A
模拟即可

# include <bits/stdc++.h>
using namespace std;
# define int long long
signed main()
{
    int t;
    cin>>t;
    while(t--)
    {
    string s;
    cin>>s;
    if(s=="101")
    {
    cout<<"NO"<<endl;
    continue;
    }
    if(s.size()<=2)
    {
    cout<<"NO"<<endl;
    continue;
    }
    bool st=0;
    int n=s.size();
    string s1=s.substr(0,2);
    if(s1=="10")
    st=1;
    string s2=s.substr(2,s.size());
    if(s2[0]!='0'&&st)
    cout<<"YES";
    else cout<<"NO"; 
    cout<<endl;
    }
}

B

# include <bits/stdc++.h>
using namespace std;
# define int long long
const int N=2e5+10; 
signed main()
{
    int t;
    cin>>t;
    while(t--)
    {
    int n;
    int a[N]={0},b[N]={0};
    bool st1=0;
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    b[a[1]]=1;
    for(int i=2;i<=n;i++)
    {
    bool st=0;
    if(a[i]-1>=1)
    {
    if(b[a[i]-1]==1)
    st=1,b[a[i]]=1;
    }
        if(a[i]+1<=n&&!st)
    {
    if(b[a[i]+1]==1)
    st=1,b[a[i]]=1;
    }
    if(!st)
    {
    st1=1;
    break;
    }
    }
    if(st1)
    cout<<"NO";
    else cout<<"YES";
    cout<<endl;
    }
}

C 用结构体存值和编号然后排序遍历一遍寻找相同的记录下来编号

# include <bits/stdc++.h>
using namespace std;
# define int long long
#define x first
#define y second
const int N=2e5+10;
int n,m;
struct node
{
int w,id;
}a[N];
struct node1
{
int id;
char s;
}b[N];
bool cmp(node a,node b)
{
return a.w<b.w;
}
bool cmp1(node1 a,node1 b)
{
return a.s<b.s;
}
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int c[N]={0};
    vector<pair<int,int>>v;
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i].w,a[i].id=i,c[i]=a[i].w;
    sort(a+1,a+n+1,cmp);
    for(int i=1;i<n;i++)
    {
    if(a[i].w==a[i+1].w)
    v.push_back({a[i].id,a[i+1].id});
    }
    cin>>m;
    while(m--)
    {
    vector<pair<int,int>>v1;
    bool st1=0,st2=0;
    string s;
    cin>>s;
    if(s.size()!=n)
    {
    cout<<"NO"<<endl;
    continue;
    }
    for(auto t:v)
    {
    //cout<<t.x<<" "<<t.y<<endl;
    if(s[t.x-1]!=s[t.y-1])
    st1=1;
    }
    for(int i=0;i<s.size();i++)
    {
    b[i+1].s=s[i];
    b[i+1].id=i+1;
    }
    sort(b+1,b+1+s.size(),cmp1);
    for(int i=1;i<s.size();i++)
    {
    if(b[i].s==b[i+1].s)
    v1.push_back({b[i].id,b[i+1].id});
    }
    for(auto t:v1)
    {
    //cout<<t.x<<" "<<t.y<<endl;
    if(c[t.x]!=c[t.y])
    st2=1;
    }
    //cout<<st1<<st2<<endl;
    if(!st1&&!st2)
    cout<<"YES";
    else cout<<"NO";
    cout<<endl;
    }   
    }
}

D
考虑贪心区间覆盖的越大值越大，所以利用两个指针扫描一遍即可

# include <bits/stdc++.h>
using namespace std;
# define int long long
#define x first
#define y second
const int N=2e5+10;
int n;
struct node
{
int w;
char s;
}a[N];
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int sum=0;
    cin>>n;
    int b[N]={0};
    for(int i=1;i<=n;i++)
    cin>>a[i].w,b[i]+=b[i-1]+a[i].w;
    string s1;
    cin>>s1;
    for(int i=0;i<n;i++)
    a[i+1].s=s1[i];
    int l=1,r=n;
    while(l<r)
    {
    if(a[l].s=='L'&&a[r].s=='R')
    {
    sum+=b[r]-b[l-1],l++,r--;
    continue;
    }
    if(a[l].s=='R')l++;
    else if(a[r].s=='L')r--;
    }
    cout<<sum<<endl;
    }
}

E
把相对较高的猩猩放到被计算次数较多也就是中间的格子，具体计算多少次也要算出来，所以直接差分处理一下

# include <bits/stdc++.h>
using namespace std;
# define int long long
#define x first
#define y second
signed main()
{
    int n,m,k;
    cin>>n>>m>>k;
    int w;
    cin>>w;
    vector<int> a(w+1);
    for(int i=1;i<=w;i++)
    {
    cin>>a[i];
    }
    sort(a.begin()+1,a.end());
    int d[n+k+1][m+k+1];
    memset(d,0,sizeof(d));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            int x2=i+k-1,y2=j+k-1;
            if(x2>n or y2>m)
                continue;
            d[i][j]++;
            d[i][y2+1]--;
            d[x2+1][j]--;
            d[x2+1][y2+1]++;
        }
    }
    priority_queue<int> q;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            d[i][j]+=d[i-1][j]+d[i][j-1]-d[i-1][j-1];
            q.push(d[i][j]);
        }
    }
    int ans=0;
    for(int i=w;i>=1 and !q.empty();i--)
    {
        ans+=q.top()*a[i];
        q.pop();
    }
    printf("%lld\n",ans);

}