E - average and median

E - Average and Median( Two points )

The first question has two answers $x$, Then you can count all the numbers $-x$, That is to meet the requirements $\sum (a_i-x)\ge 0$

Sure dp.

The second question is the same , Two points answer $x$, then $b_i=[a_i\ge x]$

So that is $\sum b_i\ge 0$

It's also dp.

Time complexity ：$O(nlogV)$

The difficulty is that I didn't think of it at the beginning How to calculate quickly after two points . I can dp.

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

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 1000000007
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
int n, m, k;
int tr[MAX];
double dp[MAX][2];

bool check1(double mid){
    
    dp[0][0] = dp[0][1] = 0.0;
    for(int i = 1; i <= n; ++i){
    
        dp[i][0] = dp[i - 1][1];
        dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]) + tr[i] - mid;
    }
    if(max(dp[n][1], dp[n][0]) >= 0)return true;
    else return false;
}
bool check2(int mid){
    
    dp[0][0] = dp[0][1] = 0;
    for(int i = 1; i <= n; ++i){
    
        dp[i][0] = dp[i - 1][1];
        dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]) + (tr[i] >= mid ? 1 : -1);
    }
    if(max(dp[n][1], dp[n][0]) > 0)return true;
    else return false;
}


void work(){
    
    cin >> n;
    for(int i = 1; i <= n; ++i)cin >> tr[i];
    double l = 0, r = 1000000000.0;
    while (r - l >= 1e-6) {
    
        double mid = (l + r) / 2;
        if(check1(mid))l = mid;
        else r = mid;
    }
    int ll = 0, rr = 1000000000;
    while (ll <= rr) {
    
        int mid = (ll + rr) / 2;
        if(check2(mid))ll = mid + 1;
        else rr = mid - 1;
    }
    cout << r << '\n' << rr << endl;
}

int main(){
    
    io;
    work();
    return 0;
}