[lgr-109] Luogu may race II & windy round 6

P8344 「Wdoi-6」 Walking in the lotus field at night

P8344 「Wdoi-6」 Walking in the lotus field at night
This is to deduce a formula , Relatively simple , But at that time, the minus sign was written as a plus sign, and I didn't see it , speechless , Painted for more than ten minutes

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <stack>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%d",a) 
#define endl "\n"
#define int long long
#define mem(a,b) memset(a,b,sizeof a)
#define ull unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define LL long long 
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define x first
#define y second
const double eps = 1e-6;
const int mod = 998244353;
const int MOD = 1e9 + 7;
int x, y, z;
signed main()
{
    
	int t;
    cin >> t;
    while (t--)
    {
    
        cin >> x >> y >> z;
        int p = x * z - (x + 1) * x / 2 + z - x;

        if(x > z || p < y ) cout << "Merry" << endl;
        else cout << "Renko" << endl;
    }
    
	return 0;
}

P8345 「Wdoi-6」 Hua Xu's dream

P8345 「Wdoi-6」 Hua Xu's dream
Some people talk nonsense about graph theory , Almost believed
It's also a push question , Write directly

Pushed during the game

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <stack>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%d",a) 
#define endl "\n"
#define int long long
#define mem(a,b) memset(a,b,sizeof a)
#define ull unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define LL long long 
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define x first
#define y second
const double eps = 1e-6;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int N = 1e6 + 10;
int a[N];
int n, c, q;

signed main()
{
    
	fast;
    cin >> n >> c >> q;
    for (int i = 1; i <= n; i++) cin >> a[i];

    vector<int> v;
    while(q --)
    {
    
        v.clear();
        int x, y; cin >> x;
        int sum = 0;
        for(int i=1; i<=x; i++) 
        {
    
            cin >> y;
            v.push_back(a[y]);
            sum += a[y];
        }
        sort(v.begin(), v.end());

        cout << (x - 1) * c - sum - v[x - 1] + 2 * v[0] << endl;
    }
	return 0;
}

P8346 「Wdoi-6」 The clearest air and sea

P8346 「Wdoi-6」 The clearest air and sea

The main idea of this question is


Find out some hidden dangers of my code


In the valley of Los Angeles ,vector You can't empty it like this
1e6 You're almost ready to use fast reading , Be careful STL, Here we use set Want to remove the heavy edge , to T 了 , Split

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <stack>
#include<bitsdc++.h>
using namespace std;
#define ll long long
#define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%d",a) 
#define endl "\n"
#define int long long
#define mem(a,b) memset(a,b,sizeof a)
#define ull unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define LL long long 
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define x first
#define y second
const double eps = 1e-6;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int N = 2e6 + 10;
int n, m, t;
vector<int> mp[N];
int in[N];
int u, v;
set<PII> s;
bool vis[N];
signed main()
{
    
	fast;
    cin >> t;
    while(t --)
    {
    
        cin >> n >> m;
        
        for(int i=1; i<=m; i++)
        {
    
            cin >> u >> v;
            v += n;
// if(s.count({u, v}) || s.count({v, u})) continue;
            mp[u].push_back(v);
            mp[v].push_back(u);
            in[u] ++ ;
            in[v] ++;
// s.insert({u, v});
        }
        
        n *= 2;
        int res = 0;
        queue<int> q;
        for(int i=1; i<=n; i++) if(in[i] == 1) q.push(i);
        while(q.size())
        {
    
            u = q.front();
            q.pop();
            if(vis[u] || in[u] != 1) continue;

            vis[u] = 1;
            in[u] --;
            for(auto i: mp[u])
            {
    
            	if(in[i] == 1 && !vis[i])
            	{
    
            		res ++;
            		in[i] --;
            		vis[i] = 1;
            	}
                else if(!vis[i]) 
                {
    
                	res ++;
                	vis[i] = 1;
                    for(auto j : mp[i])
                    {
    
						-- in[i];
						-- in[j];
                    	if(in[j] == 1 && !vis[j]) q.push(j);
					}
                    	
                }
            }
        }
        if(res == n / 2) cout << "Renko" << endl;
        else cout << "Merry" << endl;
    
// s.clear();
        for(int i=1; i<=n; i++) 
        {
    
            in[i] = 0;
            vis[i] = 0;
            mp[i].clear();
        }
    }
	return 0;
}

P8347 「Wdoi-6」 The moon on the other side

P8347 「Wdoi-6」 The moon on the other side

Questions and points for attention are in the code

The general method is , If you can find

The circled structure , It must be the first to win , There are other situations , Not one by one , See for yourself if you want to see

Reference resources 1
Reference resources 2

!((n + 1) & n) Tell me about this judgment , This is judgment n+1 Is it right? 2 To the power of , That is, it can judge the full binary tree , Feel reference 1 It shouldn't be , Yes bug, I'll think about it again

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <stack>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%d",a) 
#define endl "\n"
#define int long long
#define mem(a,b) memset(a,b,sizeof a)
#define ull unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define LL long long 
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define x first
#define y second
const double eps = 1e-6;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;
int in[N];
vector<int> v[N];
bool dfs(int u, int fa)
{
    
    for(auto i : v[u])
    {
    
        if(i == fa) continue;
        if(!(in[u] & 1) && in[i] == 1) return true;
        if(dfs(i, u)) return true;
    }
    return false;
}

signed main()
{
    
    int t;
    cin >> t;
    while (t--)
    {
    
        int n; cin >> n;
        
        
        int a, b;
        for(int i=1; i<n; i++)
        {
    
            cin >> a >> b;
            v[a].push_back(b);
            v[b].push_back(a);
            in[a] ++;
            in[b] ++;
        }
        bool flag = 0;
		for(int i=1; i<=n; i++)
			if(in[i] != 2) flag = 0;
			
        if(dfs(1, -1) || !((n + 1) & n)) cout << "Hifuu" << endl;
		//!((n + 1) & n) This determines whether it is a full binary tree , But this judgment method is not rigorous , I don't know why I should make a special judgment  
		// Students have a way of writing , The same way of thinking , But he directly judges whether there is a degree 2 The point of , I can live , There is no need for special judgment  
        else cout << "Luna" << endl;
        
        for(int i=1; i<=n; i++)
		{
    
		 	in[i] = 0;
		 	v[i].clear();
		} 
    }
    
    return 0;
}