• 如果您觉得本站非常有看点,那么赶紧使用Ctrl+D 收藏吧

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

开发技术 开发技术 2周前 (09-12) 17次浏览

1406A. Subset Mex

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

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

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=∅) 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 Produc++t

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

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

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)⋅(−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;
}

喜欢 (0)