An automobile company manufactures both a two wheeler (TW) and a four wheeler (FW)
Problem Statement – An automobile company manufactures both a two wheeler (TW) and a four wheeler (FW). A company manager wants to make the production of both types of vehicle according to the given data below:
- 1st data, Total number of vehicle (two-wheeler + four-wheeler)=v
- 2nd data, Total number of wheels = W
The task is to find how many two-wheelers as well as four-wheelers need to manufacture as per the given data.
Example :
Input :
- 200 -> Value of V
- 540 -> Value of W
Output :
- TW =130 FW=70
Explanation:
130+70 = 200 vehicles
(70*4)+(130*2)= 540 wheels
Constraints :
- 2<=W
- W%2=0
- V<W
Print “INVALID INPUT” , if inputs did not meet the constraints.
The input format for testing
The candidate has to write the code to accept two positive numbers separated by a new line.
- First Input line – Accept value of V.
- Second Input line- Accept value for W.
The output format for testing
- Written program code should generate two outputs, each separated by a single space character(see the example)
- Additional messages in the output will result in the failure of test case
C++ Coading
#include <bits/stdc++.h>
using namespace std;
int main ()
{
int v, w;
cin >> v >> w;
float x = ((4 * v) - w) / 2;
if ((w & 1) || w < 2 || w <= v)
{
cout << "INVALID INPUT";
return 0;
}
cout << "TW=" << x << " " << "FW=" << v - x;
}
Java Coading
import java.util.*;
public class Solution
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int v=sc.nextInt();
int w=sc.nextInt();
float res=((4*v)-w)/2;
if(w>=2 && (w%2==0) && v<w )
System.out.println("TW= "+(int)(res)+" FW= "+(int)(v-res));
else
System.out.println("INVALID INPUT");
}
}
Python Coading
v=int(input())
w=int(input())
if (w&1)==1 or w<2 or w<=v:
print("INVALID INPUT")
else:
x=((4*v) -w)//2
print("TW={0} FW={1}".format(x,v-x))
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