HackerEarth N girls problem solution

In this HackerEarth N girls problem solution, you are given n boxes and the size of the ith box is ai. You can select at most k boxes and change their size to a positive integer.

Select some boxes and paint them a Red color. These boxes must satisfy the following condition:

There must be no 1 <= i, j <= n such that both i, j boxes are red and ai/aj > P/Q.

Determine the maximum number of boxes that you can color in Red.

HackerEarth N girls problem solution.

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

#define ll long long
#define pb push_back

const int maxn = 2e5 + 20;

int a[maxn];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n , k , P , Q;
    cin >> n >> k >> P >> Q;

    for(int i = 0; i < n; i++)
        cin >> a[i];

    sort(a , a + n);

    int pt = 0 , res = 0;
    for(int i = 0; i < n; i++)
    {
        while(1LL * a[i] * Q > 1LL * P * a[pt])
            pt++;

        res = max(res , min(n , i - pt + 1 + k));
    }

    cout << res << endl;
}

Second solution

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif
using namespace std;
const int maxn = 200100, maxv = 100, mod = 1e9 + 7, maxa = 1005, maxs = 820, maxb = 23, base = 737, base2 = 3079, mod3 = 1e7 + 19, delt = 10513;
const long long inf = 2e14;
const int infint = 1e9 + 11;
long long max(long long x, long long y){return (x > y ? x : y);}
long long min(long long x, long long y){return (x < y ? x : y);}

long double a[maxn];

int32_t main() 
{ 
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, k;
    long double p, q;
    cin >> n >> k >> p >> q;
    for(int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    sort(a, a + n);
    long long ans = 0;
    for(int i = n - 1; i >= 0; i--)
    {
        long double x = (a[i] * q) / p;
        int ind = lower_bound(a, a + n, x) - a;
        ind = min(ind, i);
        int tmp = max(0, ind - k);
        ans = max(ans, i + 1 - tmp);
    }
    cout << ans << endl;
}

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