A

# include <bits/stdc++.h>
using namespace std;
#define int long long 
int n;
const int N=60;
int a[N];
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    cin>>n;
    int st=0;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    for(int i=1;i<n;i++)
    {
    if(abs(a[i]-a[i+1])==5)
    st=1;
    else if(abs(a[i]-a[i+1])==7)
    st=2;
    else 
    {
           st=3;
           break;
    }
    }
    if(st==3)
    cout<<"NO";
    else cout<<"YES";
    cout<<endl;
    }
}

B

# include <bits/stdc++.h>
using namespace std;
#define int long long 
int n,k;
map<int,int>ma;
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    ma.clear();
    cin>>n>>k;
    for(int i=1;i<=k;i++)
    {
    int a,b;
    cin>>a>>b;
    ma[a]+=b;
    }
    priority_queue<int> q;
    for (auto [x,y] : ma)
    {
        q.push(y);
    }
    int ans = 0;
    int cnt = 0;
    while (!q.empty() && cnt < k && cnt < n)
    {
        ans += q.top();
        q.pop();
        cnt++;
    }
    cout<<ans<<endl;
    }
}

C
先统计字符串中”1100”的个数，对于q次询问,改变的这一个字符只影响它前面3个字符和后面3个字符,每次判断即可

# include <bits/stdc++.h>
using namespace std;
#define int long long 
typedef long long ll;

string s;
int q,sum,m;
void check1(string s)
{
        for(int i=1;i<=m;i++)
        {
        if(i+3<=m)
        {
        if(s[i+1]=='1'&&s[i]=='1'&&s[i+2]=='0'&&s[i+3]=='0')
        sum++;        
        }
        }
}
bool check2(int x,int a)
{
        //int cnt=0;
        int start=max(x-3,(ll)1),end=min(x+3,m);
        for(int i=start;i<=end;i++)
        {
        if(i+3<=end)
        if(s[i+1]=='1'&&s[i]=='1'&&s[i+2]=='0'&&s[i+3]=='0')
        sum--;
        }
        if(a==1)
        s[x]='1';
        else s[x]='0';
        //cout<<s<<endl;
        //cout<<start<<" "<<end<<endl;
        for(int i=start;i<=end;i++)
        {
        if(i+3<=end)
        if(s[i+1]=='1'&&s[i]=='1'&&s[i+2]=='0'&&s[i+3]=='0')
        sum++;
        }
        //cout<<cnt<<endl;
        if(sum>0)
        return true;
        return false ;
}
signed main()
{
        int T;
        cin>>T;
        while(T--)
        {
        cin>>s>>q;
        m=s.size();
        s=" "+s;
        //cout<<m<<endl;
        sum=0;
        check1(s);
        //cout<<sum<<endl;
        while(q--)
        {
        int a,b;
        cin>>a>>b;
        if(check2(a,b))
        cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
        }
        }
}

D
每次考虑边界将一个顺时针环所有点加入,每次都计算一遍累加

#include<bits/stdc++.h>
using namespace std;


void solve(){
    int n,m;
    cin>>n>>m;
    string s[n+10];
    for(int i=1;i<=n;i++)
    {
        cin>>s[i];
        s[i]=" "+s[i];
    }
    int t=1,x=n,y=m;
    int ans=0;
    while(t<=x&&t<=y)
    {
        vector<char>v;
        for(int i=t;i<=y;i++)
        {
            v.push_back(s[t][i]);
        }
        for(int i=t+1;i<=x;i++)
        {
            v.push_back(s[i][y]);
        }
        for(int i=y-1;i>=t;i--)
        {
            v.push_back(s[x][i]);
        }
        for(int i=x-1;i>=t+1;i--)
        {
            v.push_back(s[i][t]);
        }

        int len=v.size();
        for(int i=0;i<len;i++)
        {
            if(v[i]=='1'&&v[(i+1)%len]=='5'&&v[(i+2)%len]=='4'&&v[(i+3)%len]=='3'){
                ans++;
            }
        }
        t++,x--,y--;
    }
    cout<<ans<<"\n";
}
signed main() 
{
    int _=1;
    cin>>_;
    while(_--) {
        solve();
    }
    return 0;
}