\(100+40+20+8=168\)，拿到了大众分，至少没挂分吧
#A. 矩阵交换
一个 \(m\) 维偏序，可以使用 \(m-1\) 维树状数组解决
以第 \(i\) 作为第 \(i\) 关键字，进行排序，这样一定最优。排完之后直接判断是否满足条件即可
// BLuemoon_
#include <bits/stdc++.h>using namespace std;const int kMaxN = 100 + 5;struct P {int e[kMaxN];
};int t, n, m;
bool flag = 1;
P a[kMaxN];void pr(bool pr) {cout << (pr ? "YES" : "NO") << '\n';
}int main() {ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);freopen("exchange.in", "r", stdin), freopen("exchange.out", "w", stdout);for (cin >> t; t; t--, flag = 1) {cin >> n >> m;for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {cin >> a[i].e[j];}}sort(a + 1, a + n + 1, [](P l, P r) {for (int i = 1; i <= m; i++) {if (l.e[i] != r.e[i]) {return l.e[i] < r.e[i];}}return 0 == 1;});for (int j = 1; j <= m; j++) {for (int i = 2; i <= n; i++) {if (a[i].e[j] < a[i - 1].e[j]) {flag = 0;break;}}if (!flag) {break;}}pr(flag);}return 0;
}
#B. 砖块摆放/原题
定义 A 为 \(0\)，B 为 \(1\)，C 为 \(2\)
所以可以把每一个值 \(d\) 被其下面两个数 \(d_1,d_2\) 表示：
\[d=-(d_1+d_2)\bmod 3
\]
那么最顶层就可以表示成这样：
可以发现这很像杨辉三角，但是模数只有 \(3\)，没有逆元。只能用 Lucas 解决
// BLuemoon_
#include <bits/stdc++.h>using namespace std;
using LL = long long;const int kMaxN = 2e5 + 5;int t, n, a[kMaxN], ans;
char ch;int C(int p, int k, int ret = 1) {if (p < k) {return 0;}for (int i = p - k + 1; i <= p; ret *= i, i++);for (int i = 1; i <= k; ret /= i, i++);return ret % 3;
}
int Lucas(int p, int k) {if (p < k) {return 0;}if (p <= 10) {return C(p, k);}return Lucas(p / 3, k / 3) * C(p % 3, k % 3) % 3;
}int main() {ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);freopen("brick.in", "r", stdin), freopen("brick.out", "w", stdout);for (cin >> t; t; t--, ans = 0) {cin >> n;for (int i = 1; i <= n; i++) {cin >> ch, a[i] = ch - 'A';}for (int i = 1; i <= n; i++) {(ans += Lucas(n - 1, i - 1) * a[i] % 3) %= 3;}(n & 1 ^ 1) && (ans = (3 - ans) % 3);cout << (char)(ans + 'A') << '\n';}return 0;
}
#C. 学习 LIS
考虑状压 dp，令 \(dp_{i,j,s}\) 为：正在考虑 \(i\) 填哪里，正在考虑从后向前数第 \(j\) 位是否填 \(i\)，填入的集合是 \(s\) 的方案数
转移直接分是否选 \(i\) 分类讨论即可，具体可以看代码
// BLuemoon_
#include <bits/stdc++.h>using namespace std;
using LL = long long;const int kMaxN = 20, kL = 1 << kMaxN, kMaxM = 3e3 + 5;
const LL kP = 998244353;int n, m, a[kMaxN + 1], mx[kL], ans[kMaxN + 1], dp[2][kMaxN + 1][kL], cur, nxt = 1, cnt;
LL c[kMaxM][kMaxM];int main() {ios::sync_with_stdio(0), cin.tie(0), cout.tie(0), c[0][0] = c[1][0] = c[1][1] = 1;freopen("lis.in", "r", stdin), freopen("lis.out", "w", stdout);cin >> n >> m;for (int i = 0; i < n; i++) {cin >> a[i];}for (int i = 2; i < kMaxM; i++) {c[i][0] = 1;for (int j = 1; j <= i; j++) {c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % kP;}}for (int j = 0; j < (1 << n); j++) {for (int i = 0; i < n; i++) {if (j & (1 << i)) {mx[j] = max(mx[j], a[i]);}}}dp[0][n - 1][0] = 1;for (int u = 1; u <= n; u++) {for (int i = n - 1; ~i; i--) {for (int j = 0; j < (1 << n); j++)if (dp[cur][i][j]) {int tmp = dp[cur][i][j];(dp[i ? cur : nxt][(i ? i : n) - 1][j] += tmp) %= kP;if (!(j & (1 << i)) && mx[j & ((1 << i) - 1)] + 1 == a[i]) {(dp[i ? cur : nxt][(i ? i : n) - 1][j | (1 << i)] += tmp) %= kP;}dp[cur][i][j] = 0;}}ans[u] = dp[nxt][n - 1][(1 << n) - 1], swap(cur, nxt);}for (int i = 1; i <= n; i++) {for (int j = i - 1; j >= 1; j--) {ans[i] = (ans[i] + (-1 * c[i][j] * ans[j] % kP) + kP) % kP;}cnt = (cnt + ans[i] * c[m][i] % kP) % kP;}cout << cnt << '\n';return 0;
}
#D. 战略轰炸
人机线段树二分，先咕咕咕
