“Hard work beats talent when talent doesn’t work hard”

Contest

You can find the contest here.

The solutions will be uploaded after the contest.

The upsolving contest is here.

Fifty-Fifty

Is you sort the string you dont need to do a lot of checks.

#include <bits/stdc++.h>

#define all(A) begin(A), end(A)
#define rall(A) rbegin(A), rend(A)
#define sz(A) int(A.size())
#define pb push_back
#define mp make_pair

using namespace std;

typedef long long ll;
typedef pair <int, int> pii;

int main () {
  ios::sync_with_stdio(false); cin.tie(0);
  string s;
  cin >> s;
  sort(all(s));
  if (s[0] == s[1] and s[2] == s[3] and s[1] != s[2]) cout << "Yes\n";
  else cout << "No\n";
  return (0);
}

Handstand 2

We can fix the first and last digit and generate the number that hold the conditions.

#include <bits/stdc++.h>
     
#define all(A) begin(A), end(A)
#define rall(A) rbegin(A), rend(A)
#define sz(A) int(A.size())
#define pb push_back
#define mp make_pair
     
using namespace std;
     
typedef long long ll;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef vector <ll> vll;
     
int intPower (int a, int b) {
  int ret = 1;
  while (b--) {
    ret *= a;
  }
  return ret;
}

int main () {
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  ll ans = 0;
  for (int a = 1; a <= 9; a++) {
    for (int b = 1; b <= 9; b++) {
      set <int> left, right;
      if (a == b and a <= n) {
        left.insert(a);
        right.insert(a);
      }
      for (int d = 0; d <= 4; d++) {
        int p = intPower(10, d);
        for (int center = 0; center < p; center++) {
          int num1 = a * 10 * p + 10 * center + b;
          int num2 = b * 10 * p + 10 * center + a;
          if (num1 <= n) left.insert(num1);
          if (num2 <= n) right.insert(num2);
        }
      }
      ans += 1LL * sz(left) * sz(right);
    }
  }
  cout << ans << '\n';
  return (0);
}

Count Order

We can just generate all possible permutations and find the indices of the given permutations.

#include <bits/stdc++.h>
     
#define all(A) begin(A), end(A)
#define rall(A) rbegin(A), rend(A)
#define sz(A) int(A.size())
#define pb push_back
#define mp make_pair
     
using namespace std;
     
typedef long long ll;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef vector <ll> vll;
     
int main () {
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  vector <int> p(n);
  vector <int> q(n);
  for (int& elem: p) cin >> elem;
  for (int& elem: q) cin >> elem;
  vector <int> perm(n);
  iota(all(perm), 1);
  int a = -1;
  int b = -1;
  int index = 1;
  do {
    if (perm == p) a = index;
    if (perm == q) b = index;
    index++;
  } while (next_permutation(all(perm)));
  cout << abs(a - b) << '\n';
  return (0);
}

Next Alphabet

A char is stored as an int that represents its ascii representation and the letters “abc..z” are continuous. We can use that fact to void a lot of ifs.

#include <bits/stdc++.h>

#define all(A) begin(A), end(A)
#define rall(A) rbegin(A), rend(A)
#define sz(A) int(A.size())
#define pb push_back
#define mp make_pair

using namespace std;

typedef long long ll;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef vector <ll> vll;

int main () {
  ios::sync_with_stdio(false); cin.tie(0);
  char ch;
  cin >> ch;
  cout << char(ch + 1) << '\n';
  return (0);
}

Equal Sums

For a sequence \(seq_i = a_1, a_2, \dots, a_n\) let \(sum = a_1 + a_2 + \dots + a_n\).

We can store in a map \((sum - a[j]) \to [(i, j)]\)

Then if there are two sequence that holds the conditions, they must have been added to the map.

#include <bits/stdc++.h>
     
#define all(A) begin(A), end(A)
#define rall(A) rbegin(A), rend(A)
#define sz(A) int(A.size())
#define pb push_back
#define mp make_pair
     
using namespace std;
     
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef vector <pii> vpii;
typedef vector <pll> vpll;

int main () {
  ios::sync_with_stdio(false); cin.tie(0);
  int k;
  cin >> k;
  map <int, vpii> mp;
  for (int i = 1; i <= k; i++) {
    int n;
    cin >> n;
    vi arr(n);
    for (int& elem: arr) cin >> elem;
    int sum = accumulate(all(arr), 0);
    for (int j = 0; j < n; j++) {
      if (mp[sum - arr[j]].empty() or
          mp[sum - arr[j]].back().first != i) {
        mp[sum - arr[j]].pb(pii(i, j + 1));
      }
    }
  }
  for (auto row: mp) {
    if (sz(row.second) >= 2) {
      auto pp = row.second[0];
      auto qq = row.second[1];
      cout << "YES\n";
      cout << pp.first << ' ' << pp.second << '\n';
      cout << qq.first << ' ' << qq.second << '\n';
      return (0);
    }
  }
  cout << "NO\n";
  return (0);
}

RGB Boxes

Fix the number of red and green boxes, then you can compute how many blue boxes you need.

#include <bits/stdc++.h>

#define all(A) begin(A), end(A)
#define sz(A) int(A.size())

using namespace std;

typedef long long ll;

int main () {
  ios::sync_with_stdio(false); cin.tie(0);
  int r, g, b, n;
  cin >> r >> g >> b >> n;
  int ans = 0;
  for (int i = 0; i <= n / r + 1; i++) {
    for (int j = 0; j <= n / g + 1; j++) {
      int k = n - i * r - j * g;
      if (k >= 0 and k % b == 0) {
        ans++;
      }
    }
  }
  cout << ans << endl;
  return (0);
}

Next

Previous