51Nod - 1013 3的幂的和

#include <iostream> 
#include <vector> 
using namespace std;
const int MOD = 1e9 + 7; 

vector<vector<long long> > multiple(const vector<vector<long long> >& a, const vector<vector<long long> >& b){
	vector<vector<long long> > ans(a.size(), vector<long long>(b[0].size(), 0) ); 
	for(int i=0; i<a.size(); ++i){
		for(int j=0; j<b[0].size(); ++j){
			for(int k=0; k<a[0].size(); ++k){
				ans[i][j] += (a[i][k]*b[k][j]) % MOD; 
				ans[i][j] = ans[i][j] % MOD; 
			}
		}
	}
	return ans; 
}

vector<vector<long long> > Pow(vector<vector<long long> > M, int p){
	vector<vector<long long> > ans = M; 
	p--; 
	while(p > 0){
		if(p%2 == 1){
			ans = multiple(ans, M); 
		}
		M = multiple(M, M); 
		p = p/2; 
	} 
	return ans; 
}



int main(){
//	freopen("in.txt", "r", stdin); 

	int N, ans;
	vector<vector<long long> > t(2, vector<long long>(2, 0)); 
	t[0][0] = 3;   t[0][1] = 1; 
	t[1][0] = 0;   t[1][1] = 1; 
	while(scanf("%d", &N) != EOF){
		if(N <= 0){
			ans = 1; 
		}else{
			vector<vector<long long> > tg = Pow(t, N); 
			ans = int((tg[0][0]*1 + tg[0][1]*1) % MOD); 
		}
		printf("%d\n", ans ); 

	}
	return 0; 
}