Problem p02549 (C++) — buggy submission, verdict: Runtime Error

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)

const int mod = 998244353;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  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; }
};
istream &operator>>(istream &is, mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

int n, k;
int l[11], r[11];
mint dp[200005]; // マスNに行く方法
mint sum[200005];

int main() {
  cin >> n >> k;

  rep(i, k) { cin >> l[i] >> r[i]; }
  dp[1] = 1;
  sum[1] = 1;
  for (int i = 2; i <= n; i++) {
    mint cnt = 0;
    rep(j, k) {
      if (i - r[j] - 1 > 0) {
        cnt += sum[i - l[j]] - sum[i - r[j] - 1];
      } else if (i - l[j] > 0) {
        cnt += sum[i - l[j]];
      }
    }
    dp[i] = cnt;
    sum[i] = sum[i - 1] + cnt;
  }

  cout << dp[n] << endl;
  return 0;
}

Error: 0

Fixed submission:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)

const int mod = 998244353;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  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; }
};
istream &operator>>(istream &is, mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

int n, k;
int l[11], r[11];
mint dp[200005]; // マスNに行く方法
mint sum[200005];

int main() {
  cin >> n >> k;

  rep(i, k) { cin >> l[i] >> r[i]; }
  dp[1] = 1;
  sum[1] = 1;
  for (int i = 2; i <= n; i++) {
    mint cnt = 0;
    rep(j, k) {
      if (i - r[j] - 1 > 0) {
        cnt += sum[i - l[j]] - sum[i - r[j] - 1];
      } else {
        cnt += sum[i - l[j]];
      }
    }
    dp[i] = cnt;
    sum[i] = sum[i - 1] + cnt;
  }

  cout << dp[n] << endl;
  return 0;
}