取石子

思路:
简单博弈.每次只取一个即可.
因此胜利先手胜利条件:$n_1>n_2$

AC Code：

#include<bits/stdc++.h>
#pragma optimize(2)
#define endl '\n'
#define ll() to_ullong() 
#define string() to_string()
using namespace std;
typedef long long ll;
typedef pair<int,int>PII; 
typedef unsigned long long ull;
const int M=2010;
const int P=13331;
const ll  llinf=0x3f3f3f3f3f3f3f3f;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int N=2e6+10;
int dx[4]={0,1,0,-1};
int dy[4]={-1,0,1,0};
void solve()
{
       int n1,n2,k1,k2;
       cin>>n1>>n2>>k1>>k2;
       if(n2>=n1)cout<<"Second"<<endl;
       else cout<<"First"<<endl;
       return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
      //  freopen("test.in","r",stdin);
        solve();
        return 0;
}

卡牌

思路:
贪心的考虑,将能变的更小的先变小.
前提是$k$张朝上.

AC Code:

#include<bits/stdc++.h>
#pragma optimize(2)
#define endl '\n'
#define ll() to_ullong() 
#define string() to_string()
using namespace std;
typedef long long ll;
typedef pair<int,int>PII; 
typedef unsigned long long ull;
const int M=2010;
const int P=13331;
const ll  llinf=0x3f3f3f3f3f3f3f3f;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int N=2e6+10;
int dx[4]={0,1,0,-1};
int dy[4]={-1,0,1,0};
struct node{
        int x,y;
        int dif,id;
        bool operator<(const node &a)const
        {
                return a.dif<dif;
        }
};
bool vis[N];
void solve()
{
       int n,k;
       cin>>n>>k;
       vector<node>a(n+1);
       for(int i=1;i<=n;i++)cin>>a[i].x;
       for(int i=1;i<=n;i++)cin>>a[i].y;
       for(int i=1;i<=n;i++)
       {
               if(a[i].x>a[i].y)a[i].dif=a[i].x-a[i].y;
               else a[i].dif=0;
               a[i].id=i;
       }
       sort(a.begin()+1,a.end());
       int cur=n;
       for(int i=1;i<=n;i++)
       {
               if(a[i].dif>0&&cur>k)vis[i]=true,cur--;
       }
       ll ans=0;
       for(int i=1;i<=n;i++)
        if(vis[i])ans+=a[i].y;
        else ans+=a[i].x;

        cout<<ans<<endl;
        return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     //   freopen("test.in","r",stdin);
        solve();
        return 0;
}

查询字符串

思路：哈希表/字符串.
由于字符串长度较短.枚举单个字符串所有的可能字串即可.
不过需要注意的是:一个子串可能在,一个完整的字符串中多次出现,但是只会做一次计数.

AC Code：

#include<bits/stdc++.h>
#pragma optimize(2)
#define endl '\n'
#define ll() to_ullong() 
#define string() to_string()
using namespace std;
typedef long long ll;
typedef pair<int,int>PII; 
typedef unsigned long long ull;
const int M=2010;
const int P=13331;
const ll  llinf=0x3f3f3f3f3f3f3f3f;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int N=2e6+10;
int dx[4]={0,1,0,-1};
int dy[4]={-1,0,1,0};
map<string,PII>mp;
map<string,bool>Q;
void solve()
{
       int n;cin>>n;
       vector<string>a(n+1);
       for(int i=1;i<=n;i++)
       {
               string s;cin>>s;
               a[i]=s;
               Q.clear();
               for(int j=0;j<s.size();j++)
                for(int k=j;k<s.size();k++)
                { 
                        string tmp=s.substr(j,k-j+1);
                        if(!Q[tmp])mp[tmp].first=i,mp[tmp].second++,Q[tmp]=true;//只计算一次 
                }//枚举所有子串 
       }
       int q;cin>>q;
       while(q--)
       {
               string s;cin>>s;
            //   cout<<s<<endl;
               if(mp[s].second!=0)cout<<mp[s].second<<' '<<a[mp[s].first]<<endl;
               else cout<<0<<' '<<"-"<<endl;
       }
       return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     //   freopen("test.in","r",stdin);
        solve();
        return 0;
}