Plant fast power matrix ,,,, somewhat forgotten

Topic links: https://codeforces.com/contest/185/problem/A

After the number is subject to the effect of n - views triangular upward direction

Previously wrote this question, but forgot how to do a ,,, and back again and found that the n-th after the total number is 2 ^ n + (2 ^ n +!) / 2 Ge ,, but make life difficult part of the data, may the card is a long long, then looked at the others wrote.

Each upward-pointing triangle exploded law can be decomposed into three new upward triangle and a downward triangle, the downward triangle can be decomposed into three new triangle and a downward triangle upwardly

所以  b(n)=3*b(n-1)+a(n-1)....　　a(n)=3*a(n-1)+b(n-1);

Structure matrix 

3  1

1  3     

 

b(n-1)  0;

a(n-1)  0

OVER

//n  b(shang)   s(xia)
//0    1            0
//1         3            1
//2    10            6
//b(1)=1
//a(1)=3
//bn=3b(n-1)+a(n-1);
//an=3a(n-1)+b(n-1);
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
struct stu{
    ll arr[4][4];
};
stu mul(stu x,stu y){
    stu ans;
    memset(ans.arr,0,sizeof(ans.arr));
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++){
                ans.arr[i][j]=(ans.arr[i][j]%mod+(x.arr[i][k]%mod*y.arr[k][j]%mod)%mod)%mod;
            }
    }
    return ans;
} 
stu ksm(stu x ,ll y){
    stu res;
    memset(res.arr,0,sizeof(res.arr));
    for(int i=0;i<4;i++){
        res.arr[i][i]=1;
    }
    while(y){
        if(y&1)    res=mul(res,x);
        x=mul(x,x);
        y>>=1;
    }
    return res;
}

int main(){
    stu ans,a;
    a.arr[0][0]=3;
    a.arr[0][1]=1;
    a.arr[1][0]=1;
    a.arr[1][1]=3;
    memset(ans.arr,0,sizeof(ans.arr));
    ans.arr[0][0]=1;
    ll n;
    cin>>n;
    if(n==0){
        cout<<1<<endl;
    }
    else {
        ans=ksm(a,n-1);
        ans=mul(ans,a);
        cout<<ans.arr[0][0]%mod<<endl;
    }
    return 0;
}