Codeforces-Round#569 Div.2

Front exam week, I'm sorry it took so long.

A题 Alex and a Rhombus

The basic idea

Recursive solution, n when 1 =, Count (1) = 1; n> when 1, Count (n) = Count (n-1) + (n-1) * 4

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;cin>>n;
    long long res=1;
    for(int i=1;i<=n;i++)
    {
        res+=4*(i-1);
    }
    cout<<res;
    return 0;
 } 

B题 Nick and Array

The basic idea

In absolute terms, negative> positive, so long as the number of revolutions of all negative, if n is divisible by 2, the result is positive, if more than n% 2 1, will be converted into the smallest negative positive.

Note that the subject request output sequence

#include<bits/stdc++.h>
using namespace std;
int num[100005];
int MINN = 99999,pos=0;
int main()
{
    int n,ncount=0;cin>>n;
    for(int i=0;i<n;i++)
    {
        cin >> num[i];
        if(num[i]>=0)
            num[i] = -num[i] - 1;
        if(num[i]<=MINN&&num[i]!=-1) 
        {
            pos=i;
            MINN=num[i];
        }
    }
    if(n%2!=0)
    {
        num[pos]=-num[pos]-1;
    }
    for(int i=0;i<n;i++)
    {
        cout<<num[i]<<" ";
    }
    return 0;
}

C title

The basic idea

When deque q [0] is the maximum value will fall into the loop, therefore, the number of recursive steps need not exceed \ (10 ^ 5 \) the second number, the back die may be used to select the output array subscript of . (Remember to pop_front modulo n-1)

#include<bits/stdc++.h>
#include<deque>
using namespace std;
struct{
    int a,b;
}qu[100010];
int main()
{
    deque<int>q;int n,query,maxn=-99999;
    cin>>n>>query;
    for(int i=0;i<n;i++)
    {
        int num;cin>>num;
        q.push_back(num);
        maxn=max(maxn,num);
    }
    long long cnt=0;
    while(q.front()!=maxn)
    {
        cnt++;
        int n1=q.front();
        q.pop_front();int n2=q.front();q.pop_front();

        q.push_front(max(n1,n2));
        q.push_back(min(n1,n2));
        qu[cnt].a=n1;qu[cnt].b=n2;
    }
    q.pop_front();
    for(int i=0;i<query;i++)
    {
        long long m;
        cin>>m;
        if(m<=cnt)
        {
            cout<<qu[m].a<<" "<<qu[m].b<<endl;
        }
        else
        {
            int pos=(m-cnt-1)%(n-1);
            cout<<maxn<<" "<<q[pos]<<endl;
        }
    }
    return 0;
}