一、覆盖点集

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int x,y;
    int min_x=INT_MAX,max_x=INT_MIN;
    int min_y=INT_MAX,max_y=INT_MIN;
    while(cin>>x>>y)
    {
        if(x==0&&y==0)
            break;
        min_x=min(min_x,x),max_x=max(max_x,x);
        min_y=min(min_y,y),max_y=max(max_y,y);
    }
    printf("%d %d %d %d",min_x,min_y,max_x,max_y);
    return 0;
}

二、回文串

递归

#include<bits/stdc++.h>
using namespace std;
bool judge(string s,int st,int ed)
{
    if(st>=ed)
        return true;
    if(s[st]==s[ed])
        return judge(s,st+1,ed-1);
    return false;
}
int main()
{
    string s;
    getline(cin,s);
    if(judge(s,0,s.size()-1))
        cout<<"yes"<<endl;
    else
        cout<<"no"<<endl;
    return 0;
}

非递归

#include<bits/stdc++.h>
using namespace std;
bool judge(string str)
{
    stack<char>s;
    for(char x : str)
        s.push(x);
    for(char x : str)
    {
        if(x!=s.top())
            return false;
        s.pop();
    }
    return true;
}
int main()
{
    string s;
    getline(cin,s);
    if(judge(s))
        cout<<"yes"<<endl;
    else
        cout<<"no"<<endl;
    return 0;
}

三、逆序对的数量(暴力可得大部分分)

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int n,a[N],w[N];
typedef long long LL;
LL num(int l,int r)
{
    if(l>=r)return 0;
    int mid=l+r>>1;
    LL res=0;
    res+=num(l,mid)+num(mid+1,r);
    int i=l,j=mid+1,k=0;
    while(i<=mid&&j<=r)
    {
        if(a[i]<=a[j])w[k++]=a[i++];
        else
            w[k++]=a[j++],res+=mid-i+1;
    }
    while(i<=mid)w[k++]=a[i++];
    while(j<=r)w[k++]=a[j++];
    for(int i=l,j=0;j<k;i++,j++)
        a[i]=w[j];
    return res;
}
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i];
    cout<<num(0,n-1);
    return 0;
}