Codeforces Round 1050 (Div. 4)补题

补题链接：Codeforces Round 1050 (Div. 4)

A. Sublime Sequence

如果 $n$ 为奇数，输出 $x$；否则输出 $0$.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;const int N = 200010, M = N * 2, mod = 1e9 + 7;ll n, x;void solve()
{cin >> x >> n;if (n & 1) cout << x << endl;else cout << 0 << endl;
}int main()
{cin.tie(0);ios::sync_with_stdio(false);ll Case = 1;cin >> Case;while (Case--) solve();return 0;
}

B. Lasers

输出 $n+m$.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;const int N = 200010, M = N * 2, mod = 1e9 + 7;ll n, m, x, y;
ll a[N], b[N];void solve()
{cin >> n >> m >> x >> y;for (int i = 1; i <= n; i++) cin >> a[i];for (int i = 1; i <= m; i++) cin >> b[i];cout << n + m << endl;
}int main()
{cin.tie(0);ios::sync_with_stdio(false);ll Case = 1;cin >> Case;while (Case--) solve();return 0;
}

C. Pacer

记录前一次的位置 $pre$ 和时间点 $cur$，设 $d$ 为 $a-cur$，对答案的贡献最大为 $d$ 或 $d-1$，取值取决于状态的改变和 $d$ 的奇偶性。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;const int N = 200010, M = N * 2, mod = 1e9 + 7;ll n, m;void solve()
{cin >> n >> m;ll res = 0, cur = 0, pre = 0, a, b;while (n--){cin >> a >> b;ll d = a - cur;if (b != pre) res += (d & 1) ? d : d - 1;else res += (d & 1) ? d - 1 : d;cur = a;pre = b;}res += m - a;cout << res << endl;
}int main()
{cin.tie(0);ios::sync_with_stdio(false);ll Case = 1;cin >> Case;while (Case--) solve();return 0;
}

D. Destruction of the Dandelion Fields

如果存在奇数，加上所有偶数，并从大到小加上一半的奇数；否则输出 $0$.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;const int N = 200010, M = N * 2, mod = 1e9 + 7;ll n;void solve()
{cin >> n;vector<ll> even, odd;for (int i = 1; i <= n; i++){ll t;cin >> t;if (t & 1) odd.push_back(t);else even.push_back(t);}if (odd.size() == 0){cout << 0 << endl;return;}ll res = 0;for (auto i : even) res += i;sort(odd.begin(), odd.end());for (int i = odd.size() - 1; i >= (int)(odd.size() / 2); i--)res += odd[i];cout << res << endl;
}int main()
{cin.tie(0);ios::sync_with_stdio(false);ll Case = 1;cin >> Case;while (Case--) solve();return 0;
}

E. Split

使用双指针，从 $1$ 到 $n$ 枚举右指针 $r$，同时记录最小的左指针，满足 $[l,r]$ 区间内每个数的数量不超过上界，每次在答案上加上 $r-l+1$.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;const int N = 200010, M = N * 2, mod = 1e9 + 7;ll n, k;
ll a[N];
ll up[N], cnt[N];void solve()
{cin >> n >> k;for (int i = 1; i <= n; i++) cin >> a[i], up[i] = cnt[i] = 0;bool flag = true;for (int i = 1; i <= n; i++) up[a[i]]++;for (int i = 1; i <= n; i++)if (up[i] % k != 0){flag = false;break;}else up[i] /= k;if (!flag){cout << 0 << endl;return;}ll l = 1, res = 0;for (int i = 1; i <= n; i++){if (cnt[a[i]] + 1 > up[a[i]]){while (l < i && a[l] != a[i]) cnt[a[l]]--, l++;cnt[a[l]]--, l++;}cnt[a[i]]++;res += i - l + 1;}cout << res << endl;
}int main()
{cin.tie(0);ios::sync_with_stdio(false);ll Case = 1;cin >> Case;while (Case--) solve();return 0;
}

F. Gravity Falls

每次从第 $k$ 个位置找到字典序最小的数组，然后将该数组的所有数加入答案，更新 $k$ 为数组长度 + 1.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;const int N = 200010, M = N * 2, mod = 1e9 + 7;ll n;
ll s[N];void solve()
{cin >> n;vector<vector<ll>> a(n + 1);set<ll> have;for (int i = 1; i <= n; i++) have.insert(i);ll mx = 0;for (int i = 1; i <= n; i++){cin >> s[i];for (int j = 0; j < s[i]; j++){ll t;cin >> t;a[i].push_back(t);}mx = max(mx, s[i]);}vector<ll> res;for (int i = 0; i < mx; i++){auto tmp = have;ll id, mn = LLONG_MAX;for (auto j : tmp){if (s[j] < i + 1) have.erase(j);else if (a[j][i] < mn){mn = a[j][i];id = j;}else if (a[j][i] == mn){bool flag = true, Equal = true;for (int k = i; k < min(s[j], s[id]); k++){if (a[j][k] != a[id][k]) Equal = false;if (a[j][k] > a[id][k]){flag = false;break;}else if (a[j][k] < a[id][k]) break;}if (Equal && s[j] > s[id]) flag = false;if (flag) id = j;}}for (int j = i; j < s[id]; j++) res.push_back(a[id][j]);have.erase(id);i = s[id] - 1;}for (auto i : res) cout << i << ' ';cout << endl;
}int main()
{cin.tie(0);ios::sync_with_stdio(false);ll Case = 1;cin >> Case;while (Case--) solve();return 0;
}

G. Farmer John’s Last Wish

在 $O(n\sqrt{n})$ 的时间复杂度内求出第 $i$ 个位置前所有数的所有因子的个数，该位置的答案是数量小于 $i$ 的因子的数量最大值。使用 $set$ 的时间复杂度会达到 $O(n\sqrt{n}logn)$，会超时。

技巧：第 $i$ 个位置的答案一定大于等于前一个位置的答案，如果答案比前一个位置大，取到答案的因子一定是 $a[i-1]$ 或 $a[i]$ 的因子。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
#define x first
#define y secondconst int N = 200010, M = N * 2, mod = 1e9 + 7;int n;
int a[N];
int mp[N];
vector<int> fac[N + 10];void init()
{for (int i = 1; i <= N; i++)for (int j = 1; j * j <= i; j++)if (i % j == 0){if (j != 1) fac[i].push_back(j);if (j * j != i) fac[i].push_back(i / j);}
}void solve()
{cin >> n;for (int i = 1; i <= n; i++) cin >> a[i], mp[i] = 0;vector<int> res;int mx = 0;for (int i = 1; i <= n; i++){for (auto j : fac[a[i]]) mp[j]++;for (auto j : fac[a[i]])if (mp[j] != i)mx = max(mx, mp[j]);for (auto j : fac[a[i - 1]])if (mp[j] != i)mx = max(mx, mp[j]);res.push_back(mx);}for (auto i : res) cout << i << ' ';cout << endl;
}int main()
{cin.tie(0);ios::sync_with_stdio(false);init();int Case = 1;cin >> Case;while (Case--) solve();return 0;
}