[Template] [power] matrix quickly find the n-th term Fibonacci

Note that a array initialization, call Pow (a, n, w), n is the first of several, in seeking Fibonacci default w 2

ll tmp[2][2],res[2][2];
void multi(ll a[][2],ll b[][2],int n)
{
    memset(tmp,0,sizeof(tmp));
    for(ll i=0;i<n;i++)
    {
       for(ll j=0;j<n;j++)
       {
        for(ll k=0;k<n;k++)
        {
        tmp[i][j]+=(a[i][k]*b[k][j])%p;
        }
        tmp[i][j]=tmp[i][j]%p;
      }
   }
    for(ll i=0;i<n;i++)
        for(ll j=0;j<n;j++)
        a[i][j]=tmp[i][j];
}
void Pow(ll a[][2],ll m,int n)
{
    memset(res,0,sizeof(res));//m是幂，n是矩阵大小
    for(ll i=0;i<n;i++) res[i][i]=1;
    while(m)
    {
        if(m&1)
        multi(res,a,n);//res=res*a;复制直接在multi里面实现了；
        multi(a,a,n);//a=a*a
        m>>=1;
    }
}
ll qm(ll a,ll b)
{
    ll ret=1;
    while(b)
    {
        if(b&1)
        {
            ret=ret*a%p;
        }
        a=a*a%p;
        b>>=1;
    }
    return ret;
}
//Pow(a,n,2) 调用  初始化:a[0][0]=a[0][1]=a[1][0]=1,a[1][1]=0