Problem p02632 (C++) — buggy submission, verdict: Runtime Error

#pragma GCC optimize("O3")
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;
using base = complex<double>;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define M 1000000007
#define M2 998244353
#define ll long long
#define pll pair<long, long>
#define REP(i, a, b) for (ll i = a; i < b; i++)
#define REPI(i, a, b) for (ll i = b - 1; i >= a; i--)
#define ff first
#define ss second
#define pb push_back
#define db pop_back
#define mp make_pair
#define mt make_tuple
#define g(a, b) get<a>(b)
#define INF (ll)1e18 + 100
#define vl vector<ll>
#define vi vector<int>
#define vll vector<pair<ll, ll>>
#define vii vector<pair<int, int>>
#define all(v) v.begin(), v.end()
#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))
#define INT_SIZE 32
/*SOME BITMASK KNOWLEDGE
1)x & (x - 1):sets the last one bit of x to zero
power of two exactly when x & (x − 1) = 0.
2)x & -x:sets all the one bits to zero, except last one bit
3)x | (x - 1):inverts all the bits after the last one bit*/
#define endl "\\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define o_set                                                                  \\
  tree<int, null_type, less<int>, rb_tree_tag,                                 \\
       tree_order_statistics_node_update>
#define o_setll                                                                \\
  tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
// member functions :
// 1. order_of_key(k) : number of elements strictly lesser than k
// 2. find_by_order(k) : k-th element in the set
ll powM(ll x, ll y, ll m) {
  ll ans = 1, r = 1;
  x %= m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= m;
    }
    r <<= 1;
    x *= x;
    x %= m;
  }
  return ans;
}
ll modI(ll a, ll m) {
  ll m0 = m, y = 0, x = 1;
  if (m == 1)
    return 0;
  while (a > 1) {
    ll q = a / m;
    ll t = m;
    m = a % m;
    a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0)
    x += m0;
  return x;
}

ll logint(ll x, ll y) {
  ll ans = 0;
  ll a = 1;
  for (ll i = 0; i <= x; i++) {
    if (x < a) {
      return ans;
    }
    ans++;
    a *= y;
  }
  return -1;
}
ll extended(ll a, ll b, ll &x, ll &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  ll x1, y1;
  ll d = extended(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
const int N = 2e6 + 5;
ll fact[N];
ll inv(ll i) {
  if (i < 0) {
    return 1;
  } else {
    return modI(fact[i], M);
  }
}
void solve() {
  ll k;
  cin >> k;
  string s;
  cin >> s;
  ll n = s.size();
  fact[0] = 1;
  REP(i, 1, N) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= M;
    // inv[i] = modI(fact[i], M);
  }
  // ll dp[n + 1];
  // REP(i, 0, n + 1) {
  //     dp[i] = 0;
  // }
  // dp[0] = 1;
  // REP(i, 0, n + k) {
  //     dp[n] *= 26;
  //     dp[n] %= M;
  //     for (ll j = n - 1; j >= 0; j--) {
  //         dp[j] *= 25;
  //         dp[j] %= M;
  //         if (j > 0) {
  //             dp[j] += dp[j - 1];
  //             dp[j] %= M;
  //         }
  //     }
  // }
  // cout << dp[n] % M;
  ll ans = 0;
  REP(i, 0, k + 1) {
    // ans += ((powM(26, k - i, M) * powM(25, i, M)) % M * (fact[n + i - 1] *
    // (((inv(i - 1) * inv(n)) % M) % M)) % M;
    ll temp = (powM(26, k - i, M) * powM(25, i, M)) % M;
    temp *= fact[n + i - 1];
    temp %= M;
    temp *= inv(n - 1) % M;
    temp %= M;
    temp *= inv(i) % M;
    temp %= M;
    ans += temp;
    ans %= M;
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  // freopen("input.txt", "r", stdin);
  // freopen("input.txt", "w", stdout);
  ll ntc;
  // cin >> ntc;
  ntc = 1;
  REP(tc, 1, ntc + 1) {
    // cout << "Case #" << tc << ": ";
    solve();
  }
}

Error: WA

Fixed submission:

#pragma GCC optimize("O3")
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;
using base = complex<double>;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define M 1000000007
#define M2 998244353
#define ll long long
#define pll pair<long, long>
#define REP(i, a, b) for (ll i = a; i < b; i++)
#define REPI(i, a, b) for (ll i = b - 1; i >= a; i--)
#define ff first
#define ss second
#define pb push_back
#define db pop_back
#define mp make_pair
#define mt make_tuple
#define g(a, b) get<a>(b)
#define INF (ll)1e18 + 100
#define vl vector<ll>
#define vi vector<int>
#define vll vector<pair<ll, ll>>
#define vii vector<pair<int, int>>
#define all(v) v.begin(), v.end()
#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))
#define INT_SIZE 32
/*SOME BITMASK KNOWLEDGE
1)x & (x - 1):sets the last one bit of x to zero
power of two exactly when x & (x − 1) = 0.
2)x & -x:sets all the one bits to zero, except last one bit
3)x | (x - 1):inverts all the bits after the last one bit*/
#define endl "\\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define o_set                                                                  \\
  tree<int, null_type, less<int>, rb_tree_tag,                                 \\
       tree_order_statistics_node_update>
#define o_setll                                                                \\
  tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
// member functions :
// 1. order_of_key(k) : number of elements strictly lesser than k
// 2. find_by_order(k) : k-th element in the set
ll powM(ll x, ll y, ll m) {
  ll ans = 1, r = 1;
  x %= m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= m;
    }
    r <<= 1;
    x *= x;
    x %= m;
  }
  return ans;
}
ll modI(ll a, ll m) {
  ll m0 = m, y = 0, x = 1;
  if (m == 1)
    return 0;
  while (a > 1) {
    ll q = a / m;
    ll t = m;
    m = a % m;
    a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0)
    x += m0;
  return x;
}

ll logint(ll x, ll y) {
  ll ans = 0;
  ll a = 1;
  for (ll i = 0; i <= x; i++) {
    if (x < a) {
      return ans;
    }
    ans++;
    a *= y;
  }
  return -1;
}
ll extended(ll a, ll b, ll &x, ll &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  ll x1, y1;
  ll d = extended(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
const int N = 1e6 + 5;
ll fact[N];
ll inv(ll i) {
  if (i < 0) {
    return 1;
  } else {
    return modI(fact[i], M);
  }
}
void solve() {
  ll k;
  cin >> k;
  string s;
  cin >> s;
  ll n = s.size();
  fact[0] = 1;
  REP(i, 1, N) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= M;
    // inv[i] = modI(fact[i], M);
  }
  // ll dp[n + 1];
  // REP(i, 0, n + 1) {
  //     dp[i] = 0;
  // }
  // dp[0] = 1;
  // REP(i, 0, n + k) {
  //     dp[n] *= 26;
  //     dp[n] %= M;
  //     for (ll j = n - 1; j >= 0; j--) {
  //         dp[j] *= 25;
  //         dp[j] %= M;
  //         if (j > 0) {
  //             dp[j] += dp[j - 1];
  //             dp[j] %= M;
  //         }
  //     }
  // }
  // cout << dp[n] % M;
  ll ans = 0;
  REP(i, 0, k + 1) {
    // ans += ((powM(26, k - i, M) * powM(25, i, M)) % M * (fact[n + i - 1] *
    // (((inv(i - 1) * inv(n)) % M) % M)) % M;
    ll temp = (powM(26, k - i, M) * powM(25, i, M)) % M;
    temp *= fact[n + i - 1];
    temp %= M;
    temp *= inv(n - 1) % M;
    temp %= M;
    temp *= inv(i) % M;
    temp %= M;
    ans += temp;
    ans %= M;
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  // freopen("input.txt", "r", stdin);
  // freopen("input.txt", "w", stdout);
  ll ntc;
  // cin >> ntc;
  ntc = 1;
  REP(tc, 1, ntc + 1) {
    // cout << "Case #" << tc << ": ";
    solve();
  }
}