「题解」Codeforces Round 888 (Div. 3)

A. Escalator Conversations

Problem

题目

Sol & Code

签到

#include <bits/stdc++.h>

typedef long long ll;

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int T, n, m, k, h;

int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d %d %d", &n, &m, &k, &h);
    int ans = 0;
    for (int i = 1, x; i <= n; ++i) {
      scanf("%d", &x);
      if (x == h) continue;
      int d = std::abs(x - h);
      if (d % k == 0 && d / k < m) ++ans;
    }
    printf("%d\n", ans);
  }
  return 0;
}

B. Parity Sort

Problem

题目

Sol & Code

只能奇数之间交换，偶数之间交换，就看奇偶数分别排序后在插回去是否有序。

#include <bits/stdc++.h>
#define N 200001

typedef long long ll;

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int T, n, a[N], b[N], c[N];

int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      if (a[i] & 1) b[++cnt1] = a[i];
      else c[++cnt2] = a[i];
    }
    std::sort(b + 1, b + cnt1 + 1);
    std::sort(c + 1, c + cnt2 + 1);
    bool okay = true;
    int cnt1_ = 1, cnt2_ = 1;
    for (int i = 1; i <= n; ++i) {
      if (a[i] & 1) a[i] = b[cnt1_++];
      else a[i] = c[cnt2_++];
      if (a[i] < a[i - 1]) { okay = false; break; }
    }
    puts(okay ? "YES" : "NO");
  }
  return 0;
}

C. Tiles Comeback

Problem

题目

Sol & Code

贪心，看首尾两个数是否出现了足够多次且不相交（抽象）。

#include <bits/stdc++.h>
#define N 200001

typedef long long ll;

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int T, n, k, a[N];

int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    int p1 = 0, p2 = 0;
    for (int i = 1, cnt = 0; i <= n; ++i) {
      if (a[i] == a[1]) ++cnt;
      if (cnt == k) { p1 = i; break; }
    }
    for (int i = n, cnt = 0; i >= 1; --i) {
      if (a[i] == a[n]) ++cnt;
      if (cnt == k) { p2 = i; break; }
    }
    if (a[1] == a[n]) {
      if (!p1) puts("NO");
      else puts("YES");
    } else {
      if (!p1 || !p2 || p1 >= p2) puts("NO");
      else puts("YES");
    }
  }
  return 0;
}

D. Prefix Permutation Sums

Problem

题目

Sol & Code

看相邻两个数之间的差值，以及最大值是否是 \(1+2+\dots+n\)。

差值要求 \(1\sim n\) 每个数都出现一次。判断是否满足即可。

#include <bits/stdc++.h>
#define N 200001

typedef long long ll;

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

ll a[N], d[N];
int T, n, vis[N];

int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) vis[i] = 0;
    int cnt1 = 0, cnt2 = 0, ret = 0;
    bool okay = true;
    for (int i = 1; i < n; ++i) scanf("%lld", &a[i]);
    for (int i = 1; i < n; ++i) {
      d[i] = a[i] - a[i - 1];
      if (d[i] > 2 * n - 1) { okay = false; break; }
      if (d[i] > n) {
        if (cnt1 || cnt2) { okay = false; break; }
        ++cnt1, ret = d[i];
      } else {
        if (vis[d[i]]) {
          if (cnt2 || cnt1) { okay = false; break; }
          ++cnt2;
        }
        ++vis[d[i]];
      }
    }
    if (!okay) { puts("NO"); continue; }
    else {
      int res = 0, cnt = 0;
      for (int i = 1; i <= n; ++i) {
        if (!vis[i]) res += i, ++cnt;
      }
      if (cnt == 1 || vis[res] == 2 || res == ret) puts("YES");
      else puts("NO");
    }
  }
  return 0;
}

E. Nastya and Potions

Problem

题目

Sol & Code

若 \(a\) 能和其他的混合成 \(b\) 就由 \(a\) 向 \(b\) 连线。

拓扑序 dp

#include <bits/stdc++.h>
#define N 200001

typedef long long ll;

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int cnt, head[N];
std::queue<int> q;
int T, n, k, f[N], du[N], res[N];
struct Edge {
  int v, nxt;
}e[N << 1];

void add(int u, int v) {
  e[++cnt].v = v, e[cnt].nxt = head[u], head[u] = cnt;
}

int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%d", &f[i]), res[i] = 0, head[i] = 0;
    for (int i = 1, x; i <= k; ++i) {
      scanf("%d", &x);
      f[x] = 0;
    }
    cnt = 0;
    for (int i = 1, x; i <= n; ++i) {
      scanf("%d", &x);
      for (int j = 1, u; j <= x; ++j) {
        scanf("%d", &u);
        add(u, i);
        ++du[i];
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (du[i] == 0) q.push(i);
    }
    while (!q.empty()) {
      int u = q.front();
      for (int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].v;
        res[v] += f[u];
        if (res[v] >= f[v]) res[v] = f[v];
        --du[v];
        if (du[v] == 0) {
          f[v] = min(f[v], res[v]);
          q.push(v);
        }
      }
      q.pop();
    }
    for (int i = 1; i <= n; ++i) printf("%d ", f[i]);
    puts("");
  }
  return 0;
}

F. Lisa and the Martians

Problem

题目

Sol & Code

使 \((a_i\oplus x)\&(a_j \oplus x)\) 最大。

找两两之间不同的位的最高位最低的两个数。然后构造 \(x\) 使相同的位都为 \(1\) 即可。

#include <bits/stdc++.h>
#define N 200001

typedef long long ll;

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int T, n, k;
struct qwq {
  int x, id;
  friend bool operator < (qwq q1, qwq q2) {
    return q1.x < q2.x;
  }
}a[N];

int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i].x);
      a[i].id = i;
    }
    std::sort(a + 1, a + n + 1);
    int ans, minn = 2147483647;
    for (int i = 1; i < n; ++i) {
      int flag = a[i].x ^ a[i + 1].x;
      if (flag < minn) {
        minn = flag;
        ans = i;
      }
    }
    printf("%d %d ", a[ans].id, a[ans + 1].id);
    int x = 0;
    for (int i = 0; i < k; ++i) {
      if (minn & (1 << i)) continue;
      else {
        if (a[ans].x & (1 << i)) x += 0;
        else x += (1 << i);
      }
    }
    printf("%d\n", x);
  }
  return 0;
}