Vaccine Distribution[VACCINE 2](Solution)

Finally, a COVID vaccine is out on the market and the Chefland government has asked you to form a plan to distribute it to the public as soon as possible. There are a total of N people with ages a1,a2,…,aN.

There is only one hospital where vaccination is done and it is only possible to vaccinate up to D people per day. Anyone whose age is ≥80 or ≤9 is considered to be at risk. On each day, you may not vaccinate both a person who is at risk and a person who is not at risk. Find the smallest number of days needed to vaccinate everyone.

Input

• The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
• The first line of each test case contains two space-separated integers NN and D.
• The second line contains N space-separated integers a1,a2,…,aN.

Output

For each test case, print a single line containing one integer ― the smallest required number of days.

Constraints

• 1≤T≤10
• 1≤N≤10⁴
• 1≤D≤10⁵
• 1≤ai≤100 for each valid i

Subtasks

Subtask #1 (100 points): original constraints

Example Input

2
10 1
10 20 30 40 50 60 90 80 100 1
5 2
9 80 27 72 79

Example Output

10
3

Explanation

Example case 1: We do not need to worry about how the people are grouped since only one person can be vaccinated in a single day. We require as many days as there are people.

Example case 2: There are two people at risk and three people who are not at risk. One optimal strategy is to vaccinate the two people at risk on day 1 and the remaining three on the next 2 days.

SOLUTION:

#include <iostream>
using namespace std;
#include<cmath>
#define ll long long int
int main() {
    //ios_base::sync_with_stdio(false);
    //cin.tie(NULL);
    
 // your code goes here
 ll t, D, N;
        
        cin >> t;
        
 while(t--)
 {   N=0;D=0;
     cin >> N >> D;
     ll arr[N], Atrist = 0, Notrisk = 0;
     for( int i = 0; i < N; i++)
     {
         cin >> arr[i];
         
         if(arr[i] >= 80 || arr[i] <= 9)
            Atrist++;
         else
             Notrisk++;
     }
       ll total = ceil((double)Atrist/(double)D) + ceil((double)Notrisk/(double)D);
       
       cout << total <<"\n";
         
     }
 return 0;
}

LINK OF SOLUTION:

Solution: 40000159 | CodeChef

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