Codeforces Round #563 (Div. 2)

Posted 2021-12-28 cycleke

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A	B	C	D	E	F
构造	排序	贪心，筛法	异或，前缀和	计数问题	点分治
1000	1200	1300	1900	2500	2300

A. Ehab Fails to Be Thanos

将数组排序，判断前半部分和后半部分的和是否相同。

int main() 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);

  int n;
  cin >> n;
  n *= 2;
  vi a(n);
  FOR(i, 0, n)  cin >> a[i]; 
  sort(ALL(a));
  ll sum = 0, sum_half = 0;
  FOR(i, 0, n) 
    sum += a[i];
    if (i < n / 2) 
      sum_half += a[i];
    
  
  if (sum_half * 2 < sum) 
    for (int x : a) 
      cout << x << " ";
    
    cout << "\n";
   else 
    cout << "-1\n";
  

  return 0;

B. Ehab Is an Odd Person

若数组有奇数有偶数，那么我们总可以通过交换将数组排序，判断一下奇偶就好了

int main() 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);

  int n;
  cin >> n;
  vi a(n);
  bool odd = false, even = false;
  FOR(i, 0, n) 
    cin >> a[i];
    odd |= a[i] & 1;
    even |= ~a[i] & 1;
  
  if (odd && even) 
    sort(ALL(a));
  
  for (int x : a) 
    cout << x << " ";
  
  cout << "\n";

  return 0;

C. Ehab and a Special Coloring Problem

将每个质数的倍数赋予相同的值就好了

const int MAXN = 1e5 + 3;
int a[MAXN];

int main() 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);

  int n;
  cin >> n;
  int cur = 1;
  FOR(i, 2, n + 1) 
    if (a[i]) 
      continue;
    
    for (int j = i; j <= n; j += i) 
      a[j] = cur;
    
    ++cur;
  
  FOR(i, 2, n + 1)  cout << a[i] << " "; 
  cout << "\n";

  return 0;

D. Ehab and the Expected XOR Problem

将子串通过前缀和就可以变为两个数的异或值。现在的要求就是求一个集合，使得其中没有两个数的异或值为 x

const int MAXS = 1 << 18 | 1;

bool mark[MAXS];

int main() 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);

  int n, x;
  cin >> n >> x;
  mark[0] = true;
  vi a(0);

  FOR(i, 1, (1 << n)) 
    if (!mark[i ^ x]) 
      mark[i] = true;
      a.pb(i);
    
  
  cout << a.size() - 1 << "\n";
  FOR(i, 1, a.size())  cout << (a[i] ^ a[i - 1]) << " "; 
  cout << "\n";
  return 0;

E. Ehab and the Expected GCD Problem

不懂，官方题解说所有符合条件的序列的 gcd 为\(2^x \times 3^y\)，因为和相同时将数拆分为 2 和 3（最多 1 个 3）可
以获得最多的因数。

定义 dp[i][x][y]，表示到第 i 位，gcd 为\(2^x \times 3^y\)的方案数。

const int MAXN = 1e6 + 3;
int dp[MAXN][21][2];

int main() 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);

  int n;
  cin >> n;
  int p = 0;
  while ((1 << p) <= n) 
    ++p;
  

  auto gao = [&](const int &x, const int &y) 
    ll t = n >> x;
    return y ? t / 3 : t;
  ;
  auto add = [&](int &x, const int &y) 
    x += y;
    if (x >= MOD) 
      x -= MOD;
    
  ;

  dp[1][p - 1][0] = 1;
  dp[1][p - 2][1] = gao(p - 2, 1);
  FOR(i, 1, n) 
    FOR(x, 0, p) 
      FOR(y, 0, 2) 
        add(dp[i + 1][x][y], dp[i][x][y] * (gao(x, y) - i) % MOD);
        if (x) 
          add(dp[i + 1][x - 1][y],
              dp[i][x][y] * (gao(x - 1, y) - gao(x, y)) % MOD);
        
        if (y) 
          add(dp[i + 1][x][y - 1],
              dp[i][x][y] * (gao(x, y - 1) - gao(x, y)) % MOD);
        
      
    
  
  cout << dp[n][0][0] << "\n";

  return 0;

F. Ehab and the Big Finale

点分治，每次判断是在重心的上方还是下方。

const int MAXN = 2e5 + 3;

int interact(char oper, int u) 
  printf("%c %d\n", oper, u);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;


vi adj[MAXN];
bool deleted[MAXN];
int fa[MAXN], size[MAXN], dep[MAXN], depx;

void dfs(int u) 
  for (int v : adj[u]) 
    if (v == fa[u]) 
      continue;
    
    fa[v] = u;
    dep[v] = dep[u] + 1;
    dfs(v);
  


int sub_size, root, root_key;

void get_root(int u, int from) 
  int key = 0;
  size[u] = 1;
  for (int v : adj[u]) 
    if (deleted[v] || v == from) 
      continue;
    
    get_root(v, u);
    size[u] += size[v];
    key = max(key, size[v]);
  
  key = max(key, sub_size - key);
  if (key < root_key) 
    root = u;
    root_key = key;
    // debug("root = %d root_key = %d\n", root, root_key);
  
  // debug("u = %d key = %d\n", u, key);


int gao(int u) 
  if (size[u] == 1) 
    return u;
  
  sub_size = size[u];
  root = u;
  root_key = sub_size;
  get_root(u, 0);
  int dist = interact('d', root);
  if (dist == 0) 
    return root;
  
  deleted[root] = true;
  if (dep[root] + dist == depx) 
    return gao(interact('s', root));
   else 
    return gao(fa[root]);
  


int main() 
  int n;
  scanf("%d", &n);

  FOR(i, 1, n) 
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].pb(v);
    adj[v].pb(u);
  
  dfs(1);

  depx = interact('d', 1);
  size[1] = n;
  printf("! %d\n", gao(1));

  return 0;