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# Car Range CodeChef Solution

## Problem – Car Range CodeChef Solution

The fuel economy of a car is the distance which it can travel on one litre of fuel.

The base fuel economy (i.e, its fuel economy when there is only one person – the driver – in the car) of a certain car is M kilometres per litre. It was also observed that every extra passenger in the car decreases the fuel economy by 1 kilometre per litre.

P people want to take this car for a journey. They know that the car currently has V litres of fuel in its tank.

What is the maximum distance this car can travel under the given conditions, assuming that all P people always stay in the car and no refuelling can be done?

Note that among the P people is also a driver, i.e, there are exactly P people in the car.

### Input Format

• The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
• Each test case consists of a single line of input, containing three space-separated integers P,M, and V – the number of people, base fuel economy, and amount of fuel present in the car, respectively.

### Output Format

For each test case, output a single line containing one integer – the maximum distance that the car can travel.

• 1≤T≤3⋅10^4
• 1≤P≤5
• 6≤M≤56
• 1≤V≤100

### Sample 1:

``````Input:
3
5 10 20
1 6 10
4 6 1
Output:
120
60
3``````

### Explanation:

Test Case 1: There are 5 people in the car, and its base fuel economy is 10. Thus, its effective fuel economy is 10−4=6, since there are 4 people present other than the driver. There are 20 litres of fuel in the car, so the maximum distance that can be travelled is 6⋅20=120.

Test Case 2: The effective fuel economy is 6, and the maximum distance that can be travelled is 6⋅10=60.

Test Case 3: The effective fuel economy is 3, and the maximum distance that can be travelled is 3⋅1=3.

## Car Range CodeChef Solution in C++17

``````#include <iostream>
using namespace std;

int main() {
int t;
cin>>t;
while(t--)
{
int p,m,v;
cin>>p>>m>>v;

if(p==1)
cout<<m*v<<endl;
else cout<<(m-p+1)*v<<endl;
}
return 0;
}
``````

## Car Range CodeChef Solution in Java

``````/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int p=sc.nextInt();
int m=sc.nextInt();
int v=sc.nextInt();
if(p==1)System.out.println(m*v);
else System.out.println((m-p+1)*v);
}
}
}
``````

## Car Range CodeChef Solution in Pyth 3

``````# cook your dish here
for i in range(int(input())):
p,m,v=map(int,input().split())
o=p-1
if p==1:
print(m*v)
else:
print((m-o)*v)``````
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