CF #488 Div2

A

模拟

#include <bits/stdc++.h>

using namespace std;

int a[100];

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int mask = 0;
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    mask |= 1 << x;
  }
  for (int i = 0; i < n; i++) {
    if ((mask >> a[i]) & 1) {
      printf("%d ", a[i]);
    }
  }
  printf("\n");
  return 0;
}

B

开个堆搞一搞。

#include <bits/stdc++.h>

using namespace std;

const int N = 100010;

int n, k;
int s[N], v[N], o[N];
long long res[N];

int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &s[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  if (k == 0) {
    for (int i = 0; i < n; i++) {
      printf("%d%c", v[i], " \n"[i == n - 1]);
    }
    return 0;
  }
  iota(o, o + n, 0);
  sort(o, o + n, [](int x, int y) {
    return s[x] < s[y];
  });
  multiset<int> heap;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    int ii = o[i];
    res[ii] = sum + v[ii];
    if ((int) heap.size() < k) {
      sum += v[ii];
      heap.emplace(v[ii]);
    } else if (v[ii] > *(heap.begin())) {
      sum += v[ii] - *(heap.begin());
      heap.erase(heap.begin());
      heap.emplace(v[ii]);
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%lld%c", res[i], " \n"[i == n - 1]);
  }
  return 0;
}

C

枚举一下平面上的点，看看有没有公共的。

#include <bits/stdc++.h>

using namespace std;

const int inf = 2e9;

struct square {
  int xmin;
  int xmax;
  int ymin;
  int ymax;

  square() {
    xmin = inf;
    xmax = -inf;
    ymin = inf;
    ymax = -inf;
    for (int i = 0; i < 4; i++) {
      int x, y;
      scanf("%d %d", &x, &y);
      xmin = min(xmin, x);
      xmax = max(xmax, x);
      ymin = min(ymin, y);
      ymax = max(ymax, y);
    }
  }

  bool inside(int x, int y) {
    return xmin <= x && x <= xmax && ymin <= y && y <= ymax;
  }

  bool inside2(int x, int y) {
    if (x < xmin || xmax < x) {
      return false;
    }
    int diff = abs(x - (xmin + xmax) / 2);
    return ymin + diff <= y && y <= ymax - diff;
  }
};

int main() {
  square a = square();
  square b = square();
  for (int x = -100; x <= 100; x++) {
    for (int y = -100; y <= 100; y++) {
      if (a.inside(x, y) && b.inside2(x, y)) {
        puts("yes");
        return 0;
      }
    }
  }
  puts("no");
  return 0;
}

D

沙雕文字游戏题。

#include <bits/stdc++.h>

using namespace std;

const int N = 20;

int n, m;
pair<int, int> a[N], b[N];

int match(pair<int, int>& a, pair<int, int>& b) {
  if (a == b) {
    return -1;
  }
  if (a.first == b.first) {
    return a.first;
  }
  if (a.first == b.second) {
    return a.first;
  }
  if (a.second == b.first) {
    return a.second;
  }
  if (a.second == b.second) {
    return a.second;
  }
  return -1;
}

int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &a[i].first, &a[i].second);
    if (a[i].first > a[i].second) {
      swap(a[i].first, a[i].second);
    }
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &b[i].first, &b[i].second);
    if (b[i].first > b[i].second) {
      swap(b[i].first, b[i].second);
    }
  }
  vector< set<int> > sa(n), sb(m);
  set<int> v;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int res = match(a[i], b[j]);
      if (res == -1) {
        continue;
      }
      v.emplace(res);
      sa[i].emplace(res);
      sb[j].emplace(res);
    }
  }
  if ((int) v.size() == 1) {
    printf("%d\n", *(v.begin()));
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if ((int) sa[i].size() > 1) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < m; i++) {
    if ((int) sb[i].size() > 1) {
      puts("-1");
      return 0;
    }
  }
  puts("0");
  return 0;
}

E

钦定某些船对射，统计下每个 \(y\) 的贡献然后拼一拼。

#include <bits/stdc++.h>

using namespace std;

map< int, pair<long long, long long> > mp;
int a[100], b[100];

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &b[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      mp[a[i] + b[j]].first |= 1ll << i;
      mp[a[i] + b[j]].second |= 1ll << j;
    }
  }
  int ans = 0;
  for (auto x : mp) {
    for (auto y : mp) {
      ans = max(ans, __builtin_popcountll(x.second.first | y.second.first) + __builtin_popcountll(x.second.second | y.second.second));
    }
  }
  printf("%d\n", ans);
  return 0;
}

常识和思考

• E 在处理小数时要非常小心，\(-1 / 2 = 1 / 2\) 卡了半天。。。

原文地址：https://www.cnblogs.com/arg-53/p/9196450.html