https://ac.nowcoder.com/acm/contest/134
This one is simpler than the last one .

triangle 【 thinking enumeration 】

You'll find that , Since it's the longest . So these three sides must be connected together .

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long int LL;
typedef pair<int,int> pii;
LL n,m;
vector<pii>ve;
int main(void)
{
    
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
    
		int x; cin>>x;
		ve.push_back({
    x,i+1});
	}
	sort(ve.begin(),ve.end());
	while(m--) 
	{
    
		int id; cin>>id;
		LL a=0,b=0,c=0,flag=0;// Article 1 with a , Second , Article 3 the .
		for(int i=n-1;i>=0;i--)
		{
    
			if(ve[i].second==id) continue;
			c=b,b=a,a=ve[i].first;
			if(a+b>c&&c!=0)// Satisfy triangle   And all three sides have 
			{
    
				flag=1;
				break;
			}
		}
		if(flag) cout<<a+b+c<<'\n';
		else puts("-1");
	}
	return 0;
}

Game theory 【 Violence enumeration 】

#include<bits/stdc++.h>
using namespace std;
string s;
int main(void)
{
    
	int n; cin>>n;
	while(n--) 
	{
    
		int x;cin>>x;
		s+=to_string(x);
	}
	for(int i=0;i<=2000;i++)
	{
    
		string a=to_string(i);
		if(s.find(a)==-1)
		{
    
			cout<<i;
			break;
		}
	}
	return 0;
}

Bacterial infection 【 simulation 】

#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int st[N][N],n,m;
int dx[4]={
    -1,0,0,1};
int dy[4]={
    0,-1,1,0};
bool check(int x,int y)
{
    
	int cnt=0;
	for(int i=0;i<4;i++)
	{
    
		int tempx=x+dx[i],tempy=y+dy[i];
		if(tempx<=0||tempx>n||tempy<=0||tempy>n) continue;
		if(st[tempx][tempy]) cnt++;
	}
	return cnt>=2;
}
int main(void)
{
    
	cin>>n>>m;
	queue< pair<int,int> >q;
	while(m--)
	{
    
		int x,y; cin>>x>>y;
		q.push({
    x,y});
		st[x][y]=1;
	}
	while(q.size())
	{
    
		auto temp=q.front(); q.pop();
		int x=temp.first,y=temp.second;
		for(int i=0;i<4;i++)
		{
    
			int tempx=x+dx[i],tempy=y+dy[i];
			if(tempx<=0||tempx>n||tempy<=0||tempy>n) continue;
			if(st[tempx][tempy]) continue;
			if(check(tempx,tempy)) q.push({
    tempx,tempy}),st[tempx][tempy]=1;
		}
	}
	bool flag=1;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++) if(!st[i][j]) flag=0;
	if(flag) puts("YES");
	else puts("NO");
	return 0;
}

Suburban spring outing 【floyd + DP】

Floating point output 【 Sign in 】

#include<bits/stdc++.h>
using namespace std;
const int N=1e5*2+10;
typedef long long int LL;
int main(void)
{
    
	string s; cin>>s;
	cout<<s;
	return 0;
}

Equivalent string 【 thinking 】

Black and white 【 simulation 】

Adjacent candy 【 greedy 】

Chorus formation 【 greedy 】

First pretreatment , Compress , Will be adjacent to 1/0 Press together .

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
vector<pii>ve;
int n,cnt;
string s;
int main(void)
{
    
	cin>>n>>s;
	for(int i=0;i<s.size();i++)
	{
    
		int j=i;
		while(j+1<s.size()&&s[j+1]==s[i]) j++;
		ve.push_back({
    s[i]-'0',j-i+1});
		i=j;
	}
	for(int i=0;i<s.size();i++) if(s[i]=='0') cnt++;
	int ans=0;
	for(int i=0;i<ve.size();i++)
	{
    
		if(ve[i].first==0) ans=max(ans,ve[i].second);// yes 0
		if(ve[i].first==1)// yes 1
		{
    
			if(ve[i].second==1)// only one 
			{
    
				int temp=0;
				if(i-1>=0&&ve[i-1].first==0) temp+=ve[i-1].second;
				if(i+1<ve.size()&&ve[i+1].first==0) temp+=ve[i+1].second;
				if(temp<cnt) ans=max(ans,temp+1);// You can exchange 
			}
			else// Only in a certain period 0 After add 1 individual 0.
			{
    
				int temp=0;
				if(i-1>=0&&ve[i-1].first==0) temp=max(temp,ve[i-1].second);
				if(i+1<ve.size()&&ve[i+1].first==0) temp=max(temp,ve[i+1].second);
				if(temp<cnt) ans=max(ans,temp+1);
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}

Obsessive compulsive disorder 【 greedy 】

Obviously, the least operand is the total number of numbers minus the total number of numbers . Because you can combine the repeated and the maximum number at a time . At this time, the quantity is less than 1.

#include<bits/stdc++.h>
using namespace std;
int main(void)
{
    
	int n; cin>>n;
	map<int,int>mp;
	for(int i=0;i<n;i++)
	{
    
		int x; cin>>x;
		mp[x]++;
	}
	cout<<n-mp.size()<<endl;
}