이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, k, nrec;
int lx[N], ly[N], rx[N], ry[N];
long long ans = 1e18;
int get(int t, int l, int r) {
// count the number of x such that l <= x <= r and (x / k) % 2 == t
if (l / k == r / k) {
if ((l / k) % 2 == t) return r - l + 1; else return 0;
}
int ret = 0;
if ((l / k) % 2 == t) ret += ((l / k) + 1) * k - l;
if ((r / k) % 2 == t) ret += r - (r / k) * k + 1;
l = (l / k) + 1;
r = (r / k) - 1;
if (l % 2 != t) ++l;
if (r % 2 != t) --r;
ret += max(0, (r - l) / 2 + 1) * k;
return ret;
}
long long calc(int color, int lx, int ly, int rx, int ry) {
// count the number of cells with (x + y) == color (modulo 2)
int row[2] = {0, 0}, col[2] = {0, 0};
for (int i = 0; i < 2; ++i) {
row[i] = get(i, lx, rx);
col[i] = get(i, ly, ry);
}
long long ret = 0;
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) if ((i + j) % 2 == color) {
ret += 1LL * row[i] * col[j];
}
}
return ret;
}
void solve() {
for (int t = 0; t < 2; ++t) {
// black cells: (x, y) such that (x + y) == t (modulo 2)
long long cur = calc(t, 0, 0, n - 1, n - 1); // number of black cells
for (int i = 1; i <= nrec; ++i) {
cur -= calc(t, lx[i], ly[i], rx[i], ry[i]); // black cells
cur += calc(t ^ 1, lx[i], ly[i], rx[i], ry[i]); // white cells
}
ans = min(ans, cur);
}
}
int main() {
ios_base::sync_with_stdio(false); cin.tie(0);
cin >> n >> nrec;
for (int i = 1; i <= nrec; ++i) {
cin >> lx[i] >> ly[i] >> rx[i] >> ry[i];
--lx[i]; --ly[i]; --rx[i]; --ry[i];
}
for (k = 1; k < n; ++k) if (n % k == 0) {
solve();
}
cout << ans << endl;
}
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