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;
}
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;
}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;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;
}