A

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#include<bits/stdc++.h>
using namespace std;
using i64=long long;
void solve(){
	int n;cin>>n;
	string s="";
	for(int i=1;i<=n;i++){
		s+=to_string(n);
	}
	cout<<s;
}
int main(){
	int _=1;
	while(_--)solve();
	return 0;
}

B

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#include <iostream>
using namespace std;

int sa (char x,char y){
		if(x > y) swap(x, y);
		if(y-x == 1 || y-x == 4){
		  return 1;
		}
		else return 2;
}

int main() {
	char a, b, c, d;
	cin >> a >> b >> c >> d;
	
	if(sa(a,b)==sa(c,d)){
	  cout<<"Yes"<<endl;
	}
	else cout<<"No"<<endl;

}

C

暴力算完去重就行

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#include <bits/stdc++.h>
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;

    long long now = 1;
    vector<long long> rep;
    for(int i = 0; i < 15; i++){
        rep.push_back(now);
        now = now * 10 + 1;
    }

    vector<long long> ans;
    int m = rep.size();
    for(int i = 0; i < m; i++){
        for(int j = 0; j < m; j++){
            for(int k = 0; k < m; k++){
                ans.push_back(rep[i] + rep[j] + rep[k]);
            }
        }
    }

    sort(ans.begin(), ans.end());
    ans.erase(unique(ans.begin(), ans.end()), ans.end());
    cout << ans[n-1] << endl;
}

D

这场D比较简单

就是先建图，找到1节点最大的子树，然后用a节点所有连通节点的数量减去最大子树的大小后加1

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#include<bits/stdc++.h>
using namespace std;
using i64=long long;
const int N=3e5+10;
vector<int>g[N];
bool vis[N];
queue<int>q;
int bfs(int st,int k){
	memset(vis,0,sizeof vis);
	q.push(st);
	vis[1]=1;
	vis[st]=1;
	k++;
	while(q.size()){
		int f=q.front();
		q.pop();
		for(auto &i:g[f]){
			if(!vis[i]){
				q.push(i);
				k++;
				vis[i]=1;
			}
		}
	}
	return k;
}
void solve(){
	int n;cin>>n;
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	if(g[1].size()==1){
		cout<<"1";
		return;
	}
	int ll=0;
	int ans=0;
	for(auto &i:g[1]){
		int res=bfs(i,0);
		ll+=res;
		ans=max(ans,res);
	}
	cout<<ll-ans+1;
}
int main(){
	int _=1;
	while(_--)solve();
	return 0;
}