New Dimensions

我们假设枚举 \(a, b\) 那么我们显然可以发现 \(a^2 + b^2 + c^2 - ab - ab - bc\) 中 \(c\) 越大越好

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 5e3 + 5;

int n, v[N], ans;

signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  sort(v + 1, v + n + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      ans = max(ans, v[i] * v[i] + v[j] * v[j] + v[n] * v[n] - v[i] * v[n] - v[i] * v[j] - v[j] * v[n]);
    }
  }
  cout << ans;
  return 0;
}

Typographic Kaleidoscope

先别急着说是大水题,我给个 \(HACK\)

###..
.#...
.####
.#.#.
####.
.#.#.
.#.#.
.#...
.#...

你很难将其判成 \(A\),但是他是三个 \(T\).
那么我们正着匹配不行了,考虑一下从右至左,从下至上,你会惊奇的发现,对了!我们如何证明呢我们容易 把\(P + T\) 判成 \(A\), \(T + T\) 判成 \(A 或 P\)......不难发现,都是 \(T\)在捣乱,然而 \(T\) 有一个很明显的特征,就是 \(T\) 的尾巴很长,在判的过程中一定会露出一个尾巴,我们反正匹配,那么就一定会先把尾巴长的拆掉,然后就无法组成 \(A\) 或 \(P\) 了

#include <bits/stdc++.h>

using namespace std;

using Pii = pair<int, int>;

#define int long long

const int N = 1e3 + 10;

int n, m, cnt[4], len[4];

char a[N][N];

Pii d[4][N];

void check(int x, int y, int op) {
  for (int i = 1; i <= len[op]; i++) {
    int nx = x + d[op][i].first, ny = y + d[op][i].second;
    if (nx < 1 || ny < 1) {
      return ;
    }
    if (a[nx][ny] != '#') {
      return ;
    }
  }
  cnt[op]++;
  for (int i = 1; i <= len[op]; i++) {
    int nx = x + d[op][i].first, ny = y + d[op][i].second;
    a[nx][ny] = '.';
  }
}

signed main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  d[1][1] = {0, -1};
  d[1][2] = {-1, -1};
  d[1][3] = {-2, -1};
  d[1][4] = {-3, -1};
  d[1][5] = {-4, 0};
  d[1][6] = {-4, -1};
  d[1][7] = {-4, -2};
  len[1] = 7;
  d[2][1] = {0, 0};
  d[2][2] = {-1, 0};
  d[2][3] = {0, -2};
  d[2][4] = {-1, -2};
  d[2][5] = {-2, 0};
  d[2][6] = {-3, 0};
  d[2][7] = {-4, 0};
  d[2][8] = {-2, -2};
  d[2][9] = {-3, -2};
  d[2][10] = {-4, -2};
  d[2][11] = {-2, -1};
  d[2][12] = {-4, -1};
  len[2] = 12;
  d[3][1] = {0, -2};
  d[3][2] = {-1, -2};
  d[3][3] = {-2, 0};
  d[3][4] = {-3, 0};
  d[3][5] = {-4, 0};
  d[3][6] = {-2, -2};
  d[3][7] = {-3, -2};
  d[3][8] = {-4, -2};
  d[3][9] = {-2, -1};
  d[3][10] = {-4, -1};
  len[3] = 10;
  for (int j = m; j >= 1; j--) {
    for (int i = n; i >= 1; i--) {
      for (int k = 1; k <= 3; k++) {
        check(i, j, k);
      }
    }
  }
  cout << cnt[1] << " " << cnt[2] << " " << cnt[3];
  return 0;
}