Editorial-Longest 6 |Encoding Junior November 21

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Author: Srinjoy Pal
Tester: Abhisekh Paul ,Sayan Poddar
Editorialist: Srinjoy Pal

DIFFICULTY:

CAKEWALK

PREREQUISITES:

Basic Math , Conditional Statements

PROBLEM:

Finding max element of the array. Then performing simple math operations we have to find the cost of each steroid . find the minimum cost.

EXPLANATION:

First we have to find the max element of the array. Subtracting N from the max element and adding 1 we get the required length to cover. Next, We have to find the number of each steroid required to cover the distance. Then , we have to find the minimum cost for buying out of the two.

TIME COMPLEXITY:

O(n)

SPACE COMPLEXITY:

O(1)

SOLUTION:

Setter’s Solution
C++

#include <bits/stdc++.h>
using namespace std;
int main(){
ll t;
cin>>t;
while(t--)
{
    ll m,p,x,q,y,r,st1,st2;
    cin>>m>>p>>x>>q>>y>>r;
    ll a[p];
    for(int i=0;i<p;i++)
    {
        cin>>a[i];
    }
    ll z=*max_element(a, a + p);
    ll req= z-m+1;
    if(req%x==0)
    st1=req/x;
    else if(req%x!=0) 
    st1=req/x+1;
    if(req%y==0)
    st2=req/y;
    else if(req%y!=0)
    st2=req/y+1;
    cout<<min(st1*q,st2*r)<<endl;
 }
return 0;
}

Tester’s Solution
Python

import math
d = int(input())
for i in range(0, d):
m,p,x,q,y,r= list(map(int, input().split()))
a = list(map(int, input().split()))
z=max(a)
n=z-m+1
print(min(((math.ceil(n/x))*q),((math.ceil(n / y)) * r)))

Tester’s Solution
Java

//package myplace;
import java.util.*;
class exam {
public static void  main (String[] args) {
    Scanner sc=new Scanner(System.in);
    int t=Integer.parseInt(sc.nextLine());
    while(t>0)
    {
        String s[]=sc.nextLine().split(" ");
       int m=Integer.parseInt(s[0]);
       int p=Integer.parseInt(s[1]);
       int x=Integer.parseInt(s[2]);
       int q=Integer.parseInt(s[3]);
       int y=Integer.parseInt(s[4]);
       int r=Integer.parseInt(s[5]);
       String s1[]=sc.nextLine().split(" ");
       int max=0;
       for(int i=0;i<s1.length;i++)
       {
           if(Integer.parseInt(s1[i])>max)
           {
           max=Integer.parseInt(s1[i]);
           }
       }
       
       double diff=max-m+1;
       double diffx=diff/x;
       double diffy=diff/y;
      int a1= (int) (Math.ceil(diffx)*q);
      int a2=(int) (Math.ceil(diffy)*r);
       double cost=Math.min(a1,a2);
       System.out.println((int)cost);
       
        t--;
            
    }
}
}
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