Problem p02635 (C++) — buggy submission, verdict: Runtime Error

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")
// #pragma warning(disable : 4996)

#ifdef _MSC_VER
#include <intrin.h>
#define __builtin_popcount __popcnt
#define __builtin_popcountll __popcnt64
#endif

#include <limits.h>
#include <math.h>
#include <time.h>

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, m, n) for (int i = m; i < n; ++i)
#define FORR(i, m, n) for (int i = m - 1; i >= n; --i)
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v, n) reverse(v, v + n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define print(x) cout << (x) << '\\n'
#define pe(x) cout << (x) << " "
#define DEBUG(x) cout << #x << ": " << x << endl
#define lb(v, n) lower_bound(v.begin(), v.end(), (n))
#define ub(v, n) upper_bound(v.begin(), v.end(), (n))
#define int long long
#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v)                                                         \\
  REP(i, v.size()) cout << v[i] << ((i == v.size() - 1) ? "\\n" : " ")
template <typename T1, typename T2> inline void chmin(T1 &a, T2 b) {
  if (a > b)
    a = b;
}
template <typename T1, typename T2> inline void chmax(T1 &a, T2 b) {
  if (a < b)
    a = b;
}
void fail() {
  cout << -1 << endl;
  exit(0);
}

typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 998244353;
constexpr int MAX = 3030;
const double pi = acos(-1);
constexpr double EPS = 1e-8;
constexpr ll LINF = 1e18 + 1;
constexpr int INF = 1e9 + 1;
// const int dx[4] = { 0,0,-1,1 }, dy[4] = { 1,-1,0,0 };

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
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;
}
ll add(ll x, ll y) {
  x += y;
  if (x >= MOD)
    return x - MOD;
  return x;
}
ll sub(ll x, ll y) {
  x -= y;
  if (x < 0)
    return x + MOD;
  return x;
}
ll mult(ll x, ll y) { return (x * y) % MOD; }
ll bin_pow(ll x, ll p) {
  if (p == 0)
    return 1;
  if (p & 1)
    return mult(x, bin_pow(x, p - 1));
  return bin_pow(mult(x, x), p / 2);
}

// const int mod = 1000000007;
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, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

// int cnt[303];
mint dp[303][303][303][3]; // i番目で、j回使用済みでk個1を持っている
vector<int> cnt;
void solve() {
  string S;
  cin >> S;
  int K;
  cin >> K;
  VREVERSE(S);
  S += '0';
  int N = S.size();
  int m = 0;
  int c = 0;
  int sum = 0;
  REP(i, N) {
    if (S[i] == '1') {
      sum++;
      c++;
    } else {
      cnt.push_back(c);
      c = 0;
    }
  }
  K = min(K, sum);
  int n = cnt.size();
  dp[0][0][0][0] = 1;
  REP(i, n + 1) {
    // if (i < n) { pe(i); print(cnt[i]); };
    REP(j, K + 1)
    REP(k, K + 1) dp[i][j][k][0] += dp[i][j][k][1] + dp[i][j][k][2];
    REP(j, K + 1) {
      REP(k, j + 1) {
        // 1を取る
        if (i < n && cnt[i]) {
          dp[i + 1][j + 1][k + 1][1] += dp[i][j][k][0];
          int mn = min(j + cnt[i], K);
          int d = mn - j;
          dp[i + 1][j + d + 1][k + d + 1][1] -= dp[i][j][k][0];
        }
        // 1を置く
        dp[i + 1][j][0][2] += dp[i][j][k][0];
        dp[i + 1][j][k + 1][2] -= dp[i][j][k][0];
      }
      // REP(k, K + 1)
    }
    REP(j, K + 1) {
      REP(k, K + 1) {
        dp[i + 1][j + 1][k + 1][1] += dp[i + 1][j][k][1];
        dp[i + 1][j][k + 1][2] += dp[i + 1][j][k][2];
      }
    }
  }
  mint ans = 0;
  REP(j, K + 1) ans += dp[n][j][0][0];
  print(ans);
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  // mt.seed((int)(time(0)));
  // int q;
  // cin >> q;
  // while (q--)
  solve();
}


Error: -11

Fixed submission:

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")
// #pragma warning(disable : 4996)

#ifdef _MSC_VER
#include <intrin.h>
#define __builtin_popcount __popcnt
#define __builtin_popcountll __popcnt64
#endif

#include <limits.h>
#include <math.h>
#include <time.h>

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, m, n) for (int i = m; i < n; ++i)
#define FORR(i, m, n) for (int i = m - 1; i >= n; --i)
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v, n) reverse(v, v + n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define print(x) cout << (x) << '\\n'
#define pe(x) cout << (x) << " "
#define DEBUG(x) cout << #x << ": " << x << endl
#define lb(v, n) lower_bound(v.begin(), v.end(), (n))
#define ub(v, n) upper_bound(v.begin(), v.end(), (n))
#define int long long
#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v)                                                         \\
  REP(i, v.size()) cout << v[i] << ((i == v.size() - 1) ? "\\n" : " ")
template <typename T1, typename T2> inline void chmin(T1 &a, T2 b) {
  if (a > b)
    a = b;
}
template <typename T1, typename T2> inline void chmax(T1 &a, T2 b) {
  if (a < b)
    a = b;
}
void fail() {
  cout << -1 << endl;
  exit(0);
}

typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 998244353;
constexpr int MAX = 3030;
const double pi = acos(-1);
constexpr double EPS = 1e-8;
constexpr ll LINF = 1e18 + 1;
constexpr int INF = 1e9 + 1;
// const int dx[4] = { 0,0,-1,1 }, dy[4] = { 1,-1,0,0 };

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
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;
}
ll add(ll x, ll y) {
  x += y;
  if (x >= MOD)
    return x - MOD;
  return x;
}
ll sub(ll x, ll y) {
  x -= y;
  if (x < 0)
    return x + MOD;
  return x;
}
ll mult(ll x, ll y) { return (x * y) % MOD; }
ll bin_pow(ll x, ll p) {
  if (p == 0)
    return 1;
  if (p & 1)
    return mult(x, bin_pow(x, p - 1));
  return bin_pow(mult(x, x), p / 2);
}

// const int mod = 1000000007;
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, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

// int cnt[303];
mint dp[303][303][303][3]; // i番目で、j回使用済みでk個1を持っている
vector<int> cnt;
void solve() {
  string S;
  cin >> S;
  int K;
  cin >> K;
  VREVERSE(S);
  S += '0';
  int N = S.size();
  int m = 0;
  int c = 0;
  REP(i, N) {
    if (S[i] == '1')
      c++;
    else {
      cnt.push_back(c);
      c = 0;
    }
  }
  int n = cnt.size();
  dp[0][0][0][0] = 1;
  REP(i, n + 1) {
    // if (i < n) { pe(i); print(cnt[i]); };
    REP(j, K + 1)
    REP(k, K + 1) dp[i][j][k][0] += dp[i][j][k][1] + dp[i][j][k][2];
    REP(j, K + 1) {
      REP(k, j + 1) {
        // 1を取る
        if (i < n && cnt[i]) {
          dp[i + 1][j + 1][k + 1][1] += dp[i][j][k][0];
          int mn = min(j + cnt[i], K);
          int d = mn - j;
          dp[i + 1][j + d + 1][k + d + 1][1] -= dp[i][j][k][0];
        }
        // 1を置く
        dp[i + 1][j][0][2] += dp[i][j][k][0];
        dp[i + 1][j][k + 1][2] -= dp[i][j][k][0];
      }
      // REP(k, K + 1)
    }
    REP(j, K + 1) {
      REP(k, K + 1) {
        dp[i + 1][j + 1][k + 1][1] += dp[i + 1][j][k][1];
        dp[i + 1][j][k + 1][2] += dp[i + 1][j][k][2];
      }
    }
  }
  mint ans = 0;
  REP(j, K + 1) ans += dp[n][j][0][0];
  print(ans);
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  // mt.seed((int)(time(0)));
  // int q;
  // cin >> q;
  // while (q--)
  solve();
}