Codeforces Round #670 (Div. 2) 深夜掉分（A

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

Note

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

In the second test case, $A=\{0,1,2\},B=∅$ 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

Note

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

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

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

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

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 $ - * - $ 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;

}