 # MILEAGE - Editorial

Testers: utkarsh_25dec, iceknight1093

831

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')
``````