矩阵快速幂模板 参考bjfu1440

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<map>
#include<set>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff                                            //INT_MAX
#define inf 0x3f3f3f3f                                            //int??????????????????
#define FOR(i,a) for((i)=0;i<(a);(i)++)                            //[i,a);
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
#define LL long long
const double PI = acos(-1.0);
template<class T> T gcd(T a, T b) {return b ? gcd(b, a % b) : a;}
template<class T> T lcm(T a, T b) {return a / gcd(a, b) * b;}
template<class T> inline T Min(T a, T b) {return a < b ? a : b;}
template<class T> inline T Max(T a, T b) {return a > b ? a : b;}
using namespace std;
LL n,m;
#define MOD 1000000007							//要取余的数
#define STAND 2									//行标
#define LIST 2									//列标
struct Matrix{
	LL M[STAND][LIST];
	friend Matrix operator * (Matrix a, Matrix b);
};
Matrix operator * (Matrix a, Matrix b){
	Matrix tmp;
	for(int i = 0; i < STAND; ++i){
		for(int j = 0; j < LIST; ++j){
			tmp.M[i][j] = 0;
			for(long long k = 0; k < 2; ++k){
				tmp.M[i][j] = (tmp.M[i][j] + a.M[i][k] * b.M[k][j]) % MOD;
			}
		}
	}
	return tmp;
}
LL Matrix_fast_mod(LL Power){					//Power ：幂
	Matrix base,ans;							//base ：要乘的矩阵		ans ： 答案矩阵
	//以下根据实际情况赋值
	base.M[0][0] = base.M[1][0] = 2;
	base.M[0][1] = 1;
	base.M[1][1] = 0;
	ans.M[0][0] = 3;
	ans.M[0][1] = 1;
	ans.M[1][0] = ans.M[1][1] = 0;
	//以上根据实际情况赋值
	while(Power){
		if(Power & 1){
			ans = ans * base;
		}
		base = base * base;
		Power >>= 1;
	}
	return ans.M[0][1];
}
int main() {
#ifndef ONLNE_JUDGE
//	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
#endif
	while(~sfI(n)){
		pfI(Matrix_fast_mod(n));
	}
	return 0;
}