Submission #1192394

#	Time	Username	Problem	Language	Result	Execution time	Memory
1192394		stefanopulos	Chessboard (IZhO18_chessboard)	C++20	100 / 100	385 ms	9284 KiB

chessboard

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;

#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
 
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const ll inf = (ll)1e15 + 5;
const int mxN = 100005; 

int n, k;

vector<pii> in[mxN];
vector<pii> out[mxN];

ll resi(int d){
    if(d == n)return inf;

    ll ans1 = 0, ans2 = 0; int cnt[2] = {0, 0};
    ff(i,0,n - 1){
        for(auto c : in[i]){
            int l = c.fi;
            int r = c.se;

            if(l / d == r / d){
                int m = l / d;
                cnt[m % 2] += r - l + 1;
                continue;
            }

            if(l % d != 0){
                int m = l / d;
                cnt[m % 2] += (m + 1) * d - l;
                l = (m + 1) * d;
            }

            if(r % d != d - 1){
                int m = r / d;
                cnt[m % 2] += r - m * d + 1;
                r = m * d - 1;
            }

            if(l <= r){
                int mL = l / d;
                int mR = r / d;
                int len = mR - mL + 1;
                if(mL % 2 == 0){
                    cnt[0] += ((len + 1) / 2) * d;
                    cnt[1] += (len / 2) * d;
                }
                else{
                    cnt[0] += (len / 2) * d;
                    cnt[1] += ((len + 1) / 2) * d;
                }
            }

        }

        int m = i / d;
        if(m % 2 == 0){
            ans1 += cnt[0] + ((n / d) / 2) * d - cnt[1];
            ans2 += ((n / d + 1) / 2) * d - cnt[0] + cnt[1];
        }
        else{   
            ans1 += ((n / d + 1) / 2) * d - cnt[0] + cnt[1];
            ans2 += cnt[0] + ((n / d) / 2) * d - cnt[1];
        }

        for(auto c : out[i]){
            int l = c.fi;
            int r = c.se;

            if(l / d == r / d){
                int m = l / d;
                cnt[m % 2] -= r - l + 1;
                continue;
            }

            if(l % d != 0){
                int m = l / d;
                cnt[m % 2] -= (m + 1) * d - l;
                l = (m + 1) * d;
            }

            if(r % d != d - 1){
                int m = r / d;
                cnt[m % 2] -= r - m * d + 1;
                r = m * d - 1;
            }

            if(l <= r){
                int mL = l / d;
                int mR = r / d;
                int len = mR - mL + 1;
                if(mL % 2 == 0){
                    cnt[0] -= ((len + 1) / 2) * d;
                    cnt[1] -= (len / 2) * d;
                }
                else{
                    cnt[0] -= (len / 2) * d;
                    cnt[1] -= ((len + 1) / 2) * d;
                }
            }

        }

    }

    return min(ans1, ans2);

}

int main(){
    cin.tie(0)->sync_with_stdio(0);

    cin >> n >> k;
    ff(i,1,k){
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2; x1--,y1--,x2--,y2--;
        in[x1].pb({y1, y2});
        out[x2].pb({y1, y2});
    }

    ll rez = inf;
    for(int i = 1; i * i <= n; i++){
        if(n % i == 0){
            rez = min(rez, resi(i));
            if(n / i != i)rez = min(rez, resi(n / i));
        }
    }

    cout << rez << '\n';

    return 0;
}
/*

2 0

6 8
3 3 3 3
1 2 1 2
3 4 3 4
5 5 5 5
4 3 4 3
4 4 4 4
2 1 2 1
3 6 3 6

4 1
4 1 4 4

// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 8 / 8
• Subtask 2: 8 / 8
• Subtask 3: 15 / 15
• Subtask 4: 16 / 16
• Subtask 5: 23 / 23
• Subtask 6: 30 / 30