A - Square

• 思维
• 按行或列排序

signed main() {int T = 1;T = read();while (T--) {vector<pii> vec;for (int i = 1; i <= 4; ++i) {int u = read(), v = read();vec.push_back({u, v});}sort(vec.begin(), vec.end());int a = abs(vec[0].second - vec[1].second);int b = abs(vec[0].first - vec[2].first);writeln(a * b);}return 0;
}

B - Arranging Cats

• 思维
• 记录两个的 $1$ 的数量。
• 相同位置同为 $1$ 则不用操作。
• 谁大则答案加上减去相同位置同为 $1$ 的数量（要么补，移，删）

signed main() {int T = 1;T = read();while (T--) {int n = read();string s1, s2; cin >> s1 >> s2;int cnt1 = count(s1.begin(), s1.end(), '1'), cnt2 = count(s2.begin(), s2.end(), '1');int ans = 0;int cnt = 0;
//        tf(cnt1),tf(cnt2);for (int i = 0; i < n; ++i) {if (s1[i] == s2[i] && s1[i] == '1') ++cnt;}if (cnt1 >= cnt2) ans += cnt1 - cnt;else ans += cnt2 - cnt;writeln(ans);}return 0;
}

C - Sending Messages

• 模拟
• 两种方事选择消耗小的一种

signed main() {int T = 1;T = read();while (T--) {int n = read(), f = read(), a = read(), b = read();vector<int> e(n + 1);bool ff = 1;int pre = 0;for (int i = 1; i <= n; ++i) e[i] = read();for (int i = 1; i <= n; ++i) {if ((e[i] - pre) * a <= b) f -= (e[i] - pre) * a;else f -= b;if (f <= 0) {ff = 0;break;}pre = e[i];}ff? puts("yes"): puts("no");}return 0;
}

D - Very Different Array

• 贪心
• $a$ 最小的与 $b$ 最大的匹配，$a$ 最大的与 $b$ 最小的匹配。

signed main() {int T = 1;T = read();while (T--) {int n = read(), m = read();deque<int> a, b;for (int i = 1; i <= n; ++i) {int u = read();a.push_back(u);}for (int i = 1; i <= m; ++i) {int u = read();b.push_back(u);}sort(a.begin(), a.end());sort(b.begin(), b.end(), greater<int>());int ans = 0;for (int i = 0; i < n; ++i) {int t1 = abs(a.front() - b.front()), t2 = abs(a.back() - b.back());if (t1 > t2) {ans += t1;a.pop_front(), b.pop_front();}else {ans += t2;a.pop_back(), b.pop_back();}}writeln(ans);}return 0;
}

E - Eat the Chip 待补

F - Sum of progression

• 根号分治
• 预处理 $d\le \sqrt{n}$ ，暴力 $d>sqrtO(n\cdot \sqrt{n})$ 。

vector<vector<int>> pref(N + 322, vector<int>(323)), sum(N + 322, vector<int>(323));
signed main() {int T = 1;T = read();while (T--) {int n = read(), q = read();int D = sqrt(n);
//        while (D * D < n) ++D;vector<int> a(n + 1);for (int i = 1; i <= n; ++i) a[i] = read();for (int i = 1; i <= D; ++i) pref[1][i] = sum[1][i] = 0;for (int d = 1; d <= D; ++d) {for (int i = 1; i <= n; ++i) {pref[i + d][d] = pref[i][d] + a[i];sum[i + d][d] = sum[i][d] + a[i] * (i / d + 1);}}while (q--) {int s = read(), d = read(), k = read();if (d <= D) {int r = s + d * k;writesp(sum[r][d] - sum[s][d] - (pref[r][d] - pref[s][d]) * (s / d));}else {int ans = 0;for (int i = 0; i < k; ++i) ans += a[s + d * i] * (i + 1);writesp(ans);}}puts("");}return 0;
}

G - Mischievous Shooter

• 枚举 + 模拟
• 枚举每一个点即可，只考虑图里第一个的情况，其余三个可以翻转矩形。

$codelearnfromjiangly$ 简洁多了

signed main() {int T = 1;T = read();while (T--) {int n = read(), m = read(), k = read();vector<string> s(n);for (int i = 0; i < n; ++i) cin >> s[i];int ans = 0;auto solve = [&] () {vector<vector<int>> pref1(n, vector<int>(m)), pref2(n, vector<int>(m));for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (i) pref1[i][j] = pref1[i - 1][j];if (i && j < m - 1) pref2[i][j] = pref2[i - 1][j + 1];pref1[i][j] += (s[i][j] == '#');pref2[i][j] += (s[i][j] == '#');}}vector<vector<int>> dp(n, vector<int>(m));for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (j) {dp[i][j] = dp[i][j - 1];if (i + j > k) {int x = i, y = j - (k + 1);if (y < 0) x += y, y = 0;dp[i][j] -= pref2[x][y];}if (i >= k + 1) dp[i][j] += pref2[i - (k + 1)][j];}dp[i][j] += pref1[i][j];if (i >= k + 1) dp[i][j] -= pref1[i - (k + 1)][j];ans = max(ans, dp[i][j]);}}};solve();reverse(s.begin(), s.end());solve();for (auto &it: s) reverse(it.begin(), it.end());solve();reverse(s.begin(), s.end());solve();writeln(ans);}return 0;
}