Problem p02549 (C++) — buggy submission, verdict: Time Limit Exceeded

//
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> pint;
typedef pair<ll, ll> pll;
typedef pair<int, pint> ppint;
typedef pair<ll, pll> ppll;
typedef vector<int> vint;
typedef vector<ll> vll;

const double pi = 3.141592653589793;
const int INF10 = 1000000001;
const ll INF15 = 1e15 + 1;
const long long INF18 = 1e18 + 1;
// const int mod = 1000000007;
const int mod = 998244353;
const double EPS = 0.00001;

#define rep(i, n) for (int i = 0; i < (ll)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (ll)(n); i++)
#define rep2(i, start, end) for (int i = (ll)start; i <= (ll)end; i++)
#define vrep(i, n) for (int i = (ll)n - 1; i >= 0; i--)
#define vrep1(i, n) for (int i = (ll)n; i > 0; i--)
#define all(n) n.begin(), n.end()
#define pb push_back
#define debug(x) cerr << #x << ": " << x << '\\n'
#define arep(it, a) for (auto it : a)

struct edge {
  edge(int s, int e, long long c) {
    to = e;
    from = s;
    cost = c;
  }
  edge(int i, long long c = 1) { edge(-1, i, c); }
  int from;
  int to;
  long cost;
};
typedef vector<edge> edges;

// bを何回足せばaを超えるか(O(a/b))
// a+b-1/bとすればよし

// 2進数表示したときの最高桁(O(log n))
int bi_max(long n) {
  int m = 0;
  for (m; (1 << m) <= n; m++)
    ;
  m = m - 1;
  return m;
}
// bi_eに二進数表示したやつを代入(O(log^2 n))
// bitset<N>
// a(n)でnの二進数表示が得られて、a[i]=0or1でi番目のfragが立ってるかわかる x^n
// mod m (nが負の時は０)(O(log n))
long myPow(long x, long n, long m = mod) {
  if (n < 0)
    return 0;
  if (n == 0)
    return 1;
  if (n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}
// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
struct mint {
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint &operator=(const mint &a) {
    x = a.x;
    return *this;
  }
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a) {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a) {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint &operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
  mint &operator++(int) { return *this += mint(1); }
  mint &operator--(int) { return *this -= mint(1); }
  bool operator!=(const int a) { return this->x != a; }
  bool operator==(const int a) { return this->x == a; }
};
istream &operator>>(istream &is, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n) : fact(n + 1), ifact(n + 1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i)
      fact[i] = fact[i - 1] * i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i)
      ifact[i - 1] = ifact[i] * i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
  mint parm(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[n - k];
  }
};
template <class T> bool maxin(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> bool minin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class M, class N> common_type_t<M, N> mygcd(M a, N b) {
  a = abs(a);
  b = abs(b);
  if (a < b)
    return mygcd(b, a);
  M r;
  while ((r = a % b)) {
    a = b;
    b = r;
  }
  return b;
}
template <class M, class N> common_type_t<M, N> mylcm(M a, N b) {
  return (a / mygcd(a, b)) * b;
}

const int N_MAX = 200005;
int n, k;
vector<pint> lr;
mint a[N_MAX], mem[N_MAX];

void Main() {
  int x = 0, y = INF10, z = 1;
  // 入力
  cin >> n >> k;
  rep(i, k) {
    cin >> x >> y;
    lr.pb({x, y});
  }
  // 処理
  sort(all(lr));
  mint X = 0;
  a[1] = 1;
  a[2] = -1;
  rep1(i, n - 1) {
    X = X + a[i];
    rep(j, k) {
      if (i + lr[j].first < N_MAX)
        a[i + lr[j].first] += X;
      if (i + lr[j].second + 1 < N_MAX)
        a[i + lr[j].second + 1] -= X;
    }
  }
  // 出力
  cout << X + a[n] << endl;
}

int main() {
  cin.tie(nullptr);
  cout << fixed << setprecision(12);
  Main();
}

Error: TLE

Fixed submission:

//
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> pint;
typedef pair<ll, ll> pll;
typedef pair<int, pint> ppint;
typedef pair<ll, pll> ppll;
typedef vector<int> vint;
typedef vector<ll> vll;

const double pi = 3.141592653589793;
const int INF10 = 1000000001;
const ll INF15 = 1e15 + 1;
const long long INF18 = 1e18 + 1;
// const int mod = 1000000007;
const int mod = 998244353;
const double EPS = 0.00001;

#define rep(i, n) for (int i = 0; i < (ll)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (ll)(n); i++)
#define rep2(i, start, end) for (int i = (ll)start; i <= (ll)end; i++)
#define vrep(i, n) for (int i = (ll)n - 1; i >= 0; i--)
#define vrep1(i, n) for (int i = (ll)n; i > 0; i--)
#define all(n) n.begin(), n.end()
#define pb push_back
#define debug(x) cerr << #x << ": " << x << '\\n'
#define arep(it, a) for (auto it : a)

struct edge {
  edge(int s, int e, long long c) {
    to = e;
    from = s;
    cost = c;
  }
  edge(int i, long long c = 1) { edge(-1, i, c); }
  int from;
  int to;
  long cost;
};
typedef vector<edge> edges;

// bを何回足せばaを超えるか(O(a/b))
// a+b-1/bとすればよし

// 2進数表示したときの最高桁(O(log n))
int bi_max(long n) {
  int m = 0;
  for (m; (1 << m) <= n; m++)
    ;
  m = m - 1;
  return m;
}
// bi_eに二進数表示したやつを代入(O(log^2 n))
// bitset<N>
// a(n)でnの二進数表示が得られて、a[i]=0or1でi番目のfragが立ってるかわかる x^n
// mod m (nが負の時は０)(O(log n))
long myPow(long x, long n, long m = mod) {
  if (n < 0)
    return 0;
  if (n == 0)
    return 1;
  if (n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}
// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
struct mint {
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint &operator=(const mint &a) {
    x = a.x;
    return *this;
  }
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a) {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a) {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint &operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
  mint &operator++(int) { return *this += mint(1); }
  mint &operator--(int) { return *this -= mint(1); }
  bool operator!=(const int a) { return this->x != a; }
  bool operator==(const int a) { return this->x == a; }
};
istream &operator>>(istream &is, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n) : fact(n + 1), ifact(n + 1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i)
      fact[i] = fact[i - 1] * i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i)
      ifact[i - 1] = ifact[i] * i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
  mint parm(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[n - k];
  }
};
template <class T> bool maxin(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> bool minin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class M, class N> common_type_t<M, N> mygcd(M a, N b) {
  a = abs(a);
  b = abs(b);
  if (a < b)
    return mygcd(b, a);
  M r;
  while ((r = a % b)) {
    a = b;
    b = r;
  }
  return b;
}
template <class M, class N> common_type_t<M, N> mylcm(M a, N b) {
  return (a / mygcd(a, b)) * b;
}

const int N_MAX = 200005;
int n, k;
vector<pint> lr;
mint dp[N_MAX];

void Main() {
  int x = 0, y = INF10, z = 1;
  // 入力
  cin >> n >> k;
  rep(i, k) {
    cin >> x >> y;
    lr.pb({x, y});
  }
  // 処理
  sort(all(lr));
  dp[1] = 1;
  bool tof = false;
  rep1(i, n - 1) {
    tof = false;
    rep(j, k) {
      rep2(l, lr[j].first, lr[j].second) {
        if (i + l > n) {
          tof = true;
          break;
        }
        dp[i + l] += dp[i];
      }
      if (tof)
        break;
    }
  }
  // 出力
  cout << dp[n] << endl;
}

int main() {
  cin.tie(nullptr);
  cout << fixed << setprecision(12);
  Main();
}