Problem:
Score : 400 points

Problem Statement

You are given an integer N. Find the number of strings of length N that satisfy the following conditions, modulo 10^9+7:

The string does not contain characters other than A, C, G and T.

The string does not contain AGC as a substring.

The condition above cannot be violated by swapping two adjacent characters once.

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

3 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of strings of length N that satisfy the following conditions, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

61

There are 4^3 = 64 strings of length 3 that do not contain characters other than A, C, G and T. Among them, only AGC, ACG and GAC violate the condition, so the answer is 64 - 3 = 61.

Sample Input 2

4

Sample Output 2

230

Sample Input 3

100

Sample Output 3

388130742

Be sure to print the number of strings modulo 10^9+7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;


//int dp[101][4][4][4];
vector<vector<vector<vector<int>>>> dp(101, vector<vector<vector<int>>>(4, vector<vector<int>>(4, vector<int>(4))));

int main() {
	int N;
	cin >> N;

	dp[0][3][3][3] = 1;
	int mod = 1e9 + 7;

	//文字列の文字数
	for (int len = 0; len < N; len++)
	{
		//最後から1文字目の文字
		for (int c1 = 0; c1 < 4; c1++)
		{
			//最後から2文字目の文字
			for (int c2 = 0; c2 < 4; c2++)
			{
				//最後から3文字目の文字
				for (int c3 = 0; c3 < 4; c3++)
				{
					//条件に当てはまるものがない場合はcontinue;
					if (dp[len][c1][c2][c3] == 0) continue;

					//新しく追加する文字
					for (int a = 0; a < 4; a++)
					{
						//ダメな5つの条件をカットする
						if (a == 2 && c1 == 1 && c2 == 0) continue;
						if (a == 2 && c1 == 0 && c2 == 1) continue;
						if (a == 1 && c1 == 2 && c2 == 0) continue;
						if (a == 2 && c1 == 1 && c3 == 0) continue;
						if (a == 2 && c2 == 1 && c3 == 0) continue;

						//ダメな条件を抜けたら、aを足した文字列が作れる

						// S = ...... c3 c2 c1
						// nextS = ...... c2 c1 a

						dp[len + 1][a][c1][c2] += dp[len][c1][c2][c3];
						dp[len + 1][a][c1][c2] %= mod;
					}
				}
			}
		}
	}

	int ans = 0;

	//長さNの全部の答えを纏める
	//最後から1文字目の文字
	for (int c1 = 0; c1 < 4; c1++)
	{
		//最後から2文字目の文字
		for (int c2 = 0; c2 < 4; c2++)
		{
			//最後から3文字目の文字
			for (int c3 = 0; c3 < 4; c3++)
			{
				ans += dp[N][c1][c2][c3];
				ans %= mod;
			}
		}
	}

	cout << ans << endl;
}