Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2)

总结:

单排成果3题,Successful hacking attempt*3,近期目标4题。
缺点:

  1. 英语阅读不行。
  2. 题目做得太少。
  3. AB都很快做完,但不敢交,一直在考虑边界和特殊情况,导致慢了好多。

A - Forgetting Things

思路:分清楚情况即可。

  1. b-a=1
  2. b=a
  3. a=9且b=1;

 #include
#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)

思路:明显求种类,经历过cf的毒打也学会了用set和map,插入前判断里面是否还有该元素存在即可。

#include
#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)

思路:同上,set和map一箭双雕。

#include
#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;
}

C. p-binary CF1247C(STL暴力)

题解

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