昨晚cf上大分了开心捏，三月了得想想新的高效一点的训练方法了不能再像个无头苍蝇乱撞了

目前的想法是1700-2000板刷cf，遇到不会的算法就学，思维还需要再提升一下，暑假开始补没有接触过的算法

CF 1937D Pinball

题目链接

手动模拟一下差不多就有思路了

如果当前是<，那么向左碰到>就会转换方向，向右碰到<就会转换方向，用pre记录下标前缀和，cnt[i]记录第i个符号的下标，sum[i]记录到第i个位置为止的符号个数，分类讨论一下，具体看代码吧

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 10;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n;cin >> n;string s;cin >> s;s = " " + s;// pre下标前缀和 cnt[i]第i个符号的下标 sum[i]到第i个位置为止的符号个数vector<int> prel(n + 1), prer(n + 1), cntl(1), cntr(1), suml(n + 1), sumr(n + 1);for (int i = 1; i <= n; i ++ ){if (s[i] == '<'){cntl.push_back(i);suml[i] = suml[i - 1] + 1;sumr[i] = sumr[i - 1];prel[i] = prel[i - 1] + i;prer[i] = prer[i - 1];}else{cntr.push_back(i);suml[i] = suml[i - 1];sumr[i] = sumr[i - 1] + 1;prel[i] = prel[i - 1];prer[i] = prer[i - 1] + i;}}for (int i = 1; i <= n; i ++ ){int ans = 0;if (s[i] == '<'){int left = sumr[i]; // 左侧的>个数int right = suml[n] - suml[i]; // 右侧的<个数if (left <= right) // 从左侧出{right = left;if (left) ans = (left * i - prer[i]) * 2;int tmp = right + suml[i];int pos = cntl[tmp];if (right) ans += (prel[pos] - prel[i] - right * i) * 2;ans += i;}else // 从右侧出{left = right + 1;int tmp = suml[i] + right;int pos = cntl[tmp];if (right) ans = (prel[pos] - prel[i] - i * right) * 2;ans += n - i + 1;tmp = sumr[i] - left + 1;pos = cntr[tmp];if (left) ans += (left * i - (prer[i - 1] - prer[pos - 1])) * 2;}}else if (s[i] == '>'){int left = sumr[i - 1]; // 左侧的>int right = suml[n] - suml[i]; // 右侧的<if (left >= right) // 从右侧出{	left = right;if (right) ans = (prel[n] - prel[i] - i * right) * 2;ans += n - i + 1;int tmp = sumr[i] - left;int pos = cntr[tmp];if (left) ans += (left * i - (prer[i - 1] - prer[pos - 1])) * 2;}else // 从左侧出{right = left + 1;if (left) ans = (left * i - prer[i - 1]) * 2;int tmp = right + suml[i];int pos = cntl[tmp];if (right) ans += (prel[pos] - prel[i] - right * i) * 2;ans += i;}}cout << ans << ' ';}cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1916D Mathematical Problem

题目链接

goodbye2023的一道题，弱智打表

打表可以发现，包含了{1, 6, 9, 0, 0, …}的集合每次都能成立

看这些集合里组成的数：

• 第一类：16900… 19600… 96100…
• 第二类：1060900… 1006009 9060100… ……

根据这个特点取数即可

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 10;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n;cin >> n;if (n == 1) cout << 1 << '\n';else if (n == 3) cout << 169 << '\n' << 196 << '\n' << 961 << '\n';else{cout << 169;for (int i = 0; i < n - 3; i ++ ) cout << 0;cout << '\n' << 196;for (int i = 0; i < n - 3; i ++ ) cout << 0;cout << '\n' << 961;for (int i = 0; i < n - 3; i ++ ) cout << 0;cout << '\n';for (int i = 1; i * 2 <= n - 3; i ++ ){cout << 1;for (int j = 0; j < i; j ++ ) cout << 0;cout << 6;for (int j = 0; j < i; j ++ ) cout << 0;cout << 9;for (int j = 0; j < n - 3 - 2 * i; j ++ ) cout << 0;cout << '\n';cout << 9;for (int j = 0; j < i; j ++ ) cout << 0;cout << 6;for (int j = 0; j < i; j ++ ) cout << 0;cout << 1;for (int j = 0; j < n - 3 - 2 * i; j ++ ) cout << 0;cout << '\n';} }
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1906M Triangle Construction

题目链接

简单思维还想了很久 不太应该

最好的情况就是所有点都能成为三角形的一个顶点，这样答案是sum/3

但是如果一条边的点数maxx太多，在同一条边的三个点不能组成三角形，所以组成三角形的个数就是sum-maxx

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 10;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n;cin >> n;vector<int> a(n);int sum = 0;for (int i = 0; i < n; i ++ ) cin >> a[i], sum += a[i];sort(a.begin(), a.end());cout << min(sum / 3, sum - a[n - 1]) << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

CF 1893B Neutral Tonality

题目链接

我一开始的想法是，把a按从小到大排序，b按从大到小排序，遍历b，在a中找到第一个小于等于b[i]的位置，然后把b[i]放到这个位置前面（这个思路有问题但是我没想明白哪里有问题）

正解思路是：b从大到小排序，双指针，遍历a，每次把大于等于当前a的b放在当前a前面

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 10;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n, m;cin >> n >> m;vector<int> a(n), b(m);for (int i = 0; i < n; i ++ ) cin >> a[i];for (int i = 0; i < m; i ++ ) cin >> b[i];sort(b.begin(), b.end(), greater<int>());int j = 0;for (int i = 0; i < n; i ++ ){while (j < m && b[j] >= a[i]){cout << b[j] << ' ';j ++ ;}cout << a[i] << ' ';}while (j < m){cout << b[j] << ' ';j ++ ;}cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1889B Doremy’s Connecting Plan

题目链接

带点trick的一题

首先可以确定的是，每次连的点 i,j 一定有一个是 1（感性理解一下，证明的话反证法也很容易）

那要想连边，必须满足 a[i] + sum[1] >= i * c （sum[1]是1所在连通块包含的a之和）

由 i 决定的就是 a[i] - i * c

因此把除了1以外的所有 a[i] - i * c 从大到小排序，每次加一个，看能不能满足要求

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 10;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n, c;cin >> n >> c;vector<int> a(n + 1);for (int i = 1; i <= n; i ++ ) cin >> a[i];vector<PII> v;for (int i = 2; i <= n; i ++ ){v.push_back({a[i] - i * c, a[i]});}sort(v.begin(), v.end(), greater<PII>());int sum = a[1];for (int i = 0; i < n - 1; i ++ ) {if (sum + v[i].first >= 0) sum += v[i].second;else{cout << "NO\n";return;}}cout << "YES\n";return;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1881F Minimum Maximum Distance

题目链接

这题先要把题意转换成：找到两个距离最远的标记点

想到这里之后，我联想到树的直径，于是可以先以任意一个标记点为起点，跑最短路，然后找到距离这个起点最远的点，以这个点为起点再跑最短路，此时最远的距离就是两个标记点之间的最远距离了

最小的f值取中点就行

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 10;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n, k;cin >> n >> k;vector<int> a(n);for (int i = 0; i < k; i ++ ) cin >> a[i];vector<vector<int>> g(n + 1);for (int i = 1; i < n; i ++ ){int u, v;cin >> u >> v;g[u].push_back(v);g[v].push_back(u);}vector<int> dist(n + 1, INF);vector<bool> st(n + 1);queue<int> q;dist[a[0]] = 0;q.push(a[0]);while (q.size()){auto t = q.front();q.pop();if (st[t]) continue;st[t] = true;for (int i = 0; i < g[t].size(); i ++ ){int j = g[t][i];if (dist[j] > dist[t] + 1){dist[j] = dist[t] + 1;q.push(j);}}}int maxx = -INF, pos;for (int i = 0; i < k; i ++ ){if (dist[a[i]] > maxx){maxx = dist[a[i]];pos = a[i];}}for (int i = 1; i <= n; i ++ ) {dist[i] = INF;st[i] = false;}dist[pos] = 0;q.push(pos);while (q.size()){auto t = q.front();q.pop();if (st[t]) continue;st[t] = true;for (int i = 0; i < g[t].size(); i ++ ){int j = g[t][i];if (dist[j] > dist[t] + 1){dist[j] = dist[t] + 1;q.push(j);}}}maxx = -INF;for (int i = 0; i < k; i ++ ){if (dist[a[i]] > maxx) {maxx = dist[a[i]];pos = a[i];}}cout << (maxx + 1) / 2 << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}