MILEAGE - Editorial

editorial
,
1

PROBLEM LINK:

Contest
Practice

Setter: inov_adm
Testers: utkarsh_25dec, iceknight1093

DIFFICULTY:

831

PREREQUISITES:

None

PROBLEM:

Given cost and mileage of petrol and diesel, find which is the cheaper option to travel N kms.

EXPLANATION:

The cost incurred if you use the petrol car, is the cost of petrol times the amount of petrol used. Which is x * (n/a). Similarly, the cost of using the diesel car is the cost of diesel times the amount of diesel used. Which is y * (n/b). We just find which one of these is smaller, and output that. If they are equal, output “Any”.

TIME COMPLEXITY:

Time complexity is O(1).

SOLUTION:

Editorialist's Solution 
#include <iostream>
#include <cmath>
using namespace std;

int T,n,x,y,a,b;

int main() {
	cin>>T;
	
	while(T--)
	{
	    cin>>n>>x>>y>>a>>b;
	    
	    double petrolCost = x * ( ((double)n)/a );
	    double dieselCost = y * ( ((double)n)/b );
	    
	    if(petrolCost < dieselCost)
	        cout<<"Petrol\n";
	    else if(petrolCost > dieselCost)
	        cout<<"Diesel\n";
	    else
	        cout<<"Any\n";
	}
	return 0;
}
Tester's Solution 
for _ in range(int(input())):
	n, x, y, a, b = map(int, input().split())
	if x*b == y*a:
	    print('any')
	elif x*b < y*a:
	    print('petrol')
	else:
	    print('diesel')