2023 Henan Mengxin League Game (5): Zhengzhou University of Light Industry

A. Buy love balloons

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 game theory: 

#include <iostream>
using namespace std;
int t,n,m;
string s1 = "Alice",s2 = "Bob";
int main() {
    cin>>t;
    while(t--){
        cin>>n>>m;
        if (n % 3 == 0) {
            cout << s1 << endl;
            continue;
        }
        if (n % 3 == 1) {
            cout << (m != 1 ? "Alice\n" : "Bob\n");
            continue;
        }
        if (n % 3 == 2) {
            cout << (m != 2 ? "Alice\n" : "Bob\n");
            continue;
        }
    }
    return 0;
}

B. Atori-My Beloved Time-

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Title:  

Idea : Simulation

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

int main() { 
    int n;
    cin >> n;
    auto useless = cin.get();
    vector<string> a(n);
    for (int i = 0 ;i  < n ;i ++) {
        getline(cin, a[i]);
    }
    string c[] = {
        "sudo pacman -S ",
        "pacman -R ",
        "pacman -Rscn ",
        "sudo rm -rf /*"
    };
    unordered_set<string> st[2];
    for (int i = 0; i < n; i++) {
        auto check = [&](int j) {
            int x = a[i].size(), y = c[j].size();
            if (x >= y && a[i].substr(0, y) == c[j]) {
                return true;
            } 
            return false;
        };

        if (check(0)) {
            auto s = a[i].substr(c[0].size());
            st[0].insert(s);
            st[1].insert(s);
        } else if (check(1)) {
            auto s = a[i].substr(c[1].size());
            st[0].erase(s);
        } else if (check(2)) {
            auto s = a[i].substr(c[2].size());
            st[0].erase(s);
            st[1].erase(s);
        } else if (check(3)) {
            cout << "wuwuwu\n";
            return 0;
        } else {
            auto o = a[i][0] - '0' - 1;
            auto s = a[i].substr(2);

            cout << (st[o].count(s) ? "yes" : "no") << '\n';
        }
    }

    return 0;
}

 D.01 Score planning

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Title:  

 Idea: In two cases, replace all with 0, or replace all with 1.

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
string s,t; int T,n;
void solve(){
    cin>>n; cin>>s;
    s=' '+s; t=s;
    for(int i=1;i<=n;i++) if(s[i]=='?') s[i]='1';
    for(int i=1;i<=n;i++) if(t[i]=='?') t[i]='0';
    int ans = 0;
    for(int i=1;i<=n;i++){
        int j =i;
        while(j<=n&&s[j]==s[i]) j++; j--;
        ans = max(ans,j-i+1);
        i=j;
    }
    for(int i=1;i<=n;i++){
        int j = i;
        while(j<=n&&t[j]==t[i]) j++; j--;
        ans = max(ans,j-i+1);
        i=j;
    }
    cout<<ans<<endl;
    return ;
}
int main(){
    cin>>T;
    while(T--)  solve();
    return 0;
}

I. Double Pointer

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Title:  

 Idea: Use the hash table to store each a/b, and then add the number of times the opposite number appeared in the map before ans

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
using namespace std;
typedef long long LL;
const int N = 2e5+10;
unordered_map<double,int> mp;
int a[N],b[N],n;
inline void solve(){
    mp.clear();
    cin>>n;
    LL ans = 0;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int j=1;j<=n;j++) cin>>b[j];
    for(int i=1;i<=n;i++){
        double x = 1.0*a[i]/b[i],y=1.0*b[i]/a[i];
        ans += mp[x];
        mp[y]++;
    }
    cout<<ans<<endl;
}
 
int main()
{
    IOS
    int _;
    cin >> _;
    while(_ --) solve();
    return 0;
}
// ai*aj =bi*bj
// ai / bi = bj / aj
// 1 2 1 1/2 3/2

J. tree dp

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Title:  

 Way of thinking: Greedy, the consideration of large weight should be placed in the place with large depth

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'

using namespace std;

typedef pair<int, int> PII;
typedef long long ll;

const int N = 200010, M = N * 2;

int h[N], e[M], ne[M], idx;
int val[N], d[N];
bool st[N];

inline void add(int a, int b)
{
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

void dfs(int u)
{
	st[u] = true;
	for(int i = h[u]; i != -1; i = ne[i])
	{
		int j = e[i];
		if(st[j])continue;
		
		d[j] = d[u] + 1;
		dfs(j);
	}
}

inline void solve()
{
	memset(h, -1, sizeof h);
	idx = 0;
	int n;
	cin >> n;
	for(int i = 1; i < n; i ++)
	{
		int a, b;
		cin >> a >> b;
		add(a, b), add(b, a);
	}
	for(int i = 1; i <= n; i ++)
	{
		cin >> val[i];
		st[i] = false;
	}
	d[1] = 1;
	dfs(1);
	
	sort(d + 1, d + 1 + n);
	sort(val + 1, val + 1 + n);
	ll ans = 0;
	for(int i = 1; i <= n; i ++)
	{
		ans += (ll)val[i] * (ll)d[i];
	}
	cout << ans << endl;
}

int main()
{
	IOS
	int _;
	cin >> _;
	while(_ --)
	{
		solve();
	}
	
	return 0;
}