Problem:
Score : 400 points

Problem Statement

There are N cells arranged in a row, numbered 1, 2, \ldots, N from left to right.

Tak lives in these cells and is currently on Cell 1. He is trying to reach Cell N by using the procedure described below.

You are given an integer K that is less than or equal to 10, and K non-intersecting segments [L_1, R_1], [L_2, R_2], \ldots, [L_K, R_K].
Let S be the union of these K segments.
Here, the segment [l, r] denotes the set consisting of all integers i that satisfy l \leq i \leq r.

When you are on Cell i, pick an integer d from S and move to Cell i + d. You cannot move out of the cells.

To help Tak, find the number of ways to go to Cell N, modulo 998244353.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq K \leq \min(N, 10)

1 \leq L_i \leq R_i \leq N

[L_i, R_i] and [L_j, R_j] do not intersect (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
L_1 R_1
L_2 R_2
:
L_K R_K

Output

Print the number of ways for Tak to go from Cell 1 to Cell N, modulo 998244353.

Sample Input 1

5 2
1 1
3 4

Sample Output 1

4

The set S is the union of the segment [1, 1] and the segment [3, 4], therefore S = \{ 1, 3, 4 \} holds.

There are 4 possible ways to get to Cell 5:

1 \to 2 \to 3 \to 4 \to 5,

1 \to 2 \to 5,

1 \to 4 \to 5 and

1 \to 5.

Sample Input 2

5 2
3 3
5 5

Sample Output 2

0

Because S = \{ 3, 5 \} holds, you cannot reach to Cell 5.
Print 0.

Sample Input 3

5 1
1 2

Sample Output 3

5

Sample Input 4

60 3
5 8
1 3
10 15

Sample Output 4

221823067

Note that you have to print the answer modulo 998244353.

Solution in C++ (Clang 10.0.0) (judge verdict: Time Limit Exceeded)


#include <iostream>
#include <vector>
#include <array>

namespace
{
  static std::array<long long, 200001> answers = {};

}

namespace ABC179
{

  class D
  {
  public:
    D() = default;
    ~D() = default;

    static constexpr long long DIV = 998244353;

    void Run()
    {
      int N,K;
      std::cin >> N;
      std::cin >> K;

      std::vector<int> S = {};
      S.reserve(N);
      for (int i = 0; i < K; ++i)
      {
        int L, R;
        std::cin >> L;
        std::cin >> R;

        for (int j = L; j <= R; ++j)
        {
          S.push_back(j);
        }
      }
      std::sort(S.begin(), S.end());

      //for (int i = 2; i <= N; ++i)
      //{
      //  long long num = 0;
      //  for (auto n : S)
      //  {
      //    auto rest = i - n;
      //    if (rest == 1) { num++; }//ゴール
      //    else if (rest < 1) { } // たどりつけない
      //    else if (rest > 1)
      //    {
      //      // restにたどり着く方法の数だけ解法がある
      //      num += answers[rest];
      //    }
      //    if (num >= DIV) { num %= DIV; }
      //  }
      //  answers[i] = num;
      //}

      answers[N] = 1;
      for (int i = N; i > 1; --i)
      {
        if (answers[i] == 0) { continue; }
        for (auto n : S)
        {
          auto d = i - n;
          if (d >= 1)
          {
            answers[d] += answers[i];
            if (answers[d] >= DIV) { answers[d] -= DIV; }
          }
          else
          {
            break;
          }
        }
      }

      std::cout << answers[1] << std::endl;
    }
  };

}
int main()
{

  ABC179::D prob;
  prob.Run();

  return 0;
}