51nod 1668 非010串（矩阵快速幂）

#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

const ll mod = 1e9+7;

mat mul(mat& A, mat &B)
{
mat C(A.size(),vec(B[0].size()));
for(int i = 0; i < A.size(); ++i)
for(int k = 0; k < B.size(); ++k)
for(int j = 0; j < B[0].size(); ++j)
C[i][j] = (C[i][j]+A[i][k]*B[k][j]+mod)%mod;
return C;
}

mat pow(mat A, ll n)
{
mat B(A.size(),vec(A.size()));
for(int i = 0; i < A.size(); ++i)
B[i][i] = 1;
while(n > 0)
{
if(n&1) B = mul(B,A);
A = mul(A,A);
n >>= 1;
}
return B;
}

void solve(ll n)
{
mat A(3,vec(3));
A[0][0] = 2;
A[0][1] = -1;
A[0][2] = 1;
A[1][0] = 1;
A[1][1] = 0;
A[1][1] = 0;
A[2][0] = 0;
A[2][1] = 1;
A[2][2] = 0;
A = pow(A,n);
cout << (A[0][0]+A[2][0]+mod)%mod << endl;
}

int main()
{
ll n;
cin >> n;
solve(n);
return 0;
}