Alternating String

我们可以设状态 \(dp[i][0/1][0/1]\) 表示当前考虑到第几个,长度为奇数还是偶数,有没有用 \(1\) 操作

#include <bits/stdc++.h>

using namespace std;

const int N = 4e5 + 5, M = 27;

int t, n, ans, a[N], dp[N][2][2];

char s;

void Solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    a[i] = (int)(s - 'a' + 1);
  }
  ans = 1e9;
  for (int i = 1; i <= 26; i++) {
    for (int j = 1; j <= 26; j++) {
      dp[0][0][0] = 0, dp[0][1][0] = 1e9;
      dp[0][0][1] = 1e9, dp[0][1][1] = 1e9;
      for (int k = 1; k <= n; k++) {
        dp[k][0][1] = min({dp[k - 1][0][0] + 1, dp[k - 1][1][0] + 1, dp[k - 1][1][1] + 1});
        dp[k][0][0] = dp[k - 1][1][0] + 1;
        dp[k][1][1] = min({dp[k - 1][1][0] + 1, dp[k - 1][0][0] + 1, dp[k - 1][0][1] + 1});
        dp[k][1][0] = dp[k - 1][0][0] + 1;
        if (a[k] == i) {
          dp[k][0][0] = min(dp[k][0][0], dp[k - 1][1][0]);
          dp[k][0][1] = min(dp[k][0][1], dp[k - 1][1][1]);
        }
        if (a[k] == j) {
          dp[k][1][0] = min(dp[k][1][0], dp[k - 1][0][0]);
          dp[k][1][1] = min(dp[k][1][1], dp[k - 1][0][1]);
        }
      }
      ans = min({ans, dp[n][0][1], dp[n][0][0]});
    }
  }
  cout << ans << "\n";
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}

Sakurako's Box

没什么好说的,就是一个最简单的概率期望

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 2e5 + 5, mod = 1e9 + 7;

int t, n, a[N], sum, ans;

int mypow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}

void Solve() {
  sum = 0, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
    sum %= mod;
  }
  for (int i = 1; i <= n; i++) {
    ans += a[i] * ((sum - a[i] + mod) % mod) % mod;
    ans %= mod;
  }
  int tmp = n * (n - 1) % mod * mypow(2, mod - 2) % mod;
  cout << ans * mypow(2, mod - 2) % mod * mypow(tmp, mod - 2) % mod << "\n";
}

signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}