A
相同字母尽可能相邻

#include <bits/stdc++.h>
using namespace std;
char a[5] = { 'a', 'e', 'i', 'o', 'u' };
int main() 
{
        int T;
        cin>>T;
        while(T--)
        {
    int n;
    cin >> n;
    string s="";
    for(int i=0;i<=n-1;i++)
    s+=a[i%5];

    sort(s.begin(),s.end());
    cout<<s<<endl;
    }
    return 0;
}

B1 B2
处理思路类似 模拟即可
复杂后用二分寻找边界

# include <bits/stdc++.h>
using namespace std;
#define int long long
int n,a,b,id,m,q;
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    cin>>n>>m>>q;
    cin>>a>>b>>id;
    if(a==id||b==id)
    {
        cout<<0<<endl;
        continue;
    }
    if(a<b)swap(a,b);

    if(id>=a||id<=b)
    {
        if(id>=a)cout<<n-a;
        else if(id<=b)
        cout<<b-1;
    }
    else if(id>b&&id<a)
    {
    int k=a-b>>1;
    cout<<k;
    }
    cout<<endl;
    }
    return 0;
}

# include <bits/stdc++.h>
using namespace std;
#define int long long
int n,a,b,id,m,q;
vector<int>v;
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    v.clear();
    cin>>n>>m>>q;
    a=n,b=1;
    for(int i=1;i<=m;i++)
    {
        int x;
        cin>>x;
        v.push_back(x);
    }
    sort(v.begin(),v.end());
    while(q--)
    {
    cin>>id;
    if(v[m-1]==id||v[0]==id)
    {
        cout<<0<<endl;
        continue;
    }
    if(id>v[m-1])cout<<n-v[m-1];
    else if(id<v[0])cout<<v[0]-1;
    else 
    {
    vector<int>::iterator iter1;
    vector<int>::iterator iter2;
    iter1=lower_bound(v.begin(), v.end(), id);
    a=(*iter1);
    iter2=lower_bound(v.begin(), v.end(), id)-1;
    //cout<<a<<" "<<b<<endl;
    b=(*iter2);
    if(id>b&&id<a)
    {
    int k=a-b>>1;
    cout<<k;
    }
    }
    cout<<endl;
    }
    }
    return 0;
}

C
dp[i][j]含义代表最大的分 i表示当前第i个字符串，j表示该匹配第j个字符，当满足匹配重置为零 有两种状态继承 一种不选取字符串 一种选取一个字符串并更改新的匹配字符

#include <bits/stdc++.h> 
using namespace std;
int dp[1005][5];
string t = "narek";
int n, m;
pair<int,int> get_score(string s, int now)
{
    int ans = 0;
    for (int i = 0; i < m; i++)
    {
        if (s[i] == t[now])
        {
        now++;
        if (now == 5) ans += 10, now = 0;
        }
        if (t.find(s[i]) != string::npos) ans--;
    }
    return {ans, now};
}
int main()
{
    ios::sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while (T--)
    {
    cin >> n >> m;  
    for (int i = 0; i <= n; i++)
    {
        for (int j = 0; j <= 4; j++)
        dp[i][j] = -2e9;
    }
    dp[0][0] = 0;

    for (int i = 1; i <= n; i++)
    {
        string s; cin >> s;
        for (int j = 0; j <= 4; j++)
        {
            dp[i][j] = max(dp[i][j], dp[i - 1][j]);
            pair<int,int> temp = get_score(s, j);
            int score = temp.first, nxt = temp.second;
            dp[i][nxt] = max(dp[i][nxt], dp[i - 1][j] + score);

        }
    }
    int mx = 0;
    for (int j = 0; j < 5; j++) mx = max(mx, dp[n][j]);
    cout << mx << endl;      
    }
}