链接：
https://ac.nowcoder.com/acm/contest/42105

A

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

//constexpr int M = 998244353;
//constexpr int M = 1000000007;

template <typename T>
class Fenwick {
 public:
  int n;
  vector<T> tree;
  Fenwick(const int &n) : n(n), tree(n) {}
  inline void modify(int pos, T x) {
    for ( ; pos <= n; pos += pos & -pos) {
      tree[pos - 1] += x;
    }
  }
  inline T get(int pos) {
    T res = 0;
    for ( ; pos > 0; pos -= pos & -pos) {
      res += tree[pos - 1];
    }
    return res;
  }
  inline T sum(int l, int r) { // (l, r]
    return get(r) - get(l);
  }
};

constexpr int N = 26;

int cnt[N];
int half[N];

void solve() {
  int n;
  string s;
  cin >> n >> s;
  s = " " + s;
  Fenwick<int> f(n);
  vector<vector<int>> pos(26);
  for (int i = 1; i <= n; i++) {
    int num = s[i] - 'a';
    half[num]++;
  }
  for (int i = 0; i < 26; i++) {
    half[i] /= 2;
  }
  vector<int> p(n + 1);
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    int num = s[i] - 'a';
    if (cnt[num] < half[num]) {
      p[i] = ++tot;
      pos[num].push_back(tot);
    } else {
      p[i] = pos[num][cnt[num] - half[num]] + n / 2;
    }
    cnt[num]++;
  }
  i64 ans = 0;
  for (int i = n; i >= 1; i--) {
    ans += f.get(p[i]);
    f.modify(p[i], 1);
  }
  cout << ans << '\n';
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(6);
  int tt = 1;
  //cin >> tt;
  for (int _ = 1; _ <= tt; _++) {
    solve();
  }
  return 0;
}

A和cf一个题差不多

链接：
https://codeforces.com/contest/1430/problem/E

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

//constexpr int M = 998244353;
//constexpr int M = 1000000007;

template <typename T>
class Fenwick {
 public:
  int n;
  vector<T> tree;
  Fenwick(const int &n) : n(n), tree(n) {}
  inline void modify(int pos, T x) {
    for ( ; pos <= n; pos += pos & -pos) {
      tree[pos - 1] += x;
    }
  }
  inline T get(int pos) {
    T res = 0;
    for ( ; pos > 0; pos -= pos & -pos) {
      res += tree[pos - 1];
    }
    return res;
  }
  inline T sum(int l, int r) { // (l, r]
    return get(r) - get(l);
  }
};

constexpr int N = 26;

int cnt[N];
int half[N];

void solve() {
  int n;
  string s;
  cin >> n >> s;
  s = " " + s;
  Fenwick<int> f(n);
  vector<vector<int>> pos(26);
  vector<int> p(n + 1);
  for (int i = 1; i <= n; i++) {
    int num = s[i] - 'a';
    pos[num].push_back(i);
  }
  i64 ans = 0;
  for (int i = 1; i <= n; i++) {
    int num = s[i] - 'a';
    p[n - i + 1] = pos[num].back();
    pos[num].pop_back();
  }
  for (int i = n; i >= 1; i--) {
    ans += f.get(p[i]);
    f.modify(p[i], 1);
  }
  cout << ans << '\n';
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(6);
  int tt = 1;
  //cin >> tt;
  for (int _ = 1; _ <= tt; _++) {
    solve();
  }
  return 0;
}

F

找三个素数乘起来就行

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

//constexpr int M = 998244353;
//constexpr int M = 1000000007;

constexpr int N = 2E6;

vector<int> a;
bool st[N + 1];

void get() {
  for (int i = 2; i <= N; i++) {
    if (!st[i])
      a.push_back(i);
    for (int j = 0; 1LL * a[j] * i <= N; j++) {
      st[i * a[j]] = 1;
      if (i % a[j] == 0)
        break;
    }
  }
}

void solve() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << 24 << '\n';
    return;
  }
  int A = *lower_bound(a.begin(), a.end(), 1 + n);
  int B = *lower_bound(a.begin(), a.end(), A + n);
  int C = *lower_bound(a.begin(), a.end(), B + n);
  i64 ans = 1;
  ans *= A;
  ans *= B;
  ans *= C;
  cout << ans << '\n';
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(6);
  int tt = 1;
  cin >> tt;
  get();
  for (int _ = 1; _ <= tt; _++) {
    solve();
  }
  return 0;
}

H

map套vector

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

//constexpr int M = 998244353;
//constexpr int M = 1000000007;

void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  map<vector<int>, int> mp;
  i64 ans = 0;
  for (int i = 0; i < m; i++) {
    vector<int> b(n);
    for (int j = 0; j < n; j++) {
      cin >> b[j];
    }
    mp[b]++;
  }
  for (auto [u, v] : mp) {
     ans += 1LL * v * (v - 1) / 2;
  }
  cout << ans << '\n';
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(6);
  int tt = 1;
  cin >> tt;
  for (int _ = 1; _ <= tt; _++) {
    solve();
  }
  return 0;
}

K

套个计算几何旋转向量公式

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

//constexpr int M = 998244353;
//constexpr int M = 1000000007;

const double Pi = acos(-1);

void solve() {
  int t;
  cin >> t;
  double l1, l2, l3;
  cin >> l1 >> l2 >> l3;
  double t1, t2, t3;
  cin >> t1 >> t2 >> t3;
  cout << (-l1 * sin(-360.0 / t1 * t * Pi / 180.0)) + 
  (-l2 * sin(-360.0 / t2 * t * Pi / 180.0)) + 
  (-l3 * sin(-360.0 / t3 * t * Pi / 180.0)) << ' ' <<
  (l1 * cos(-360.0 / t1 * t * Pi / 180.0)) +
  (l2 * cos(-360.0 / t2 * t * Pi / 180.0)) + 
  (l3 * cos(-360.0 / t3 * t * Pi / 180.0));
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(10);
  int tt = 1;
  //cin >> tt;
  for (int _ = 1; _ <= tt; _++) {
    solve();
  }
  return 0;
}