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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
//long long
using ll = long long;
 
// pair<int, int>
using PII = pair<int, int>;
 
 
 
//最大値、mod
const int MOD = 1000000007;
const int mod = 1000000007;
const int INF = 1000000000;
const long long LINF = 1e18;
const int MAX = 510000;
 
//出力系
#define print(x) cout << x << endl
#define prints(x) cout << fixed << setprecision(20) << x << endl
#define printc(x) cout << setw(2) << setfill('0') << x << endl;
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
 
// begin() end()
#define all(x) (x).begin(),(x).end()
 
//for
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define rep(i,a,b) for(int i=(a);i<(b);i++)
 
//最大公約数 
unsigned gcd(unsigned a, unsigned b) {
  if(a < b) return gcd(b, a);
  unsigned r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}
 
// 最小公倍数
unsigned lcm(unsigned a, unsigned b){
    return a / gcd(a, b) * b; 
}
 
// a = max(a, b), a = min(a, b)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
// 階乗(MODをとる)
ll pow_mod(ll num, ll pow, ll mod) {
    ll prod = 1;
    num %= mod;
    while (pow > 0) {
        if (pow & 1) prod = prod * num % mod;
        num = num * num % mod;
        pow >>= 1;
    }
    return prod;
}
 
 
 
// 二項係数(MODとる、1 ≦ k ≦ n ≦ 10^7 程度)
// COMinit()
// COM(x, y)
// とコンビで使う
 
// テーブルを作る前処理
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
 
 
// UnionFind
struct UnionFind {
    vector<int> par;
    vector<int> rank;
    vector<ll> Size;
    UnionFind(int n = 1) {
        init(n);
    }
 
    void init(int n = 1) {
        par.resize(n + 1); rank.resize(n + 1); Size.resize(n + 1);
        for (int i = 0; i <= n; ++i) par[i] = i, rank[i] = 0, Size[i] = 1;
    }
 
    int root(int x) {
        if (par[x] == x) {
            return x;
        }
        else {
            int r = root(par[x]);
            return par[x] = r;
        }
    }
 
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
 
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        Size[x] += Size[y];
        return true;
    }
 
    ll size(int x){
        return Size[root(x)];
    }
};

 
//modint構造体 
struct Mint {
    int val;
    Mint inv() const{
        int tmp,a=val,b=mod,x=1,y=0;
        while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
        return Mint(x);
    }
public:
    Mint():val(0){}
    Mint(ll x){if((val=x%mod)<0)val+=mod;}
    Mint pow(ll t){Mint res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;}return res;}
    Mint& operator+=(const Mint& x){if((val+=x.val)>=mod)val-=mod;return *this;}
    Mint& operator-=(const Mint& x){if((val+=mod-x.val)>=mod)val-=mod; return *this;}
    Mint& operator*=(const Mint& x){val=(ll)val*x.val%mod; return *this;}
    Mint& operator/=(const Mint& x){return *this*=x.inv();}
    bool operator==(const Mint& x) const{return val==x.val;}
    bool operator!=(const Mint& x) const{return val!=x.val;}
    bool operator<(const Mint& x) const{return val<x.val;}
    bool operator<=(const Mint& x) const{return val<=x.val;}
    bool operator>(const Mint& x) const{return val>x.val;}
    bool operator>=(const Mint& x) const{return val>=x.val;}
    Mint operator+(const Mint& x) const{return Mint(*this)+=x;}
    Mint operator-(const Mint& x) const{return Mint(*this)-=x;}
    Mint operator*(const Mint& x) const{return Mint(*this)*=x;}
    Mint operator/(const Mint& x) const{return Mint(*this)/=x;}
};
struct factorial {
    vector<Mint> Fact, Finv;
public:
   //factorial fact(10000010);
   //fact.nCr(a, b)
  //「fact」の部分は自由に名前変更可能
    factorial(int maxx){
        Fact.resize(maxx+1),Finv.resize(maxx+1); Fact[0]=Mint(1); rep(i,0,maxx)Fact[i+1]=Fact[i]*(i+1);
        Finv[maxx]=Mint(1)/Fact[maxx]; rrep(i,maxx,0)Finv[i-1]=Finv[i]*i;
    }
    Mint fact(int n,bool inv=0){if(inv)return Finv[n];else return Fact[n];}
    Mint nPr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[n-r];}
    Mint nCr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[r]*Finv[n-r];}
};


// 1 * 2 * 3 .... * n (mod)
ll modfact(ll n) { 
    if (n <= 1) return 1;
    return (n * modfact(n - 1)) % MOD;
}


// kが角度だった場合:cos(k * (PI / 180));
const double PI = acos(-1);


// 多次元 vector 生成   例: auto dp = make_vec<long long>(N+1, 5, 5, 5); 
template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}
template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
  return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}


int main(){
	int n; cin >>n;
	// 0: dammy, 1: 'A', 2: 'G', 3: 'C', 4: 'T'
	auto dp = make_vec<long long>(n + 1, 5, 5, 5);
	dp[0][0][0][0] = 1;
	REP(i, n)REP(a, 5)REP(b, 5)REP(c, 5)REP(d, 5){
		if(b == 1 && c == 2 && d == 3) continue;
		if(b == 2 && c == 1 && d == 3) continue;
		if(b == 1 && c == 3 && d == 2) continue;
		if(a == 1 && c == 2 && d == 3) continue;
		if(a == 1 && b == 2 && d == 3) continue;
		dp[i + 1][b][c][d] += dp[i][a][b][c];
		dp[i + 1][b][c][d] %= MOD;
	}
	ll ans = 0;
	REP(i, 5)REP(j, 5)REP(k, 5){
		if(i == 0 || j == 0 || k == 0)continue; 
		ans += dp[n][i][j][k]; ans %= MOD;
	} 
	print(ans);
	return 0;
}