hdu4990矩阵快速幂

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 10000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-9;
const int N=10+5,maxn=1<<10+5,inf=0x3f3f3f3f;

struct Node{
    ll row,col;
    ll a[N][N];
};
ll m;
Node mul(Node x,Node y)
{
    Node ans;
    ans.row=x.row,ans.col=y.col;
    memset(ans.a,0,sizeof ans.a);
    for(ll i=0;i<x.row;i++)
        for(ll j=0;j<x.col;j++)
           for(ll k=0;k<y.col;k++)
               ans.a[i][k]=(ans.a[i][k]+x.a[i][j]*y.a[j][k]+m)%m;
    return ans;
}
Node quick_mul(Node x,ll n)
{
    Node ans;
    ans.row=x.row,ans.col=x.col;
    memset(ans.a,0,sizeof ans.a);
    for(ll i=0;i<ans.col;i++)ans.a[i][i]=1;
    while(n){
        if(n&1)ans=mul(ans,x);
        x=mul(x,x);
        n/=2;
    }
    return ans;
}
int main()
{

    ios::sync_with_stdio(false);
    cin.tie(0);
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    ll n;
    while(cin>>n>>m){
        if(n==1)
        {
            cout<<1%m<<endl;
            continue;
        }
        Node A;
        A.row=3,A.col=3;
        memset(A.a,0,sizeof A.a);
        A.a[0][0]=0,A.a[0][1]=1,A.a[0][2]=0;
        A.a[1][0]=2,A.a[1][1]=1,A.a[1][2]=1;
        A.a[2][0]=0,A.a[2][1]=0,A.a[2][2]=1;
       /* for(int i=0;i<A.row;i++)
        {
            for(int j=0;j<A.col;j++)
                cout<<A.a[i][j]<<" ";
            cout<<endl;
        }*/
        Node B;
        B.row=3,B.col=1;
        B.a[0][0]=1,B.a[1][0]=2,B.a[2][0]=1;
        B=mul(quick_mul(A,n-1),B);
        cout<<B.a[0][0]<<endl;
    }
    return 0;
}