Submission #442356

#	Time	Username	Problem	Language	Result	Execution time	Memory
442356		DmitryGrigorev	Packing Biscuits (IOI20_biscuits)	C++14	100 / 100	62 ms	1352 KiB

biscuits

#include <bits/stdc++.h>
#define ll long long
#define db long double
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define all(a) a.begin(), a.end()

using namespace std;

const int mod = 1000000007;

void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}

int mult(int a, int b) {
  return a * (ll)b % mod;
}

int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}

ll dp[61][61];

ll count_tastiness(ll x, vector<ll> a) {

  int k = a.size();

  vector<ll> bounds;
  ll sum = 0;

  for (int i = 0; i < 60; ++i) {
    if (i < k) sum += (1LL<<i)*a[i];
    bounds.pb(sum / x);
  }

  for (int i = 0; i < 61; ++i) for (int j = 0; j < 61; ++j) dp[i][j] = 0;
  dp[60][0] = 1;

  for (int i = 60; i > 0; --i) {
    for (int j = 0; j < 61; ++j) {

      ll R = (1LL<<i) - 1;
      if (j > 0) R = bounds[j - 1] % (1LL<<i);
      ll C = bounds[i - 1];

      //////////////////////////////// ZERO TRANSITION


      if (R >= (1LL<<(i-1)) && C >= (1LL<<(i-1))) {
        dp[i - 1][0] += dp[i][j];
      }
      else if (R <= C) {
        dp[i - 1][j] += dp[i][j];
      }
      else {
        dp[i - 1][i] += dp[i][j];
      }

      ////////////////////////////////

      if (R < (1LL<<(i-1)) || C < (1LL<<(i-1))) {
        continue;
      }
      else if (R <= C) {
        dp[i - 1][j] += dp[i][j];
      }
      else {
        dp[i - 1][i] += dp[i][j];
      }

    }


  }

  ll ans = 0;
  for (int j = 0; j < 61; ++j) ans += dp[0][j];
  return ans;
}


#ifdef LOCAL
int main(){
	freopen("A_input.txt", "r", stdin);
	//freopen("A_output.txt", "w", stdout);
	ios_base::sync_with_stdio(0);
	cin.tie(0);

  int k;
  ll x;

  cin >> k >> x;

  vector<ll> v(k);
  for (auto &x : v) cin >> x;

  cout << count_tastiness(x, v) << endl;

}

#endif

• Subtask 1: 9 / 9
• Subtask 2: 12 / 12
• Subtask 3: 21 / 21
• Subtask 4: 35 / 35
• Subtask 5: 23 / 23