Codeforces Round 1029 (Div. 3)

A. False Alarm

思路

求最左边关着的门和最右边关着的门，如果距离大于x则说明来不及

// Author: zengyz
// 2025-06-08 22:35#include <bits/stdc++.h>using namespace std;
using i64 = long long;void solve()
{int n, x;cin >> n >> x;vector<int> a(n + 1);for(int i=0;i<n;i++)cin>>a[i];int idx1 = -1;int idx2 = n + 10;for (int i = 0; i < n; i++){if (a[i] == 1){idx1 = i;break;}}for (int i = n - 1; i >= 0; i--){if (a[i] == 1){idx2 = i;break;}}int cnt=idx2-idx1+1;if(cnt>x){cout<<"NO"<<endl;}else cout<<"YES"<<endl;return;
}int main()
{ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int _T = 1;cin >> _T;while (_T--){solve();}return 0;
}

B. Shrink

思路

观察发现先填最大的，然后从大到小依次填在右边和左边就可以满足
deque完美满足条件

// Author: zengyz
// 2025-06-08 22:45#include <bits/stdc++.h>using namespace std;
using i64 = long long;void solve()
{int n;cin >> n;deque<int> d;for (int i = n; i >= 1; i--){if (i % 2)d.push_back(i);elsed.push_front(i);}while (d.size()){cout << d.front() << " ";d.pop_front();}cout<<endl;return;
}int main()
{ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int _T = 1;cin >> _T;while (_T--){solve();}return 0;
}

C. Cool Partition

找出一个数列中最多子段，要求当前子段必须包含上一个子段的所有元素

思路1

想到第一段只选一个一定是最优的，用vector维护每个元素出现的位置，用Set记录已经出现过的所有元素，每次查询出现过的所有元素的最远位置tmp，可以用vector的队头元素取最大值，注意要将vector中存储的小于tmp的元素位置都删掉，即

for (auto it = s.begin(); it != s.end(); it++)
{while (v[*it][0] <= tmp){if (v[*it].size() == 1){flag = false;break;}v[*it].erase(v[*it].begin());}
}

// Author: zengyz
// 2025-06-08 22:52#include <bits/stdc++.h>using namespace std;
using i64 = long long;
void solve()
{int n;cin >> n;vector<int> a(n + 1);vector<vector<int>> v(n + 1);for (int i = 1; i <= n; i++){cin >> a[i];v[a[i]].push_back(i);}bool flag = true;int count = 1, idx = 1;set<int> s;s.insert(a[1]);if (v[a[1]].size() == 1){cout << 1 << endl;return;}elsev[a[1]].erase(v[a[1]].begin());int tmp = idx;while (flag){for (auto it = s.begin(); it != s.end(); it++)tmp = max(tmp, v[*it][0]);for (int j = idx; j <= tmp; j++)s.insert(a[j]);idx = tmp;count++;for (auto it = s.begin(); it != s.end(); it++){while (v[*it][0] <= tmp){if (v[*it].size() == 1){flag = false;break;}v[*it].erase(v[*it].begin());}}}cout << count << endl;return;
}int main()
{ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int _T = 1;cin >> _T;while (_T--){solve();}return 0;
}

思路2

用两个set维护,x代表上一段中出现的,y代表当前这一段出现的
在遍历时在x中删除当前元素，在y中添加当前元素
如果x删完了代表已经出现过上一段出现的所有元素，此时答案加一，并把y–即这一段出现过的所有元素赋给x进行下一段判定

// Author: zengyz
// 2025-06-08 22:52#include <bits/stdc++.h>using namespace std;
using i64 = long long;
void solve()
{int n;cin >> n;vector<int> a(n + 1);for (int i = 1; i <= n; i++){cin >> a[i];}bool flag = true;int count = 0, idx = 1;set<int> x, y;for (int i = 1; i <= n; i++){x.erase(a[i]);y.insert(a[i]);if (x.size() == 0){x = y;y.clear();count++;}}cout << count << endl;return;
}int main()
{ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int _T = 1;cin >> _T;while (_T--){solve();}return 0;
}

D. Retaliation

一个数组，每次可进行两种操作，一种是每个元素都减i,一种是每个元素都减n-i+1
求是否存在仅通过任意次上述两种操作使得数组清零

思路1

用数学方法解决，针对a[1]和a[2]可以求解一元二次方程得到x,y
$$a[1]=x+n\ast y$$
$$a[2]=2\ast x+(n-1)\ast y$$
用2*(1)-(2)得$$y=\frac{2\ast a[1]-a[2]}{(n+1)}$$ $$x=a[1]-n\ast y$$

注意x,y均不能小于0,且x,y必须为整数，然后对n>=3的所有元素带入x,y看是否满足条件，如果均满足则为yes

// Author: zengyz
// 2025-06-09 11:00#include <bits/stdc++.h>using namespace std;
using i64 = long long;void solve()
{int n;cin >> n;vector<int> a(n + 1);for (int i = 1; i <= n; i++)cin >> a[i];int a1 = a[1], a2 = a[2];int a12_a2 = a1 * 2 - a2;if (a12_a2 % (n + 1)){cout << "NO" << endl;return;}int y = a12_a2 / (n + 1);int x = a1 - n * y;if (x < 0 || y < 0){cout << "NO" << endl;return;}for (int i = 3; i <= n; i++){if (a[i] != x * i + (n - i + 1) * y){cout << "NO" << endl;return;}}cout << "YES" << endl;return;
}int main()
{ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int _T = 1;cin >> _T;while (_T--){solve();}return 0;
}

思路2

因为一次1操作和2操作会让整个数组整体减去(n+1)
所以目标是如何将这个数组的所有元素都先减为(n+1)的倍数
又因为每次操作最左边和最右边的数减的差距为(n-1) :
$$（n-1+1）-1=n-1$$
所以我们可以根据两端的差距来决定进行多少次1或2操作
注意如果有余数则说明两端元素不可能通过操作让他们相等，直接输出NO
操作完后左右两端的元素值会相等
如果此时整个数组中
所有元素均为n+1的倍数且所有元素均相等（可被相同次数的1和2操作减为0)且大于等于0(合法)那么就满足条件

// Author: zengyz
// 2025-06-08 23:59#include <bits/stdc++.h>using namespace std;
using i64 = long long;void solve()
{int n;cin >> n;vector<int> a(n + 1);for (int i = 1; i <= n; i++)cin >> a[i];if (abs(a[n] - a[1]) % (n - 1)){cout << "NO" << endl;return;}int res = abs(a[n] - a[1]) / (n - 1);if (a[n] > a[1])for (int i = 1; i <= n; i++)a[i] -= res * i;elsefor (int i = 1; i <= n; i++)a[i] -= res * (n - i + 1);// 特判1if (a[1] < 0 || a[1] % (n + 1)){cout << "NO" << endl;return;}for (int i = 2; i <= n; i++){if (a[i] % (n + 1) || a[i] < 0 || a[i] != a[i - 1]){cout << "NO" << endl;return;}}cout << "YES" << endl;return;
}int main()
{ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int _T = 1;cin >> _T;while (_T--){solve();}return 0;
}