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to sum up:
Group 2/1/1 title, should be improved, short-term goal 3/3/3.
Disadvantages:
- Reading not.
- Topic doing too little, do not know how to open a dynamic map, stuck B title.
- Not considered a special case, C boundary question and did not deal with special circumstances.
- Knowledge weak points (game theory).
A - Forgetting Things
Thinking: read the questions, erasing every odd bit, so long as the output to the even 2 * x.
#include<bits/stdc++.h>
#define ll long long
#define R register int
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
inline ll read(){
ll s=0,w=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;
}
void put1(){ puts("YES") ;}
void put2(){ puts("NO") ;}
const int manx=5e4+5;;
int a[26];
int main()
{
ll n,m;
n=read(),m=read();
if(n==9&&m==1) cout<<n<<" "<<m*10<<endl;
else if(n==m) cout<<n*10<<" "<<m*10+1<<endl;
else if(m-n==1) cout<<n*10+9<<" "<<m*10<<endl;
else puts("-1");
return 0;
}
B1 - TV Subscriptions (Easy Version)
Idea: to meet the minimum number of cells painted black cross, not because of the dynamic open the map, I have not seen this problem, ignorant, just open the map difference this off, you can simulate.
#include<bits/stdc++.h>
#define ll long long
#define R register int
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
inline ll read(){
ll s=0,w=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;
}
void put1(){ puts("Yes") ;}
void put2(){ puts("No") ;}
const int manx=2e5+5;;
ll a[manx];
map<ll,ll>b;
int main()
{
ll q;
q=read();
while(q--)
{
ll n,k,d;
n=read(),k=read(),d=read();
b.clear();
ll ans=0,res=0;
for(int i=1;i<=n;i++)
a[i]=read();
set<ll>q;
for(int i=1;i<=d;i++)
q.insert(a[i]),b[a[i]]++;
ans=q.size();
for(int i=d+1;i<=n;i++)
{
q.insert(a[i]);
b[a[i]]++;
b[a[i-d]]--;
if(!b[a[i-d]]) q.erase(a[i-d]);
ans=min(ans,(ll)q.size());
}
cout<<ans<<endl;
}
return 0;
}
B2 - TV Subscriptions (Hard Version)
Thinking: t array tag number string of letters, and then double pointer traversal s p, s and finally judged whether the length of the string and a corresponding pointer equal (less the PM judgment).
#include<bits/stdc++.h>
#define ll long long
#define R register int
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
inline ll read(){
ll s=0,w=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;
}
void put1(){ puts("Yes") ;}
void put2(){ puts("No") ;}
const int manx=2e5+5;;
ll a[manx];
map<ll,ll>b;
int main()
{
ll q;
q=read();
while(q--)
{
ll n,k,d;
n=read(),k=read(),d=read();
b.clear();
ll ans=0,res=0;
for(int i=1;i<=n;i++)
a[i]=read();
set<ll>q;
for(int i=1;i<=d;i++)
q.insert(a[i]),b[a[i]]++;
ans=q.size();
for(int i=d+1;i<=n;i++)
{
q.insert(a[i]);
b[a[i]]++;
b[a[i-d]]--;
if(!b[a[i-d]]) q.erase(a[i-d]);
ans=min(ans,(ll)q.size());
}
cout<<ans<<endl;
}
return 0;
}