Codeforces Round 972 (Div. 2) 总结

A

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

int n;
char a[]={'a','e','i','o','u'};

void solve()
{
    cin>>n;
    int x=n/5,y=n%5;
    for(int i=0;i<5;i++)
    {
        for(int j=1;j<=x;j++) cout<<a[i];
        if(y) cout<<a[i],y--;
    }
    cout<<'\n';
}
int main ()
{
    int T;
    cin>>T;
    while(T--) solve();
    return 0;
}

B1

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

const int N=1e5+5;
int n,m,q;
int b[N];

void solve()
{
    cin>>n>>m>>q;
    for(int i=1;i<=m;i++) cin>>b[i];
    sort(b+1,b+m+1);
    while(q--)
    {
        int x;
        cin>>x;
        if(x<b[1]) cout<<b[1]-1<<"\n";
        else if(x<b[2]) cout<<(b[2]-b[1])/2<<"\n";
        else cout<<n-b[2]<<"\n";
    }
}
int main ()
{
    int T;
    cin>>T;
    while(T--) solve();
    return 0;
}

B2

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

const int N=1e5+5;
int n,m,q;
int b[N];

void solve()
{
    cin>>n>>m>>q;
    for(int i=1;i<=m;i++) cin>>b[i];
    sort(b+1,b+m+1);
    while(q--)
    {
        int t;
        cin>>t;
        int x=lower_bound(b+1,b+m+1,t)-b-1;
        int y=x+1;
        if(x==0) cout<<b[y]-1<<"\n";
        else if(y==m+1) cout<<n-b[x]<<"\n";
        else cout<<(b[y]-b[x])/2<<"\n";
    }
}
int main ()
{
    int T;
    cin>>T;
    while(T--) solve();
    return 0;
}

C

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

const int N=1005,inf=1e9;
string S="narek";
int n,m;
string s;
int f[N][5];
void solve()
{
    cin>>n>>m;
    for(int i=0;i<=n;i++)
        for(int j=0;j<5;j++) 
            f[i][j]=-1e9;
    f[0][0]=0;
    for(int i=1;i<=n;i++)
    {
        cin>>s;
        for(int j=0;j<5;j++) f[i][j]=f[i-1][j];
        for(int j=0;j<5;j++)
        {
            if(f[i-1][j]==-inf) continue;
            int sum=0,ne=j;
            for(int k=0;k<m;k++)
            {
                int id=S.find(s[k]);
                if(id==-1) continue;
                if(ne==id)
                {
                    ne=(ne+1)%5;
                    sum++;
                }
                else sum--;
            }
            f[i][ne]=max(f[i][ne],f[i-1][j]+sum);
        }
    }
    int ans=0;
    for(int i=0;i<5;i++) ans=max(ans,f[n][i]-2*i);
    cout<<ans<<'\n';
}
int main ()
{
    int T;
    cin>>T;
    while(T--) solve();
    return 0;
}