HHKB Programming Contest 2023(AtCoder Beginner Contest 327) 赛后总结

又没来得及写题解。。。

赛时

A - ab

查找 ab 和 ba，只要其中一者存在就行。

#include <bits/stdc++.h>

using namespace std;

int n;
string s;

int main() {
  cin >> n >> s;
  cout << (s.find("ab") != string::npos || s.find("ba") != string::npos ? "Yes" : "No");
  return 0;
}

B - A^A

由于范围比较小，考虑枚举，注意从 \(1\) 开始。

#include <bits/stdc++.h>

using namespace std;

long long b;

long long poww(int a, int b) {
  long long ans = 1;
  for (int i = 1; i <= b; ans *= a, i++) {
  }
  return ans;
}

int main() {
  cin >> b;
  for (int i = 1; i <= 15; i++) {
    long long s = 1ll * poww(i, i);
    if (s == b) {
      cout << i;
      return 0;
    }
  }
  cout << -1;
  return 0;
}

C - Number Place

照题意模拟。

#include <bits/stdc++.h>

using namespace std;

const int pos[9][9][2] = {
  {{0, 0}, {0, 1}, {0, 2}, {1, 0}, {1, 1}, {1, 2}, {2, 0}, {2, 1}, {2, 2}},
  {{0, 3}, {0, 4}, {0, 5}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}},
  {{0, 6}, {0, 7}, {0, 8}, {1, 6}, {1, 7}, {1, 8}, {2, 6}, {2, 7}, {2, 8}},
  {{3, 0}, {3, 1}, {3, 2}, {4, 0}, {4, 1}, {4, 2}, {5, 0}, {5, 1}, {5, 2}},
  {{3, 3}, {3, 4}, {3, 5}, {4, 3}, {4, 4}, {4, 5}, {5, 3}, {5, 4}, {5, 5}},
  {{3, 6}, {3, 7}, {3, 8}, {4, 6}, {4, 7}, {4, 8}, {5, 6}, {5, 7}, {5, 8}},
  {{6, 0}, {6, 1}, {6, 2}, {7, 0}, {7, 1}, {7, 2}, {8, 0}, {8, 1}, {8, 2}},
  {{6, 3}, {6, 4}, {6, 5}, {7, 3}, {7, 4}, {7, 5}, {8, 3}, {8, 4}, {8, 5}},
  {{6, 6}, {6, 7}, {6, 8}, {7, 6}, {7, 7}, {7, 8}, {8, 6}, {8, 7}, {8, 8}}
};

int a[9][9];
int cnt[10];

bool check() {
  for (int i = 0; i < 9; i++) {
    fill(cnt + 1, cnt + 9 + 1, 0);
    for (int j = 0; j < 9; j++) {
      cnt[a[i][j]]++;
    }
    bool flag = 1;
    for (int j = 1; j <= 9; j++) {
      if (cnt[j] != 1) {
        flag = 0;
        break;
      }
    }
    if (!flag) {
      // cout << "break point:1\n";
      return 0;
    }
  }
  for (int i = 0; i < 9; i++) {
    fill(cnt + 1, cnt + 9 + 1, 0);
    for (int j = 0; j < 9; j++) {
      cnt[a[j][i]]++;
    }
    bool flag = 1;
    for (int j = 1; j <= 9; j++) {
      if (cnt[j] != 1) {
        flag = 0;
        break;
      }
    }
    if (!flag) {
      // cout << "break point:2\n";
      return 0;
    }
  }
  for (int i = 0; i < 9; i++) {
    fill(cnt + 1, cnt + 9 + 1, 0);
    for (int j = 0; j < 9; j++) {
      cnt[a[pos[i][j][0]][pos[i][j][1]]]++;
    }
    bool flag = 1;
    for (int j = 1; j <= 9; j++) {
      if (cnt[j] != 1) {
        flag = 0;
        break;
      }
    }
    if (!flag) {
      // cout << "break point:3\n";
      return 0;
    }
  }
  return 1;
}

int main() {
  for (int i = 0; i < 9; i++) {
    for (int j = 0; j < 9; j++) {
      cin >> a[i][j];
    }
  }
  cout << (check() ? "Yes" : "No");
  return 0;
}

赛后

D - Good Tuple Problem

二分图 01 染色。

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 2e5 + 5;

int n, m, c[MAXN], a[MAXN], b[MAXN], flag;
bool vis[MAXN];
vector<int> g[MAXN];

void dfs(int x, int y) {
  if (vis[x]) {
    if (c[x] != y) {
      flag = 1;
    }
    return;
  }
  vis[x] = 1, c[x] = y;
  for (int i = 0; i < g[x].size(); i++) {
    dfs(g[x][i], !y);
  }
}

int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  for (int i = 1; i <= m; i++) {
    g[a[i]].push_back(b[i]), g[b[i]].push_back(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, 0);
    }
  }
  cout << (flag ? "No" : "Yes");
  return 0;
}

E - Maximize Rating

水 DP，赛时竟然没有写出来，我太蒻了。

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 5e3 + 10, INF = 6e3;

int n, a[MAXN];

double ans = -INF, dp[MAXN][MAXN];

double getSum(int x) {
  double sum = 0;
  for (int i = 0; i < x; i++) {
    sum += pow(0.9, i);
  }
  return sum;
}

int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = n; i; i--) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = dp[i + 1][j], dp[i][j] = max(dp[i][j], dp[i + 1][j - 1] + pow(0.9, j - 1) * a[i]);
    }
  }
  for (int i = 1; i <= n; i++) {
    ans = max(ans, dp[1][i] / getSum(i) - 1200 / sqrt(i));
  }
  cout << fixed << setprecision(6) << ans;
  return 0;
}