Codeforces Round ＃723 (Div. 2)

2021.5.28

A

The question ： Enter a 2*n Sequence a,a Each element in is not equal , Requirements for a Regroup to form a sequence b, bring b Each element of is not equal to the median of its two neighbors , namely , Output sequence b

Answer key ： Just put the sequence a Sort , Then insert the larger number in the middle of the smaller number , This can ensure that each number of neighbors must be larger or smaller than this element , Large numbers can be inserted from the front or from the back , Because the sequence is even , Can guarantee the previous condition .

#include <iostream>
#include <algorithm>
using namespace std;
const int N=105;
int a[N],b[N];
int main()
{
    ios_base::sync_with_stdio(false);
    int t,n;
    cin>>t;
    while(t--)
    {
        cin>>n;
        n*=2;
        for(int i=0;i<n;i++)
            cin>>a[i];
        sort(a,a+n);
        int l=0;
        for(int i=0;i<n;i+=2)
        {
            b[i]=a[l++];
        }
        l=n-1;
        for(int i=1;i<n;i+=2)
        {
            b[i]=a[l--];
        }
        for(int i=0;i<n;i++)
            cout<<b[i]<<' ';
        cout<<'\n';
    }
    return 0;
}

B

The question ： Enter a x, Judge x Can be 11,111,………… Add up to get , among 11 This element can take any number

Answer key ： Just enumerate the above 11 The number of such elements , Then judge whether it is equal to x that will do . Because even numbers 1 You can divide them all 11, So you only need to enumerate an odd number when traversing 1, Even number 1 Remainder 11 that will do .

Game code （ There are some unnecessary ）：

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <cstring>
#define INF 111111
#define mod 998244353
#define ll long long
using namespace std;
const int N=105;
int main()
{
    ios_base::sync_with_stdio(false);
    ll t,n;
    cin>>t;
    while(t--)
    {
        cin>>n;
        ll m=n%11;
        int flag=0;
        if(m)
        {
            for(int i=0;i<=m;i++)
            {
                for(int j=0;j<=m;j++)
                {
                    for(int k=0;k<=m;k++)
                    {
                        ll sum=i*111+j*11111+k*1111111;
                        if(n>=sum&&i+j+k==m&&(n-sum)%11==0)
                        {
                            flag=1;
                            break;
                        }
                    }
                    if(flag)
                        break;
                }
                if(flag)
                    break;
            }
        }
        else
            flag=1;
        if(flag)
            cout<<"YES"<<'\n';
        else
            cout<<"NO"<<'\n';
    }
    return 0;
}

Summary after the game ：

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <cstring>
#define INF 111111
#define mod 998244353
#define ll long long
using namespace std;
const int N=105;
int main()
{
    ios_base::sync_with_stdio(false);
    ll t,n;
    cin>>t;
    while(t--)
    {
        cin>>n;
        ll m=11;
        int flag=0;
        for(int i=0;i<=m;i++)
        {
            for(int j=0;j<=m;j++)
            {
                for(int k=0;k<=m;k++)
                {
                    ll sum=i*111+j*11111+k*1111111;
                    if(n>=sum&&(n-sum)%11==0)
                    {
                        flag=1;
                        break;
                    }
                }
                if(flag)
                    break;
            }
            if(flag)
                break;
        }
        if(flag)
            cout<<"YES"<<'\n';
        else
            cout<<"NO"<<'\n';
    }
    return 0;
}

C1

C2

The question ： Enter an integer of length n Sequence a, It is required to select some numbers from the sequence and add them from left to right , The sum of each cumulative addition cannot be negative , Output the longest length of the selected sequence .

Answer key ： A priority queue is used to maintain the added negative number , If you add a negative number to make the cumulative sum negative , Then add this negative number to the queue , Then remove the smallest number in the queue , Then the cumulative sum at this moment = The previous cumulative sum + The difference between the negative number added and the minimum negative number out of the queue , The length of the sequence thus selected remains the same , If the added number cannot make the cumulative sum negative, the selected number +1.

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=2005;
ll dp[N][N],a[N];
int main()
{
    ll n,sum=0,num=0;
    cin>>n;
    priority_queue <ll,vector<ll>,greater<ll> > q;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        sum+=a[i];
        //cout<<sum<<endl;
        if(a[i]<0)
        {
            q.push(a[i]);
        }
        if(sum<0)
        {
            ll k=q.top();
            q.pop();
            sum-=a[i];
            sum+=(a[i]-k);
        }
        else
        {
            num++;
            //cout<<a[i]<<endl;
        }
        //cout<<sum<<endl;
    }
    cout<<num<<endl;
    return 0;
}

finish writing sth. C1 Find your own algorithm is O(n) Of , therefore C2 Together (●'◡'●)