10.7总结

T1

赛事拿的是 30pts 的暴力分，就直接暴力枚举就行了，复杂度 \(\mathcal O(n^3)\)。

将操作分成两部分，一部分是对 \((x,y)\) 轴同时操作的，可以一开始就用前缀 ⁡\(\max\) 统计完，将长方体变成一个底面为阶梯状的直棱柱。另一部分是对 \((x,z)\) 或 \((y,z)\) 操作的，考虑从小到大枚举 然后将 \(=z\) 的操作变成一条横线或一条竖线，容易发现这两条线也是具有单调性的，于是每层留下的点数可以直接计算。

预估：30ots

#include <fstream>

using namespace std;
using ll = long long;
using ull = unsigned long long;
using i128 = __int128;

const ll kMaxN = 3e7 + 5;

ifstream cin("A.in");
ofstream cout("A.out");

ll n, A, B, C, u[kMaxN], v[kMaxN], w[kMaxN], maxh[kMaxN];
i128 ans = 0;

ull Rand(ull &k1, ull &k2) {
  ull k3 = k1, k4 = k2;
  k1 = k4;
  k3 ^= (k3 << 23);
  k2 = k3 ^ k4 ^ (k3 >> 17) ^ (k4 >> 26);
  return k2 + k4;
}
void GetData() {
  ull x, y;
  cin >> n >> A >> B >> C >> x >> y;
  for (ll i = 1; i <= n; i++) {
    u[i] = Rand(x, y) % A + 1;
    v[i] = Rand(x, y) % B + 1;
    w[i] = Rand(x, y) % C + 1;
    if (Rand(x, y) % 3 == 0) {
      u[i] = A;
    }
    if (Rand(x, y) % 3 == 0) {
      v[i] = B;
    }
    if ((u[i] != A) && (v[i] != B)) {
      w[i] = C;
    }
  }
}
void prll(i128 x) {
  if (x >= 10) {
    prll(x / 10);
  }
  cout.put(x % 10 + '0');
}
int main() {
  GetData();
  ll maxn = 0;
  for (ll i = 1; i <= n; ++i) {
    if (u[i] == A && v[i] == B) {
      maxn = max(maxn, w[i]);
    }
  }
  ans = (i128)A * B * maxn;
  for (ll h = C; h > maxn; --h) {
    for (ll i = 1; i <= n; ++i) {
      if (w[i] == h) {
        maxh[u[i]] = max(maxh[u[i]], v[i]);
      }
    }
    for (ll i = A; i >= 1; --i) {
      maxh[i] = max(maxh[i], maxh[i + 1]);
    }
    for (ll i = 1; i <= A; ++i) {
      ans += maxh[i];
    }
  }
  prll(ans);
  return 0;
}

T2

赛事完全不会欸，期望值 DP。

正解是

#include <cstring>
#include <fstream>

using namespace std;
using ll = long long;

const int kMaxN = 305, kMod = 1e9 + 7;

ifstream cin("B.in");
ofstream cout("B.out");

ll n, p[kMaxN], pw[kMaxN][kMaxN], dp[kMaxN][kMaxN];

int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> p[i];
    pw[i][0] = 1;
    for (int j = 1; j <= n; ++j) {
      pw[i][j] = (ll)pw[i][j - 1] * (1 - p[i] + kMod) % kMod;
    }
  }
  int ans = 0;
  for (int r = 1; r <= n; ++r) {
    memset(dp, 0, sizeof(dp));
    dp[1][r] = 1;
    for (int i = 1; i < n; ++i) {
      for (int j = 1; j <= n - i + 1; ++j) {
        if (j) {
          dp[i + 1][j - 1] = (dp[i + 1][j - 1] + (ll)dp[i][j] * (1 - pw[i][j - 1] + kMod) % kMod) % kMod;
        }
        if (j < n - i + 1) {
          dp[i + 1][j] = (dp[i + 1][j] + (ll)dp[i][j] * pw[i][j] % kMod) % kMod;
        }
      }
    }
    ans = (ans + (ll)dp[n][1] * r % kMod) % kMod;
  }
  cout << ans;
  return 0;
}

T3

简单小清新分治，但是赛时没有想到显而易见的结论， 于是只能打个暴力，但是预估 30pts，实际 30pts。

其实我们可以知道如果区间 \([l, r]\) 不合法，那么它的子区间也肯定不合法，故双指针 \(i, j\) 从前和后向中间查找，如果有查找到不合法的数字 \(a_x\) 那么把子区间分成 \([l, x - 1]\) 和 \([x + 1, r]\) 两个区间操作就行了。

#include <fstream>

using namespace std;

const int kMaxN = 1e6 + 1;

ifstream cin("C.in");
ofstream cout("C.out");

int n, ans, a[kMaxN], b[kMaxN], c[kMaxN];

void Solve(int l, int r) {
  if (l >= r) {
    if (l == r) {
      c[a[l]]--;
    }
    return;
  }
  bool f = 0;
  int pos = -1;
  for (int i = l, j = r; i <= j; i++, j--) {
    // cout << i << ' ' << j << ' ' << c[a[i]] << ' ' << c[a[j]] << '\n';
    if (c[a[i]] < b[r - l + 1]) {
      pos = i;
      break;
    }
    if (c[a[j]] < b[r - l + 1]) {
      pos = j, f = 1;
      break;
    }
  }
  if (pos > 0) {
    if (f) {
      for (int i = pos; i <= r; i++) {
        c[a[i]]--;
      }
      Solve(l, pos - 1);
      for (int i = pos + 1; i <= r; i++) {
        c[a[i]]++;
      }
      Solve(pos + 1, r);
    } else {
      for (int i = l; i <= pos; i++) {
        c[a[i]]--;
      }
      Solve(pos + 1, r);
      for (int i = l; i < pos; i++) {
        c[a[i]]++;
      }
      Solve(l, pos - 1);
    }
  } else {
    for (int i = 1; i <= n; i++) {
      c[a[i]]--;
    }
    // cout << r - l + 1 << '\n';
    ans = max(ans, r - l + 1);
  }
}

int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    c[a[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
  }
  Solve(1, n);
  cout << ans << '\n';
  return 0;
}