        while(n)
        {
            if(n & 1) mul(f,m);
            mulself(m);
            n>>=1;
        }

        cout<<f[2]<<endl;
    }
    //system("pause");
    return 0;
}
posted @ 2021-04-13 15:24  Dazzling!  阅读(157)  评论(0编辑  收藏  举报