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 <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(n); i++)
#define int long long
#define double long double
#define mod 1000000007
#define F first
#define S second
#define P pair<long long,long long>
#define all(a) a.begin(),a.end()
#define INF 1000000000000000000
using namespace std;
signed main(){
    int n; cin>>n;
    
    int dp[110][4][4][4]; //[ここまでの方法の数][三文字前][二文字前][一文字前]
    
    dp[0][3][3][3] = 1; //探索時、i==1の時に　”+1だけ” するため
    
    //0...A, 1...C, 2...G, 3...T とする
    for(int i=1; i<=n; i++){//i文字目までの方法
        rep(j,4){//3文字前
            rep(k,4){//2文字前
                rep(l,4){//1文字前
                    //i文字目をXとする
                    if(j==0 and k==1 and l==2)continue; //ACGX その前の三文字がACGはだめ！
                    if(j==0 and k==2 and l==1)continue; //AGCX
                    if(j==2 and k==0 and l==1)continue; //GACX
                    
                    rep(m,4){//Xの四通り
                        
                        if(j==0 and l==2 and m==1)continue; //A?GC
                        if(j==0 and k==2 and m==1)continue; //AG?C
                        if(k==0 and l==1 and m==2)continue; //?ACG
                        if(k==0 and l==2 and m==1)continue; //?AGC
                        if(k==2 and l==0 and m==1)continue; //?GAC
                        
                        dp[i][k][l][m] += dp[i-1][j][k][l]; //OK!!!!!
                        dp[i][k][l][m] %= mod; //忘れずに余りを求めておく
                        
                    }
                }
            }
        }
    }
    
    int ans = 0; //答え
    
    //dp[n][0...3][0...3][0...3]を全て足す！
    rep(i,4){
        rep(j,4){
            rep(k,4){
                ans += dp[n][i][j][k];
                ans %= mod; //余りを求める
            }
        }
    }
    
    cout<<ans<<endl;
    
}