add ACM

docs/ACM/CF/div2_893.md

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+---
+title: CFdiv2_893
+description: (ABCDEF)
+categories:
+ - CF
+highlight_shrink: false
+abbrlink: 7fd84b65
+date: 2023-08-18 00:00:00
+mathjax: true
+copyright_author: devilran
+copyright_author_href:
+copyright_url:
+copyright_info:
+---

docs/ACM/CF/div2_899.md

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+---
+title: CFdiv2_899
+description: (ABCDEF)
+categories:
+ - CF
+highlight_shrink: false
+abbrlink: 9f0da27b
+date: 2023-09-25 00:00:00
+mathjax: true
+copyright_author: devilran
+copyright_author_href:
+copyright_url:
+copyright_info:
+---
+
+
+
+
+
+赛时写了ABCD，但C题卡了太久，b题其实完全暴力就可以，但是想歪了，最后也没时间重新写了，过了ACD
+
+
+
+## A
+
+
+
+### 题目
+
+
+
+You are given a sequence $a_{1}, a_{2}, \ldots, a_{n}$. A sequence $b_{1}, b_{2}, \ldots, b_{n}$ is called good, if it satisfies all of the following conditions:
+
+- $b_{i}$ is a positive integer for $i = 1, 2, \ldots, n$;
+- $b_{i} \neq a_{i}$ for $i = 1, 2, \ldots, n$;
+- $b_{1} \lt\; b_{2} \lt\; \ldots \lt\; b_{n}$.
+
+Find the minimum value of $b_{n}$ among all good sequences $b_{1}, b_{2}, \ldots, b_{n}$.
+
+
+
+给定a序列，求满足条件的最小$b_n$
+
+
+
+### 思路
+
+
+
+n很小，直接从t=1开始对应到b就行，如果 $b_i$==$a_i$ , t ++;
+
+
+
+### code
+
+
+
+```cpp
+#include <bits/stdc++.h>
+
+#define LL long long
+#define ULL unsigned long long
+#define x first
+#define y second
+#define endl "\n"
+
+using namespace std;
+
+typedef pair<int, int> PII;
+typedef pair<LL, LL> PLL;
+
+const int INF = 0x3f3f3f3f;
+
+void solve()
+{
+ int n;
+ cin >> n;
+
+ vector<LL> a(n + 1);
+ for (int i = 1; i <= n; i++)
+ {
+  cin >> a[i];
+ }
+
+ LL ans = 0, now = 1;
+ for (int i = 1; i <= n; i++)
+ {
+  if (now == a[i])
+  now++;
+  ans += now;
+  now++;
+ }
+
+ now--;
+ cout << now << endl;
+}
+
+int main()
+{
+ ios::sync_with_stdio(false);
+ cin.tie(0), cout.tie(0);
+
+ int t = 1;
+ cin >> t;
+
+ while (t--)
+ {
+ solve();
+ }
+
+ return 0;
+}
+```
+
+
+
+## B
+
+这道题n很小，其实直接暴力就可以，根本不需要思考方
+
+### 题目
+
+给定n个集合$S_i$, 记所有集合的并集为$S_{all}$, 求从中选取一些集合,这些集合的并集的元素数量需要严格小于$S_{all}$,求选取出来的集合的并集的元素数量的最大值
+
+
+
+### 思路
+
+挨个枚举全集中的每个元素，当这个元素不存在的时候，我可以把哪些集合进行合并。之后合并完最大值就行
+
+
+
+### code
+
+```cpp
+#include <bits/stdc++.h>
+
+#define LL long long
+#define ULL unsigned long long
+#define x first
+#define y second
+#define endl "\n"
+
+using namespace std;
+
+typedef pair<int, int> PII;
+typedef pair<LL, LL> PLL;
+
+const int INF = 0x3f3f3f3f;
+const int N = 55;
+
+set<int> mp[N], all;
+
+void solve()
+{
+ int n;
+ cin >> n;
+
+ all.clear();
+ for (int i = 1; i <= n; i++)
+ {
+  mp[i].clear();
+  int m;
+  cin >> m;
+
+  for (int j = 1; j <= m; j++)
+  {
+  int t;
+  cin >> t;
+  mp[i].insert(t);
+  all.insert(t);
+  }
+ }
+
+ int ans = 0;
+ for (auto t : all)
+ {
+  set<int> res;
+  for (int i = 1; i <= n; i++)
+  {
+  if (mp[i].find(t) != mp[i].end())
+  {
+  continue;
+  }
+
+  for (auto tt : mp[i])
+  {
+  res.insert(tt);
+  }
+  }
+
+  ans = max(ans, (int)res.size());
+ }
+
+ cout << ans << endl;
+}
+
+int main()
+{
+ ios::sync_with_stdio(false);
+ cin.tie(0), cout.tie(0);
+
+ int t;
+ cin >> t;
+
+ while (t--)
+ {
+ solve();
+ }
+
+ return 0;
+}
+```
+
+
+
+## C
+
+
+
+### 题目
+
+有n张牌堆成一堆，每张牌上写了一个价值，有以下两个操作，求操作完可以获得的最大价值。任意时候都可以停止操作。（价值中存在负数，否则就全取完就好了）
+
+操作 --- 从上向下数第几张
+
+- 拿走第奇数张，之后获得这张牌上的价值
+- 拿走第偶数张，之后直接将这张牌扔掉，并不获得这张牌的价值
+
+注意，每次拿走一站牌后，所有牌的顺序会重新计算，奇偶性可能发生改变
+
+
+
+### 思路
+
+
+
+如果一定选取第i个元素
+
+对于i后面的奇数位置的元素，可以倒着直接取完
+
+对于i后面偶数位置的元素，可以取完i，之后后这些要取的元素，就变成奇数了
+
+后缀和优化
+
+
+
+所以枚举第一个要选取的元素即可
+
+### code
+
+```cpp
+#include <bits/stdc++.h>
+
+#define LL long long
+#define ULL unsigned long long
+#define x first
+#define y second
+#define endl "\n"
+
+using namespace std;
+
+typedef pair<int, int> PII;
+typedef pair<LL, LL> PLL;
+
+const int INF = 0x3f3f3f3f;
+
+void solve()
+{
+ int n;
+ cin >> n;
+
+ vector<LL> a(n + 1), suf(n + 2);
+ for (int i = 1; i <= n; i++)
+ {
+  cin >> a[i];
+ }
+
+ suf[n + 1] = 0;
+ for (int i = n; i >= 1; i--)
+ {
+  suf[i] = suf[i + 1] + max(a[i], 0ll);
+ }
+
+ LL ans = 0;
+ for (int i = 1; i <= n; i++)
+ {
+  ans = max(ans, (i % 2 == 1) * a[i] + suf[i + 1]);
+ }
+
+ cout << ans << endl;
+}
+
+int main()
+{
+ ios::sync_with_stdio(false);
+ cin.tie(0), cout.tie(0);
+
+ int t;
+ cin >> t;
+
+ while (t--)
+ {
+ solve();
+ }
+
+ return 0;
+}
+```
+
+
+
+
+