A-rearrange
Dichotomize the answer, and then cycle one side.
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
using namespace std;
const int N=100010;
int cnt[30];
char s[N];
char p[11]="puleyaknoi";
int n;
bool check(int len)
{
if(len>n) return 1;
memset(cnt,0,sizeof cnt);
int now=0;
for(int i=1;i<=len;i++)
if(s[i]=='p'||s[i]=='u'||s[i]=='l'||s[i]=='e'||s[i]=='y'||s[i]=='a'||s[i]=='k'||s[i]=='n'||s[i]=='o'||s[i]=='i')
{
if(!cnt[s[i]-'a']) now++;
cnt[s[i]-'a']++;
}
if(now==10) return 1;
for(int i=len+1;i<=n;i++)
{
if(s[i-len]=='p'||s[i-len]=='u'||s[i-len]=='l'||s[i-len]=='e'||s[i-len]=='y'||s[i-len]=='a'||s[i-len]=='k'||s[i-len]=='n'||s[i-len]=='o'||s[i-len]=='i')
{
if(cnt[s[i-len]-'a']==1) now--;
cnt[s[i-len]-'a']--;
}
if(s[i]=='p'||s[i]=='u'||s[i]=='l'||s[i]=='e'||s[i]=='y'||s[i]=='a'||s[i]=='k'||s[i]=='n'||s[i]=='o'||s[i]=='i')
{
if(!cnt[s[i]-'a']) now++;
cnt[s[i]-'a']++;
}
if(now==10) return 1;
}
return 0;
}
int main()
{
IO;
int T=1;
cin>>T;
while(T--)
{
cin>>s+1;
n=strlen(s+1);
int l=10,r=n+1;
while(l<r)
{
int mid=l+r>>1;
if(check(mid)) r=mid;
else l=mid+1;
}
if(r==n+1) cout<<-1<<'\n';
else cout<<l<<'\n';
}
return 0;
}
B-Patchwork
Sequential automata
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=100010;
char s[N];
int ne[N][30];
char p[11]="puleyaknoi";
int n;
void build()
{
memset(ne,0,sizeof ne);
for(int i=n;i;i--)
{
for(int j=0;j<26;j++)
ne[i-1][j]=ne[i][j];
ne[i-1][s[i]-'a']=i;
}
}
int main()
{
IO;
int T=1;
cin>>T;
while(T--)
{
cin>>s+1;
n=strlen(s+1);
build();
int res=n;
for(int i=1;i<=n;i++)
{
if(s[i]==p[0])
{
int now=i;
for(int j=1;j<10;j++)
{
now=ne[now][p[j]-'a'];
if(!now) break;
}
if(now) res=min(res,now-i+1);
}
}
if(res!=n) cout<<res<<'\n';
else cout<<-1<<'\n';
}
return 0;
}
C-Mu function
The table shows that it is basically a value in the end, just need to make the very large k directly into a smaller number.
Note that the odd or even number of k should not be changed.
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int N=10000010;
int prime[N],cnt;
bool st[N];
int mob[N];
void init(int n)
{
mob[1]=1;
for(int i=2;i<=n;i++)
{
if(!st[i]) prime[++cnt]=i,mob[i]=-1;
for(int j=1;prime[j]<=n/i;j++)
{
st[i*prime[j]]=1;
if(i%prime[j]==0)
{
mob[i*prime[j]]=0;
break;
}
mob[i*prime[j]]=-mob[i];
}
}
}
int main()
{
IO;
int T=1;
cin>>T;
init(10000000);
while(T--)
{
int n;
ll k;
cin>>n>>k;
if(k>100ll)
{
if(k&1) k=101;
else k=100;
}
for(int i=1;i<=k;i++) n=n+mob[n];
cout<<n<<'\n';
}
return 0;
}
D-number tree
Note that deleting an edge means deleting all edges, not one of them . The comment code is the code to delete an edge. The number of
connected points minus the number of edges is the answer.
Because the tree has n-1 edges and n points. If there are k trees, then there must be k more points than edges.
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<map>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
map<pii,int> mp;
map<int,int> d;
int main()
{
IO;
int T=1;
//cin>>T;
while(T--)
{
int n;
cin>>n;
int cnte=0,cntv=0;
while(n--)
{
int op,u,v;
cin>>op;
if(op==1)
{
cin>>u>>v;
if(u>v) swap(u,v);
// if(!mp[{u,v}]) cnte++;
// mp[{u,v}]++;
// if(!d[u]) cntv++;
// if(!d[v]) cntv++;
// d[u]++,d[v]++;
if(mp[{
u,v}]) continue;
cnte++;
if(!d[u]) cntv++;
if(!d[v]) cntv++;
d[u]++,d[v]++,mp[{
u,v}]=1;
}
else if(op==2)
{
cin>>u>>v;
if(u>v) swap(u,v);
if(!mp[{
u,v}]) continue;
// if(mp[{u,v}]==1) cnte--;
// mp[{u,v}]--;
// d[u]--,d[v]--;
// if(!d[u]) cntv--;
// if(!d[v]) cntv--;
cnte--;
if(d[u]==1) cntv--;
if(d[v]==1) cntv--;
d[u]--,d[v]--,mp[{
u,v}]=0;
}
else
cout<<cntv-cnte<<'\n';
}
}
return 0;
}