比赛链接：Here

A - LR Constraints

赛时做这个好迷啊，英文题面解释不清楚，还是看了日语原文才搞懂

AtCoder Regular Contest 124 (A~B) - 图1 个卡牌上有两个 字符 + 数字 组合，L 的右边所有元素 + 1，R 的左边元素 + 1

最后求出现过数字的乘积，同时对 AtCoder Regular Contest 124 (A~B) - 图2 取余

注意点：开 long long！！！，今天牛客也是，没开 ll debug半天，感谢尚佬指出我的错误

  1. const int N = 1010;
  2. bool vis[N];
  3. int cnt[N];
  4. int main() {
  5.     cin.tie(nullptr)->sync_with_stdio(false);
  6.     int n, k;
  7.     cin >> n >> k;
  8.     for (int i = 1; i <= k; ++i) {
  9.         char c; int x;
  10.         cin >> c >> x;
  11.         if (c == 'L') {
  12.             vis[x] = 1;
  13.             for (int j = x + 1; j <= n; ++j)cnt[j] ++;
  14.         } else {
  15.             vis[x] = 1;
  16.             for (int j = 1; j <= x; ++j)cnt[j] ++;
  17.         }
  19.     }
  20.     ll ans = 1;
  21.     for (int i = 1; i <= n; ++i) {
  22.         if (!vis[i])ans = ans * cnt[i] % 998244353;
  23.     }
  24.     cout << ans << '\n';
  25. }

B - XOR Matching 2

对于长度为 AtCoder Regular Contest 124 (A~B) - 图3 的 AtCoder Regular Contest 124 (A~B) - 图4 两个序列，请问是否存在某种排序组合使得某个 AtCoder Regular Contest 124 (A~B) - 图5 被称为 Good

Good定义：可以对 AtCoder Regular Contest 124 (A~B) - 图6 重排序使得每一个 AtCoder Regular Contest 124 (A~B) - 图7

当 AtCoder Regular Contest 124 (A~B) - 图8 固定时，置换 AtCoder Regular Contest 124 (A~B) - 图9 使得 AtCoder Regular Contest 124 (A~B) - 图10 等价于 AtCoder Regular Contest 124 (A~B) - 图11

所以我们可以直接尝试 AtCoder Regular Contest 124 (A~B) - 图12

  • AtCoder Regular Contest 124 (A~B) - 图13#card=math&code=%5Cmathcal%7BO%7D%28N%5E2log%5C%20N%29)
  1. const int N = 2e3 + 10;
  2. int a[N], b[N];
  3. map<int, int> mp, bg;
  4. set<int>v;
  5. int main() {
  6.     cin.tie(nullptr)->sync_with_stdio(false);
  7.     int n;
  8.     cin >> n;
  9.     for (int i = 1; i <= n; ++i) cin >> a[i];
  10.     for (int i = 1; i <= n; ++i) cin >> b[i], bg[b[i]]++;
  11.     for (int i = 1; i <= n; ++i) {
  12.         int x = a[1] ^ b[i]; mp = bg;
  13.         bool f = true;
  14.         for (int j = 1; j <= n and f; ++j)if (!mp[x ^ a[j]]) f = false;
  15.         if (f) v.insert(x);
  16.     }
  17.     cout << v.size() << "\n";
  18.     for (int x : v)cout << x << "\n";
  19. }