Dashboard - The 2021 Sichuan Provincial Collegiate Programming Contest - Codeforces

过题难度：

A K D M H B L

铜奖 5  594

银奖 6 368

金奖 8 755

codeforces.com/gym/103117/problem/A

模拟出牌的过程，打表即可

// Code Start Here	
	int t;
	cin >> t;
	while(t--){
		int n;
		cin >> n;
		if(n == 1)cout << 0 << endl;
		else if(n == 2) cout << 1 <<endl;
		else if(n == 3)cout << 1 <<endl;
		else if(n == 4)cout << 2 <<endl;
		else if(n == 5)cout << 2 <<endl;
		else if(n == 6)cout << 3 <<endl;
		else if(n == 7)cout << 3 <<endl;
		else if(n == 8)cout << 3 <<endl;
		else if(n == 9)cout << 2 <<endl;
		else if(n == 10)cout << 2 <<endl;
		else if(n == 11)cout << 1 <<endl;
		else if(n == 12)cout << 1 <<endl;
		else cout << 0 << endl;
	}
	

Problem - K - Codeforces

我们发现最优解肯定是间隔是K的尽量在一起，模拟即可

	int n , k;
	cin >> n >> k;
	if(k >= n){
		for(int i = 1;i<n;i++)cout << i << " ";
		cout << n << endl;
	}
	else{
		vector<bool> vis(n + 1, false);
		vector<int> ans;
		for(int i = 1;i<=n;i++){
			if(!vis[i]){
				vis[i] = true;
				for(int j = i;j<=n;j += k){
					ans.push_back(j);
					vis[j] = true;
				}
			}
		}
		for(int i = 0;i<(int)(ans.size())-1;i++){
			cout << ans[i] << " ";	
		}
		cout << ans.back() << endl;
	}

Problem - D - Codeforces

模拟剪刀石头布即可

	int a1,b1,c1,a2,b2,c2;
	cin>>a1>>b1>>c1>>a2>>b2>>c2;
	int sum=0;
	int min1=min(a1,b2);
	sum+=min1;
	a1-=min1;
	b2-=min1;
	min1=min(b1,c2);
	sum+=min1;
	b1-=min1;
	c2-=min1;
	min1=min(c1,a2);
	sum+=min1;
	c1-=min1;
	a2-=min1;
	min1=min(a1,a2);
	a2-=min1;
	min1=min(b1,b2);
	b2-=min1;
	min1=min(c1,c2);
	c2-=min1;
	int p=0;
	p=max(a2,b2);
	p=max(p,c2);
	sum-=p;
	cout<<sum<<endl;

codeforces.com/gym/103117/problem/M

我们考虑整个扩张的过程，在0 ~ p0内的扩张肯定优先走0~p0，而在0~p0外的扩张对于1e5 * 1e5不现实，考虑优化，只保留一个最大的区间，如果这个区间可以通过，那么一定存在答案

	int n , k , x , p0;
	cin >> n >> k >> x >> p0;
	vector<int> v(n + 1);
	for(int i = 1;i<=n;i++)cin >> v[i];
	vector<int> l(k+1),r(k+1);
	for(int i = 1;i<=k;i++)cin >> l[i];
	for(int i = 1;i<=k;i++)cin >> r[i];
	int max_val = -1;
	for(int i = 1;i<=k;i++){
		if(l[i] >= p0){
			max_val = max(max_val , r[i] - l[i]);
		}
	}
	int max_time = max(p0 , max_val);
	int ans = 0;
	for(int i = 1;i<=n;i++){
		if(v[i] * max_time >= x)ans++;
	}
	cout << ans << endl;

codeforces.com/gym/103117/problem/H

大模拟

	string s;
	cin >> s;
	string now = "";
	// 构造 imasu
	if(!s.empty()) now += s.back() , s.pop_back();
	if(!s.empty()) now += s.back() , s.pop_back();
	if(!s.empty()) now += s.back() , s.pop_back();
	if(!s.empty()) now += s.back() , s.pop_back();
	if(!s.empty()) now += s.back() , s.pop_back();
	reverse(now.begin(),now.end());
	if(now != "imasu")cout << s << now << endl;
	else{
		char ch2 , ch1;
		if(!s.empty())ch2 = s.back() , s.pop_back();
		if(!s.empty())ch1 = s.back() , s.pop_back();
		if(ch1 == 'c' && ch2 == 'h'){
			cout << s << "tte" << endl;
			return;
		}
		if(ch1 == 's' && ch2 == 'h'){
			cout << s << "shite" << endl;
			return;
		}
		if(ch1 == 'i' && ch2 == 'k' && s.empty()){
			cout << s  << "itte" << endl;
			return;
		}
		if(ch1 == 'i' && ch2 == 'k' && !s.empty()){
			cout << s  << "iite" << endl;
			return;
		}
		if(ch2 == 'r'){
			cout << s << ch1 << "tte" << endl;
			return ;
		}
		if(ch2 == 'm' || ch2 == 'b' || ch2 == 'n'){
			cout << s << ch1 << "nde" << endl;
			return;
		}
		if(ch2 == 'k'){
			cout << s << ch1 << "ite" << endl;
			return;
		}
		if(ch2 == 'g'){
			cout << s << ch1 << "ide" << endl;
			return;
		}
		cout << s << ch1 << ch2 << now << endl;

Problem - B - Codeforces

考虑从奇偶性入手来找回合数，特别要注意的是如果一开始为0的话那么后面无法进行，需要特判

	int n,m,k;
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	int sum1=k/n;
	int sum2=k%n;
	vector<int>pre(m+1),last(m+1);
	map<int,int>p;
	vector<int>sum(n+1);
//	cout<<sum1<<endl;
	if(sum1==0){
		map<int,int>p3;
		for(int i=1;i<=sum2;i++){
			if(p3[a[i]]){
				sum[i]++;
				p3[a[i]]--;
				continue;
			}
			p3[a[i]]++;
		}
		for(int i=1;i<n;i++){
			cout<<sum[i]<<' ';
		}
		cout<<sum[n]<<endl;
		return ;
	}
	for(int i=1;i<=n;i++){
		p[a[i]]++;
		if(p[a[i]]==1){
			pre[a[i]]=i;
		}
		last[a[i]]=i;
	}
	map<int,int>p4;
	for(int i=1;i<=n;i++){
		p4[a[i]]++;
		if(p[a[i]]%2==0){
		if(p4[a[i]]%2==0){
			sum[i]+=sum1;
		}
		}
		else{
			if(p[a[i]]==1){
				sum[i]+=sum1/2;
			}
			else{
				if(sum1&1){
					if(p4[a[i]]&1){
						sum[i]+=sum1/2;
					}
					else{
						sum[i]+=sum1/2+1;
					}
				}
				else{
					sum[i]+=sum1/2;
				}
			}
		}
	}
	map<int,int>p2;
	for(int i=1;i<=sum2;i++){
		p2[a[i]]++;
		if(p[a[i]]==1){
			if(sum1%2==0){
				continue;
			}
			else{
				sum[i]++;
				continue;
			}
		}
			if(p[a[i]]%2==1){
				if(sum1&1){
					if(p2[a[i]]%2==1){
						sum[i]++;
						continue;
					}
					else{
						continue;
					}
				}
				else{
					if(p2[a[i]]%2==0){
						sum[i]++;
					}
					continue;
				}
			}
		if(p2[a[i]]%2==0){
			sum[i]++;
			continue;
		}
	}
	for(int i=1;i<n;i++){
		cout<<sum[i]<<' ';
	}
	cout<<sum[n];
	cout<<endl;

https://codeforces.com/gym/103117/problem/L

按照权值分类，跑 100 次bfs，将题目给出的终点作为起点，将所有其他所有点作为终点。在符合条件的权值下找到最小值

vector<int>g[N];
vector<int>val[110];
int ans[110][N];
int vis[N];

void bfs(int n){
	queue<pair<int, int>>q;
	for(int i=0; i<val[n].size(); i++){
		vis[val[n][i]] = 1;
		q.push(make_pair(val[n][i], 0));
	}
	while(q.size()){
		pair<int,int> x = q.front();
		q.pop();
		int now = x.first;
		ans[n][now] = x.second;
		for(int i=0; i<g[now].size(); i++){
			int nxt = g[now][i];
			if(vis[nxt]) continue;
			vis[nxt] = 1;
			q.push({nxt, x.second + 1});
		}
	}
}

int32_t main(){
	int n, m, q, w = 0;
	cin >> n >> m >> q;
	for(int i=1; i<=n; i++){
		int a;
		cin >> a;
		w = max(a , w);
		val[a].push_back(i);
	}
	for(int i=0; i<m; i++){
		int a, b;
		cin >> a >> b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	for(int i=0; i<=w; i++) for(int j=0; j<=n; j++) ans[i][j] = n;
	for(int i=1; i<=w; i++){
		for(int j=0; j<=n; j++) vis[j] = 0;
		bfs(i);
	}
	while(q--){
		int a, b;
		cin >> a >> b;
		int x = n;
		for(int i=1; i<=b; i++) x = ans[i][a] < x ? ans[i][a] : x;
		if(x == n)cout << -1 << endl;
		else cout << x << endl;
	}
	return 0;
}