이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//я так много думал, что опять попал
#include <bits/stdc++.h>
#define all(x) x.begin(),x.end()
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define f first
#define s second
#define left(v) v + v
#define right(v) v + v + 1
#define ub upper_bound
#define lb lower_bound
using namespace std;
typedef long long ll;
//17 SEVENTEEN
const long double Pi = acos(-1.0);
const ll dx[] = {0,0,1,-1};
const ll dy[] = {1,-1,0,0};
const ll N = (ll) 1e6 + 17;
const ll M = (ll) 5e3 + 69;
const ll inf = (ll) 1e18 + 3;
const ll mod = (ll) 1e9 + 7;
ll sq(ll x) { return x * x; }
ll zxc = 1;
vector<ll> d;
map<pair<ll, ll>, ll> cnt;
void solve() {
ll n, k;
cin >> n >> k;
for(ll i = 2; i * i <= n; i++) {
if(!(n % i)) {
d.pb(i);
if(n / i != i) d.pb(n / i);
}
}
d.pb(1);
sort(all(d));
// for(ll x : d) cout << x << "\n";
while(k--) {
ll x, y;
cin >> x >> y >> x >> y;
for(ll val : d)
cnt[{val, (x / val + y / val) % 2}]++;
}
ll ans = inf;
for(ll val : d) {
ll m = n / val, x;
if(m % 2) x = sq((m + 1) / 2) + sq(m / 2);
else x = m * m / 2;
ans = min(min(x * sq(val) - cnt[{val, 0}] + cnt[{val, 1}], sq(n) - x * sq(val) - cnt[{val, 1}] + cnt[{val, 0}]), ans);
}
cout << ans;
}
int main(/*Уверенно*/) {
//ios_base::sync_with_stdio(0);
// cin.tie(0);
/*
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
*/
// cin >> zxc;
while(zxc--) {
solve();
}
return 0;
}
// さよならさ いかなくちゃ
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