Submission #58131

#	Time	Username	Problem	Language	Result	Execution time	Memory
58131		Benq	Chessboard (IZhO18_chessboard)	C++14	100 / 100	970 ms	87968 KiB

chessboard

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

ll N,K,ans;
vector<pair<pi,pi>> v;

ll solve(int x, int y, int len) {
    int X = x%(2*len), Y = y%(2*len);
    // suppose bottom left square is black 
    ll z = (x/(2*len))*len;
    z += min(X,len);
    
    ll res = (ll)(y/(2*len))*len*x;
    res += min(Y,len)*z;
    res += max(Y-len,0)*(x-z);
    return res;
}

ll correct(pair<pi,pi> a, int len) {
    return solve(a.f.s,a.s.s,len)-solve(a.f.f-1,a.s.s,len)-solve(a.f.s,a.s.f-1,len)+solve(a.f.f-1,a.s.f-1,len);
}

ll area(pair<pi,pi> a) {
    return (ll)(a.f.s-a.f.f+1)*(a.s.s-a.s.f+1);
}

ll solve(int len) {
    ll ans = N*N-correct({{1,N},{1,N}},len);
    for (auto a: v) {
        ll x = correct(a,len);
        ans -= (area(a)-x);
        ans += x;
    }
    return ans;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> K;
    F0R(i,K) {
        pair<pi,pi> t; cin >> t.f.f >> t.s.f >> t.f.s >> t.s.s;
        v.pb(t);
    }
    FOR(i,1,N) if (N % i == 0) {
        ll t = solve(i);
        t = max(t,N*N-t);
        ans = max(ans,t);
    }
    cout << N*N-ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

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