Submission #743609

#	Time	Username	Problem	Language	Result	Execution time	Memory
743609		onebit1024	Hexagonal Territory (APIO21_hexagon)	C++17	9 / 100	1 ms	288 KiB

hexagon

#include "hexagon.h"
#define ll long long
#include <vector>
ll mod = 1e9+7;
ll ma(ll a, ll b){
  return (a+b)%mod;
}
ll mm(ll a, ll b){
  return (a*b)%mod;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y);
 
// Function to find modulo inverse of b. It returns
// -1 when inverse doesn't
ll modInverse(ll b, ll m)
{
    ll x, y; // used in extended GCD algorithm
    ll g = gcdExtended(b, m, &x, &y);
 
    // Return -1 if b and m are not co-prime
    if (g != 1)
        return -1;
 
    // m is added to handle negative x
    return (x%m + m) % m;
}
 
// Function to compute a/b under modulo m
ll modDivide(ll a, ll b, ll m)
{
    a = a % m;
    ll inv = modInverse(b, m);
    return (inv * a) % m;
}
 
// C function for extended Euclidean Algorithm (used to
// find modular inverse.
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
    // Base Case
    if (a == 0)
    {
        *x = 0, *y = 1;
        return b;
    }
 
    ll x1, y1; // To store results of recursive call
    ll gcd = gcdExtended(b%a, a, &x1, &y1);
 
    // Update x and y using results of recursive
    // call
    *x = y1 - (b/a) * x1;
    *y = x1;
 
    return gcd;
}

int draw_territory(int N, int A, int B,
                   std::vector<int> D, std::vector<int> L) {
  ll k = L[0]-2;
  if(L[0]==1){
    return ma(mm(A,3),mm(B,2));
  }
  if(L[0]==2){
    return ma(mm(A,6),mm(B,8))%mod;
  }
  ll fr = mm(L[0],3),sr=0;
  if(k%2)sr = mm((k+1)/2,k);
  else sr = mm(k/2,k+1);
  int n = L[0]+2;
  ll val = mm(n,mm(n-1,n-2));
  val = modDivide(val,6,mod);
  val = mm(val,2);
  return ma(mm(B,val),mm(A,ma(sr,fr)));
}

• Subtask 1: 3 / 3
• Subtask 2: 6 / 6
• Subtask 3: 0 / 11
• Subtask 4: 0 / 12
• Subtask 5: 0 / 15
• Subtask 6: 0 / 19
• Subtask 7: 0 / 18
• Subtask 8: 0 / 16