矩阵快速幂

#include<bits/stdc++.h>
#define ll long long 
#define scan(i) scanf("%d",&i)
#define scand(i) scanf("%lf",&i)
#define scanl(i) scanf("%lld",&i)
#define f(i,a,b) for(int i=a;i<=b;i++) 
#define pb(i) push_back(i)
#define ppb pop_back()
#define pf printf
#define debug(i,j) i==j?cout<<"correct!
":cout<<"wrong~
"<<i<<"!="<<j<<endl;return 0
using namespace std;
int t;
ll n; 
ll mi[5][5]={
    2,4,1,0,0,
    1,0,0,0,0,
    0,0,1,2,1,
    0,0,0,1,1,
    0,0,0,0,1
};
const ll mod=998244353;
const int N=5;  
ll tmp[N][N]; 
ll qpow(ll x,ll i)//ans=x^i
{
    ll ans = 1;
    while(i)
    {
        if(i&1) ans = (ans*x)%mod;
        i >>= 1;
        x = (x*x)%mod;
    }
    return ans;
}
void multi(ll a[][N],ll b[][N],ll n)//a=axb 
{  
    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];
               tmp[i][j]%=mod;
        }
    }
    for(ll i=0;i<n;i++){
        for(ll j=0;j<n;j++){
            a[i][j]=tmp[i][j];
        }
    }
}  
ll res[N][N];  
void print(){
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            cout<<res[i][j]<<" ";
        }
        cout<<endl;
    }
} 
void Pow(ll n)//res=a^n
{  

    ll a[5][5]={
    2,4,1,0,0,
    1,0,0,0,0,
    0,0,1,2,1,
    0,0,0,1,1,
    0,0,0,0,1
    };
    memset(res,0,sizeof(res));//n是幂，N是矩阵大小  
    for(int i=0;i<N;i++) res[i][i]=1;  
    while(n){  
        if(n&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
        n>>=1;  
    }  
}  

N是矩阵大小，a是做幂运算的矩阵，n是幂次，mod是取的模