• 每天的题目总结在一起，睡觉之前一定要再看一遍复习一下思路
• 尽快恢复状态，计划是先把前几场没补的牛客补一下，然后开始刷洛谷

牛客 寒假集训4F 来点每日一题

题目链接

这题首先要想到 dp

可以选很多六个数的情况比较复杂，我们首先考虑只能选六个数的情况

二维dp，使用 dp[i][j] 表示在前 i 个数里选，已经选了 j 个数的最大值，转移的时候，要不然就是前 i - 1 个数的最优情况，要不然就是前 i - 1 里选 j - 1 个数，把第 i 个数当做最后一个被选择的数的最优情况

解决了只能选六个数的问题，再回头看原题，因为能选很多个六个，所以由这里想到区间处理，maxx[i][j][k] 表示 [i, j] 中选了 k 个的最大值，minn[i][j][k] 表示 [i, j] 中选了 k 个的最小值（为什么要记录下最小值：因为有乘法，负负也能得正），之后 dp[i] 表示前 i 个数中的最佳答案

#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 + 1);for (int i = 1; i <= n; i ++ ) cin >> a[i];vector<vector<vector<int>>> maxx(n + 1, vector<vector<int>>(n + 1, vector<int>(7, -INF))), minn(n + 1, vector<vector<int>>(n + 1, vector<int>(7, INF)));vector<int> dp(n + 1);for (int i = 1; i <= n; i ++ ){for (int j = i; j <= n; j ++ ){maxx[i][j][1] = max(maxx[i][j - 1][1], a[j]);if (j - i + 1 > 1) maxx[i][j][2] = max(maxx[i][j - 1][2], maxx[i][j - 1][1] - a[j]);if (j - i + 1 > 2) maxx[i][j][3] = max({maxx[i][j - 1][3], maxx[i][j - 1][2] * a[j], minn[i][j - 1][2] * a[j]});if (j - i + 1 > 3) maxx[i][j][4] = max(maxx[i][j - 1][4], maxx[i][j - 1][3] - a[j]);if (j - i + 1 > 4) maxx[i][j][5] = max({maxx[i][j - 1][5], maxx[i][j - 1][4] * a[j], minn[i][j - 1][4] * a[j]});if (j - i + 1 > 5) maxx[i][j][6] = max(maxx[i][j - 1][6], maxx[i][j - 1][5] - a[j]);minn[i][j][1] = min(minn[i][j - 1][1], a[j]);if (j - i + 1 > 1) minn[i][j][2] = min(minn[i][j - 1][2], minn[i][j - 1][1] - a[j]);if (j - i + 1 > 2) minn[i][j][3] = min({minn[i][j - 1][3], minn[i][j - 1][2] * a[j], maxx[i][j - 1][2] * a[j]});if (j - i + 1 > 3) minn[i][j][4] = min(minn[i][j - 1][4], minn[i][j - 1][3] - a[j]);if (j - i + 1 > 4) minn[i][j][5] = min({minn[i][j - 1][5], minn[i][j - 1][4] * a[j], maxx[i][j - 1][4] * a[j]});if (j - i + 1 > 5) minn[i][j][6] = min(minn[i][j - 1][6], minn[i][j - 1][5] - a[j]);}}for (int i = 1; i <= n; i ++ ){for (int j = 0; j <= i; j ++ ){if (i - j < 6) dp[i] = max(dp[i], dp[j]);else dp[i] = max(dp[i], dp[j] + maxx[j + 1][i][6]);}}cout << dp[n] << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

牛客 寒假集训4H 数三角形（hard）

题目链接

这一题没做出来不太应该，debug的时候发现问题了但是没有想到解决办法，现在看来就是树状数组不太熟悉，区间求和思路被前缀和锁死了

大体思路就是枚举每个点为三角形顶点和底边中点，顶点时候更新答案，底边中点的时候在树状数组中加上该点的贡献，还要另外开一个数组v来记录这个点为底边中点的时候最高对哪一行产生影响，那一行计算完之后要消去这个影响

#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 = 3010;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
// const int INF = 0x3f3f3f3f3f3f3f3f;int n, m;
vector<int> tr;int lowbit(int x)
{return x & -x;
}void add(int pos, int x)
{for (int i = pos; i <= n; i += lowbit(i)){tr[i] += x;}
}int query(int x)
{int res = 0;for (; x > 0; x -= lowbit(x)){res += tr[x];}return res;
}void solve()
{cin >> n >> m;vector<string> g(n + 2);for (int i = 1; i <= n; i ++ ){string s;cin >> s;s = " " + s + " ";g[i] = s;}vector<vector<int>> l(n + 2, vector<int>(m + 2));vector<vector<int>> r(n + 1, vector<int>(m + 2));for (int i = 1; i <= n; i ++ ){for (int j = 1; j <= m; j ++ ){if (g[i][j] == '*') l[i][j] = l[i][j - 1] + 1;else l[i][j] = 0;}for (int j = m; j > 0; j -- ){if (g[i][j] == '*') r[i][j] = r[i][j + 1] + 1;else r[i][j] = 0;}}vector<vector<int>> lx(n + 2, vector<int>(m + 2)), rx(n + 2, vector<int>(m + 2));for (int i = n; i > 0; i -- ){for (int j = 1; j <= m; j ++ ){if (g[i][j] == '*'){lx[i][j] = lx[i + 1][j - 1] + 1;rx[i][j] = rx[i + 1][j + 1] + 1;}else lx[i][j] = 0, rx[i][j] = 0;}}i64 ans = 0;for (int j = 1; j <= m; j ++ ){tr = vector<int>(n + 2);vector<vector<int>> v(n + 2);for (int i = n; i > 0; i -- ){if (g[i][j] == '.'){for (auto t : v[i]) add(t, -1);continue;}// 顶点int num = min(lx[i][j], rx[i][j]);if (num >= 2) ans += query(i + num - 1) - query(i);// 底边中点num = min(l[i][j], r[i][j]);if (num > 1){add(i, 1);int lim = i - num + 1;lim = max(lim, 0);v[lim].push_back(i);}for (auto t : v[i]) add(t, -1);}}cout << ans << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

牛客 寒假集训4K 方块掉落

题目链接

果然一复杂就不会写线段树了，只能看出来是线段树但是pushup根本不知道怎么写

struct Node
{int l, r; // 区间端点int sum; // 方块总数int sum_red; // 红色方块总数int lb; // 第一个蓝色方块左边有多少红色方块int rb; // 最后一个蓝色方块右边有多少方块bool eb; // 是否有蓝色方块
} tr[N * 4];

应该怎么想到这个结点信息的定义呢？

首先端点和方块总数不用说了，想一想方块数的翻倍变化都来自于红色，所以要对红色特殊注意，合并的时候，右边结点不收红方块的影响（因为已经计算过了），收到影响的就是左边结点中，在最后一个蓝色方块右边的所有方块，所以我们要记录最后一个蓝色方块右边有多少方块，而影响它们的是右边结点中，第一个蓝色方块左边所有的红色方块，所以我们要记录第一个蓝色方块左边有多少红色方块，更新这个信息的时候需要用到红色方块总数和是否存在蓝色方块，所以记录这两个信息

#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 = 200010;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;string s;
int p[N];struct Node
{int l, r; // 区间端点int sum; // 方块总数int sum_red; // 红色方块总数int lb; // 第一个蓝色方块左边有多少红色方块int rb; // 最后一个蓝色方块右边有多少方块bool eb; // 是否有蓝色方块
} tr[N * 4];void pushup(Node& u, Node& l, Node& r)
{u.sum = (1ll * l.rb * p[r.lb] + l.sum - l.rb + r.sum) % mod;u.sum_red = (l.sum_red + r.sum_red) % mod;if (l.eb) u.lb = l.lb;else if (r.eb) u.lb = (l.sum_red + r.lb) % mod;else u.lb = (l.sum_red + r.sum_red) % mod;if (r.eb) u.rb = r.rb;else if (l.eb) u.rb = (1ll * l.rb * p[r.lb] % mod + r.sum) % mod;else u.rb = u.sum;u.eb = l.eb | r.eb;
}void pushup(int u)
{pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}void build(int u, int l, int r)
{tr[u] = {l, r};if (l == r){tr[u].sum = tr[u].rb = 1;if (s[l] == 'R') tr[u].sum_red = tr[u].lb = 1;if (s[l] == 'B') tr[u].eb = true;return;}int mid = l + r >> 1;build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);pushup(u);
}void modify(int u, int x, char c)
{if (tr[u].l == x && tr[u].r == x){tr[u].lb = tr[u].eb = tr[u].sum_red = 0;if (c == 'B') tr[u].eb = true;if (c == 'R') tr[u].lb = tr[u].sum_red = 1;return;}int mid = tr[u].l + tr[u].r >> 1;if (x <= mid) modify(u << 1, x, c);else modify(u << 1 | 1, x, c);pushup(u);
}Node query(int u, int l, int r)
{if (tr[u].l >= l && tr[u].r <= r) return tr[u];int mid = tr[u].l + tr[u].r >> 1;if (r <= mid) return query(u << 1, l, r);else if (l > mid) return query(u << 1 | 1, l, r);else{Node res;auto left = query(u << 1, l, mid);auto right = query(u << 1 | 1, mid + 1, r);pushup(res, left, right);return res;}
}void solve()
{int n, q;cin >> n >> q;p[0] = 1;for (int i = 1; i <= n; i ++ ) p[i] = p[i - 1] * 2 % mod;cin >> s;s = " " + s;build(1, 1, n);while (q -- ){int op;cin >> op;if (op == 1){int p;char c;cin >> p >> c;modify(1, p, c);}else{int l, r;cin >> l >> r;auto res = query(1, l, r);cout << res.sum % mod << '\n';}}
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

牛客 寒假集训5B tomorin的字符串迷茫值

题目链接

要比想象中简单很多，对于目标字符串mygo，因为不能删连续的两个字符，所以只有这些删除方式"mygo", "m_ygo", "my_go", "myg_o", "m_y_go", "m_yg_o", "my_g_o", "m_y_g_o"，只需要在原来的字符串中枚举是否出现过这些子串即可（暴力枚举就可），只要出现过，当前情况的方案数就是左右两边的字符串方案数相乘，所以我们先预处理出长度为i的字符串有多少种删除方案，dp即可，dp[i] = dp[i - 1] + dp[i - 2]

#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()
{string s;cin >> s;int n = s.size();s = " " + s + "enfikojlrnvol";vector<int> cnt(n + 1);cnt[0] = 1, cnt[1] = 2;for (int i = 2; i <= n; i ++ ) cnt[i] = (cnt[i - 1] + cnt[i - 2]) % mod;vector<string> v = {"mygo", "m_ygo", "my_go", "myg_o", "m_y_go", "m_yg_o", "my_g_o", "m_y_g_o"};auto check = [&](string s1, string s2){for (int i = 0; i < s1.size(); i ++ ){if (s1[i] == s2[i]) continue;else if (s1[i] == '_') continue;else return false;}return true;};int sum = 0;for (int i = 1; i <= n; i ++ ){for (auto t : v){auto j = s.substr(i, t.size());if (check(t, j)){sum = (sum + cnt[i - 1] * cnt[n - (i + t.size() - 1)]) % mod;}}}cout << sum << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

牛客 寒假集训5E soyorin的数组操作（easy）

题目链接

偶数是一定满足情况的，只需要判定奇数

我们可以计算出每一个偶数位最多能进行多少次操作，进行最多次操作判断即可

#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;vector<int> a(n + 1);for (int i = 1; i <= n; i ++ ) cin >> a[i];if (n % 2 == 0){cout << "YES\n";return;}int op = 0;for (int i = n - 1; i >= 1; i -= 2){if (a[i] + op * i > a[i + 1]){cout << "NO\n";return;}a[i] += op * i;a[i - 1] += op * (i - 1);int diff = a[i + 1] - a[i];int cnt = diff / i;op += cnt;a[i] += cnt * i;a[i - 1] += cnt * (i - 1);if (a[i] < a[i - 1]){cout << "NO\n";return;}}cout << "YES\n";
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

牛客 寒假集训5F soyorin的数组操作（hard）

题目链接

利用E题代码判断能否实现，如果能实现的话，答案就是相邻两数最大差值

#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 + 1);for (int i = 1; i <= n; i ++ ) cin >> a[i];int ans = 0;if (n % 2 == 0){for (int i = n - 1; i >= 1; i -- ){if (a[i] > a[i + 1]) ans = max(ans, a[i] - a[i + 1]);}}else{for (int i = n - 1; i >= 1; i -- ){if (a[i] > a[i + 1]) ans = max(ans, a[i] - a[i + 1]);}int op = 0;for (int i = n - 1; i >= 1; i -= 2){if (a[i] + op * i > a[i + 1]){cout << "-1\n";return;}a[i] += op * i;a[i - 1] += op * (i - 1);int diff = a[i + 1] - a[i];int cnt = diff / i;op += cnt;a[i] += cnt * i;a[i - 1] += cnt * (i - 1);if (a[i] < a[i - 1]){cout << "-1\n";return;}}}cout << ans << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

牛客 寒假集训5J rikki的数组陡峭值

题目链接

贪心，每次取最优，记录一下可以取的左右端点

#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> l(n), r(n);for (int i = 0; i < n; i ++ ) cin >> l[i] >> r[i];int ltl = l[0], ltr = r[0];int ans = 0;for (int i = 1; i < n; i ++ ){if (ltl <= r[i] && ltr >= l[i]){ltl = max(ltl, l[i]);ltr = min(ltr, r[i]);}else if (ltr >= l[i] && ltl <= r[i]){ltl = max(ltl, l[i]);ltr = min(ltr, r[i]);}else{if (r[i] < ltl){ans += abs(r[i] - ltl);ltl = r[i];ltr = r[i];}else if (l[i] > ltr){ans += abs(l[i] - ltr);ltl = l[i];ltr = l[i];}}}cout << ans << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}