Problem:
Score : 800 points

Problem Statement

Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.

Constraints

1 \leq N \leq 3000

1 \leq X \leq 2N

N and X are integers.

Input

Input is given from Standard Input in the following format:

N X

Output

Print the number, modulo 998244353, of sequences that satisfy the condition.

Sample Input 1

3 3

Sample Output 1

14

14 sequences satisfy the condition: (0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(0,2,0),(0,2,2),(1,0,0),(1,0,1),(1,1,0),(2,0,0),(2,0,2),(2,2,0) and (2,2,2).

Sample Input 2

8 6

Sample Output 2

1179

Sample Input 3

10 1

Sample Output 3

1024

Sample Input 4

9 13

Sample Output 4

18402

Sample Input 5

314 159

Sample Output 5

459765451

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;


// chmax, chmin
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;
}

template <typename T>
function<T(T, T)> op_max = [](T a, T b) -> T { return max(a, b); };
template <typename T>
function<T(T, T)> op_min = [](T a, T b) -> T { return min(a, b); };

template <typename T>
function<T(T, T)> op_sum = [](T a, T b) -> T { return a + b; };

namespace detail {
template <typename T, std::size_t NDIMS>
struct vector_builder {
  using type = std::vector<typename vector_builder<T, NDIMS - 1>::type>;

  static type make(std::vector<std::size_t> dims, const T& v = {}) {
    const auto vec = std::vector<T>(dims.empty() ? 0 : dims.back(), v);
    if (!dims.empty()) dims.pop_back();
    return vector_builder<std::vector<T>, NDIMS - 1>::make(dims, vec);
  }
};

template <typename T>
struct vector_builder<T, 1> {
  using type = std::vector<T>;

  static type make(std::vector<std::size_t> dims, const T& v = {}) {
    return type(dims.empty() ? 0 : dims.back(), v);
  }
};
}  // namespace detail

template <typename T, typename... SIZE_T>
auto exvector(const T& v, SIZE_T... dims) {
  static_assert(sizeof...(dims) != 0, "invalid dimension");
  std::vector<std::size_t> vec_dims{dims...};
  return detail::vector_builder<T, sizeof...(dims)>::make(vec_dims, v);
}

using namespace std;
//********************  dumps ************************//

template <typename T>
void dump(const T& data, vector<int>& iter) {
  cout << data << " ";
}
template <typename T>
void dump(const vector<T>& data, vector<int>& iter) {
  for (auto elem : data) dump(elem, iter);
  cout << endl;
}

template <typename T>
void dump(const vector<vector<T>>& data, vector<int>& iter) {
  for (auto elem : iter) {
    cout << "[" << elem << "]";
  }
  cout << endl;
  for (int i = 0; i < data.size(); i++) {
    iter.push_back(i);
    dump(data[i], iter);
    iter.pop_back();
  }
  cout << endl;
}

template <typename T>
void dump(const vector<T>& data, int dummy) {
  for (int i = 0; i < data.size(); i++) {
    cout << "[" << i << "] " << data[i] << endl;
  }
}
template <typename T>
void dump(const T& data) {  // T : data, U = base type val.
  vector<int> iter;
  dump(data, iter);
}
///////////////////////////////////////////////

const long long MOD = 998244353;
long long mul(long long a, long long b, const long long MOD) {
  return b ? (mul(a * 2, b / 2, MOD) + (b & 1 ? a : 0)) % MOD : 0;
}

long long bpow(long long a, long long b, const long long MOD) {
  return (b ? bpow(a * a % MOD, b / 2, MOD) * (b & 1 ? a : 1) : 1) % MOD;
}

long long inv(long long a, const long long MOD) {
  return bpow(a, MOD - 2, MOD);
}

vector<long long> MODS = {998244353};
template <int kind = 0>
class mint {
 public:
  long long v;

  mint() : v(0) {}
  mint(long long v) : v((v % MODS[kind] + MODS[kind]) % MODS[kind]) {}
  long long get_mod() { return MODS[kind]; }
  long long get_val() { return v; }
};

template <int kind>
ostream& operator<<(ostream& os, const mint<kind>& x) {
  return os << x.v;
}

template <int kind>
bool operator==(mint<kind>& a, mint<kind> b) {
  return a.v == b.v;
}
template <int kind>
mint<kind>& operator+=(mint<kind>& a, mint<kind> b) {
  return a = a.v + b.v;
}
template <int kind>
mint<kind>& operator-=(mint<kind>& a, mint<kind> b) {
  return a = a.v - b.v;
}
template <int kind>
mint<kind>& operator*=(mint<kind>& a, mint<kind> b) {
  return a = a.v * b.v;
}
template <int kind>
mint<kind>& operator/=(mint<kind>& a, mint<kind> b) {
  return a = a.v * inv(b.v, a.get_mod());
}
template <int kind>
mint<kind> operator+(mint<kind> a, mint<kind> b) {
  return a += b;
}
template <int kind>
mint<kind> operator-(mint<kind> a, mint<kind> b) {
  return a -= b;
}
template <int kind>
mint<kind> operator*(mint<kind> a, mint<kind> b) {
  return a *= b;
}
template <int kind>
mint<kind> operator/(mint<kind> a, mint<kind> b) {
  return a /= b;
}
template <int kind>
mint<kind>& operator+=(mint<kind>& a, long long b) {
  return a = a.v + b;
}
template <int kind>
mint<kind>& operator-=(mint<kind>& a, long long b) {
  return a = a.v - b;
}
template <int kind>
mint<kind>& operator*=(mint<kind>& a, long long b) {
  return a = a.v * b;
}
template <int kind>
mint<kind>& operator/=(mint<kind>& a, long long b) {
  return a = a.v * inv(b, a.get_mod());
}
template <int kind>
mint<kind> operator+(mint<kind> a, long long b) {
  return a += b;
}
template <int kind>
mint<kind> operator-(mint<kind> a, long long b) {
  return a -= b;
}
template <int kind>
mint<kind> operator*(mint<kind> a, long long b) {
  return a *= b;
}
template <int kind>
mint<kind> operator/(mint<kind> a, long long b) {
  return a /= b;
};

vector<mint<>> kaijo;
void init() {
  kaijo.push_back(mint<>(1));
  for (int i = 1; i <= 400000; i++) {
    kaijo.push_back(kaijo.back() * i);
  }
}

mint<> comb(LL N, LL K) {
  if (K < 0) return mint<>(0);
  if (K > N) return mint<>(0);
  return kaijo[N] / kaijo[N - K] / kaijo[K];
}


void solve(long long N, long long X){
  init();
  mint<> zero = 0;
  auto dp = exvector(zero, 2*N+5, N+5);
  dp[0][1] = 1;

  for(int i=1; i<=2*N; i++) {
    for(int j=0; j<=N+1; j++) {
      dp[i][j] = 0;
      if(j>0) dp[i][j] += dp[i-1][j-1];
      if(i>1 && j>0) dp[i][j] += dp[i-2][j-1];
    }
  }
  // in_1.txt

  mint<> ans = 0;
  for(int i=0; i<X-1; i++) {
    for(int k=0; k<=N+1; k++)
      ans += dp[i][k] * comb(N, k-1);

  }
  for(int i=0; i<=N; i++) {
    for(int k=0; k<=N+1; k++) {
      if(X-1-i*2 >= 0) 
        ans += dp[X-1-i*2][k] * comb(N, k-1 + i*2);
    }
  }
  if(X%2==1) {
    for(int i=X; i<=N; i++) {
      int hako = i+1;
      int nokori = N-i;
      ans += comb(N, nokori);

    }
  }

  cout << ans << endl;



}

int main(){ 
    cout << setprecision(10);
    long long N;
    scanf("%lld",&N);
    long long X;
    scanf("%lld",&X);
    solve(N, X);
    return 0;
}