Lifeline - Editorial

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PROBLEM LINK:

Author and Editorialist: viresh_dev

DIFFICULTY:

SIMPLE-EASY

PREREQUISITES:

None

PROBLEM:

Chef has been tested for Covid Positive. Based on Chef’s age(A), No. of beds available(N) and days to recover, we have to find whether the Chef could be saved or not if he has ‘d’ days left to get saved.

QUICK EXPLANATION:

Just Check When a bed will get vacant and then find days remaining for the Chef to get saved. If it comes out to be more than days to recover print “Yes” else print “NO”.

EXPLANATION:

Step wise solution as follows:

  • Check whether any bed is vacant or not,
    if vacant admit the Chef
    If not,
  • Check whether a child is admitted or not,
    Then, After 10 days Chef can be admitted
    If not,
  • Check whether an adult is admitted or not,
    Then, The Chef can be admitted after 14 days
    If not,
  • Chef can only be admitted after 21 days
  • After admitting Check if there are enough days left for the Chef to be cured
    If true - print “YES”
    else print “NO”

TIME COMPLEXITY:

O(1)

SOLUTIONS:

Editorialist's Solution 
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ll a,n,d,x,y,z,min;
        cin>>a>>n>>d>>x>>y>>z;   // (x+y+z)<=n

        // defining minimum days to recover
        if(a<19)
            min=10;
        else if(a>18 && a<51)
            min=14;
        else
            min=21;

        ll diff =0;
        if(x+y+z<n)    // a bed is already vacant
        {
            if(d>=min)
                cout<<"YES\n";
            else
                cout<<"NO\n";
        }
        else
        {
            diff = d-10;   // there is a child on bed
            if( diff < min)
                cout<<"NO\n";
            else
            {
                if(x>0)
                    cout<<"YES\n";
                else
                {
                    diff = diff-4;   // when there is no child, checking for adult
                    if(diff >=min)
                    {
                        if(y>0)
                            cout<<"YES\n";
                        else
                        {
                            diff = diff - 7;    //when there is no child or adult, checking for old
                            if(diff>=min)
                                cout<<"YES\n";
                            else
                                cout<<"NO\n";
                        }
                    }
                    else
                        cout<<"NO\n";
                }
            }
        }
    }
    return 0;
}
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