Codeforces Round #670 (Div. 2) 深夜掉分（A - C题补题）

1406A. Subset Mex

https://codeforces.com/contest/1406/problem/A

Example

input

4
6
0 2 1 5 0 1
3
0 1 2
4
0 2 0 1
6
1 2 3 4 5 6

output

5
3
4
0

Note

In the first test case,$A=\{0,1,2\},B=\{0,1,5\}$ is a possible choice.

In the second test case, $A=\{0,1,2\},B=\varnothing$ is a possible choice.

In the third test case, $A=\{0,1,2\},B=\{0\}$ is a possible choice.

In the fourth test case,$A=1,3,5,B=2,4,6$ is a possible choice.

题意：

给定一个集合，并定义 $mex$ 操作：集合中的最小非负数。

如：$mex(\{1,4,0,2,2,1\})=3$

求 集合分为两部分的最大值：$max(mex(A)+mex(B))$

思路：

通过维护两个变量从0开始，如果有0、1、2、3...这样的直接慢慢向上叠加

#include<bits/stdc++.h>
#define ms(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long ll;
const int N = 1e5 + 100;
ll n, a[N];
void solve() {
	cin >> n; 
	for (int i = 0; i < n; ++i)cin >> a[i];
	sort(a, a + n);
	ll m = 0, k = 0;
	for (int i = 0; i < n; ++i) {
		if (a[i] == m)m++;
		else if (a[i] == k)k++;
	}
	cout << m + k << endl;//m、k相当于两个集合中的非负最小值
}

int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	ll _; cin >> _;
	while (_--)solve();
}

1406B. Maximum Product

https://codeforces.com/contest/1406/problem/B

Example

input

4
5
-1 -2 -3 -4 -5
6
-1 -2 -3 1 2 -1
6
-1 0 0 0 -1 -1
6
-9 -7 -5 -3 -2 1

output

-120
12
0
945

Note

In the first test case, choosing $a1,a2,a3,a4,a5$ is a best choice: $(-1)\cdot (-2)\cdot (-3)\cdot (-4)\cdot (-5)=-120$.

In the second test case, choosing $a1,a2,a3,a5,a6$ is a best choice: $(-1)\cdot (-2)\cdot (-3)\cdot 2\cdot (-1)=12$.

In the third test case, choosing$a1,a2,a3,a4,a5$ is a best choice: $(-1)\cdot 0\cdot 0\cdot 0\cdot (-1)=0$.

In the fourth test case, choosing $a1,a2,a3,a4,a6$ is a best choice: $(-9)\cdot (-7)\cdot (-5)\cdot (-3)\cdot 1=945$.

题意：

给定 大小为n的一个数组，求下标 $(i,j,k,l,t)(i<j<k<l<t).$ 使得$a1,a2,a3,a4,a5$ 最大

思路：

一开始以为不能排序，搞得卡了好久。

先对所给数组进行排序，这样必然倒数5个最大，又因为存在负数的关系，所以也许 $-\ast -$ 反而最大。详情见代码

#include<bits/stdc++.h>
#define ms(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long ll;
const int N = 1e5 + 100;
ll n, a[N];
void solve() {
	cin >> n; ll ans[4] = {};
	for (int i = 1; i <= n; i++) cin >> a[i];
	sort(a + 1, a + n + 1);
	ans[0] = a[n] * a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4];
	ans[1] = a[n] * a[n - 1] * a[n - 2] * a[1] * a[2];
	ans[2] = a[n] * a[1] * a[2] * a[3] * a[4];
	sort(ans, ans + 3); cout << ans[2] << endl;
}

int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	ll _; cin >> _;
	while (_--)solve();
}

1406C. Link Cut Centroids

https://codeforces.com/contest/1406/problem/C

题目太长这里不粘贴了。

题意：

思路：

DFS搜索，详细待补。请先阅读代码

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e5 + 10;
int n;
int p1, p2, p3, c[N];
vector<int>g[N];
void dfs(int u, int fa) {
	c[u] = 1;
	for (auto v:g[u]) if (v != fa) {
		dfs(v, u), c[u] += c[v];
	}
	if (!p3) {
		if (c[u] == 1) p1 = fa, p2 = u;
		if (n - c[u] == c[u]) p3 = fa;
	}
}
signed main() {
	ios::sync_with_stdio(false);
	int t;
	cin >> t;
	while (t--) {
		cin >> n;
		p1 = p2 = p3 = 0;
		for (int i = 1; i <= n; i++) g[i].clear(), c[i] = 0;
		for (int i = n; --i;) {
			int u, v;
			cin >> u >> v;
			g[u].push_back(v);
			g[v].push_back(u);
		}
		dfs(1, -1);
		cout << p1 << ' ' << p2 << '\n' << p2 << ' ' << (p3 ? p3 : p1) << '\n';
	}
	return 0;
}