A. Division?

思路:分类/输入/输出

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 x;cin>>x;
       if(x<=1399)cout<<"Division 4"<<endl;
       else if(x>=1400&&x<=1599)cout<<"Division 3"<<endl;
       else if(x>=1900)cout<<"Division 1"<<endl;
       else cout<<"Division 2"<<endl;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
//        freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}

B. Triple

思路:记录并输出第一个出现第三次的值即可

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<int,int>mp;
void solve()
{
   int n;cin>>n;
   int ans=-1;
   mp.clear();
   for(int i=1;i<=n;i++)
   {
            int x;cin>>x;
            mp[x]++;
            if(mp[x]>=3)ans=x;
   }
   if(ans!=-1)cout<<ans<<endl;
   else cout<<ans<<endl;
   return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     //   freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}

C. Odd/Even Increments

思路:
找下是否奇数和偶数都出现在了 奇数位置或者偶数位置.

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};
int n;
bool check(vector<int>a,int sta)
{
        if(sta)
        {
                bool ok1=false,ok2=false;
                for(int i=2;i<=n;i+=2)
                 if(a[i]%2!=0)ok1=true;
                 else ok2=true;

                if(ok1&&ok2)return true;
                else return false;
        }else
        {
                bool ok1=false,ok2=false;
                for(int i=1;i<=n;i+=2)
                 if(a[i]%2!=0)ok1=true;
                 else ok2=true;

                 if(ok1&&ok2)return true;
                 else return false;
        }
}
void solve()
{
   cin>>n;
   vector<int>a(n+1);
   for(int i=1;i<=n;i++)cin>>a[i];

   if(check(a,1)||check(a,0))cout<<"NO"<<endl;
   else cout<<"YES"<<endl;
   return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
      //  freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}

D. Colorful Stamp

思路:
由于每次只能搞出两个字符,且无法消掉一个.因此可以染色的部分长度必须$>1$

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 n;cin>>n;
       string s;cin>>s;

       int l=0,r;
       while(s[l]=='W'&&l<s.size())l++;
       r=l;
       while(l<s.size())
       {

               while(s[r]!='W'&&r<s.size())r++;
               set<char>st;
               for(int i=l;i<r;i++)st.insert(s[i]);
               if(st.size()==1)
               {
                       cout<<"NO"<<endl;
                       return ;
               }
               while(s[r]=='W'&&r<s.size())r++;
               l=r;
       }
       cout<<"YES"<<endl;
       return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
 //       freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}

E. 2-Letter Strings

思路:由于字符串的实际种类非常小.因此每次只需要遍历所有种类的字符串和当前的进行比较即可.

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,int>mp;
set<string>st;
int g[26][26];
void solve()
{
       memset(g,0,sizeof g);
       int n;cin>>n;
       ll ans=0;
       for(int i=1;i<=n;i++)
       {
               string s;cin>>s;
               for(int i=0;i<26;i++)
                for(int j=0;j<26;j++)
                {
                        string cur;
                        cur+=char(i+'a');
                        cur+=char(j+'a');
                        if((i==s[0]-'a'&&j!=s[1]-'a')||(i!=s[0]-'a'&&j==s[1]-'a'))ans+=g[i][j];
                }
               g[s[0]-'a'][s[1]-'a']++;
       }
       cout<<ans<<endl;
       return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     //   freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}

F. Eating Candies

思路:枚举$A$的可能情况,在前缀上二分出$B$即可

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};
ll pre[N];
void solve()
{
       int ans=0;
       int n;cin>>n;
       vector<int>a(n+1);
       for(int i=1;i<=n;i++)cin>>a[i],pre[i]=pre[i-1]+a[i];

       for(int i=1;i<n;i++)
       {
               ll A=pre[i];
               if(pre[n]-pre[i]<A)continue;
               else
               {
                       int l=i+1,r=n;
                       while(l<r)
                       {
                               int mid=l+r+1>>1;
                           //    if(i==1)cout<<l<<' '<<r<<' '<<mid<<endl;
                               if(pre[n]-pre[mid-1]>=A)l=mid;
                               else r=mid-1;
                       }
                       if(pre[n]-pre[r-1]==A)ans=max(ans,n-r+1+i);
               }
       }
       cout<<ans<<endl;
       return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
    //    freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}

G. Fall Down

思路:按列模拟下下坠即可.

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};
char g[60][60];
void solve()
{
       int n,m;
       cin>>n>>m;
       for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)cin>>g[i][j];

     /*  for(int i=1;i<=n;i++)
       {
               for(int j=1;j<=m;j++)cout<<g[i][j]<<' ';
               cout<<endl;
       }
       cout<<endl;*/
       for(int j=1;j<=m;j++)
       {
               int l=1,r;
               while(g[l][j]=='o'&&l<=n)l++;//找到起点
               r=l;
               while(l<n)
               {
                        int cnt=0;
                        while(g[r][j]!='o'&&r<=n)
                        {
                                if(g[r][j]=='*')cnt++;
                                r++;
                        }
                        for(int i=r-1;i>=r-cnt;i--)g[i][j]='*';
                        for(int i=r-cnt-1;i>=l;i--)g[i][j]='.'; 
                        while(g[r][j]=='o'&&r<=n)r++;
                        l=r;   
               }
       }
       for(int i=1;i<=n;i++)
       {
               for(int j=1;j<=m;j++)cout<<g[i][j];
               cout<<endl;
       }
       cout<<endl;
       return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     //   freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}

H. Maximal AND

思路:
考虑从高位往低位构造答案.因为最后求的是$AND$运算,假如某一个二进制位对$ans$有贡献必须保证每一个$a_i$都在该位上是1.且应该贪心的构造尽可能先满足高位变成1的情况.

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};
int cnt[31];
void solve()
{
       int n,k;
       cin>>n>>k;
       memset(cnt,0,sizeof cnt);
       vector<int>a(n+1);
       for(int i=1;i<=n;i++)cin>>a[i];

       for(int i=1;i<=n;i++)
        for(int j=0;j<=30;j++)
         if((a[i]>>j)&1)cnt[j]++;

        ll ans=0;
        for(int i=30;i>=0;i--)
        {
                if(cnt[i]!=n&&k>=n-cnt[i])k-=n-cnt[i],ans+=(1<<i);
                else if(cnt[i]==n)ans+=(1<<i);
        }
        cout<<ans<<endl;
        return ;
}
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
    //    freopen("test.in","r",stdin);
        int t;cin>>t;
        while(t--)solve();
        return 0;
}