Game (2) of 2022 Henan Mengxin League: solution to supplementary questions of Henan University of Technology

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Just one question , The two questions of the three check-in questions are all right, but they are not written , I can only say that I'm still too delicious , Need to work on ！

Address of the competition

F Garage Kit

I'm a fool , I have considered the meaning of the topic , But why didn't you count the enumeration and it won't timeout , Sorry, I will take time complexity seriously in the future
This question is from 1 Start enumerator to the third power , Look, that's n Constraints , Because it is exponential growth , So there's no overtime , Fortunately, I'm still silly where I count all 1e9 The qualified rational number within ran for a long time without writing it , What a fool

#include<bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;

int T;
int n, m;

int main()
{
    
	cin>>T;
	while(T -- ){
    
		cin>>n;
		int res = 0;
		for(int i = 1; i * i * i <= n; i ++ ){
    
			if(n % (i * i * i) == 0) res ++ ;
		}
		
		cout<<res<<endl;
	}
	return 0;
}

J Sign in

Another question to prove that I am a fool , The idea is the same as the positive solution , But my own consideration is too complicated , Pay attention to simplifying your code in the future , The greatest truths are the simplest , It's a coincidence , Great sound is hard to hear , The great form has no shape
use n^2 The complexity of all a+b Count in map in , after n ^2 enumeration d-c, seek map Whether this number exists in , Don't add a bunch of restrictions to your imagination , Finally, the card timed out

#include<bits/stdc++.h>

using namespace std;

const int N = 1e3 + 10;

int a[N];
int n;
map<int, int>mp;

int main()
{
    
	cin>>n;
	for(int i = 1; i <= n; i ++ ) cin>>a[i];
	
	sort(a + 1, a + 1 + n);
	
	for(int i = 1; i <= n; i ++ ){
    
		for(int j = 1; j <= n; j ++ ){
    
			mp[a[i] + a[j]] = 1;
		}
	}
	
	for(int i = 1; i <= n; i ++ ){
    
		for(int j = 1; j <= n; j ++ ){
    
			if(mp[a[j] - a[i]] == 1){
    
				cout<<"Yes"<<endl;
				return 0;
			}
		}
	}
	cout<<"No"<<endl;
	return 0;
}

L HPU

Simple check-in

#include<bits/stdc++.h>

using namespace std;

string str;
int ans;

int main()
{
    
	cin>>str;
	for(int i = 0; i < str.length(); i ++ ){
    
		if(str[i] == 'H' && str[i + 1] == 'P' && str[i + 2] == 'U'){
    
			ans ++ ;
		}
	}
	cout<<ans<<endl;
	return 0;
}