12_Question.cpp

// Problem Statement

// Particulate matters are the biggest contributors to Delhi pollution. The main reason behind the increase in the 
// concentration of PMs include vehicle emission by applying Odd Even concept for all types of vehicles. 
// The vehicles with the odd last digit in the registration number will be allowed on roads on odd dates and those 
// with even last digit will on even dates.

// Given an integer array a[], contains the last digit of the registration number of N vehicles traveling on 
// date D(a positive integer). The task is to calculate the total fine collected by the traffic police department 
// from the vehicles violating the rules.

// Note : For violating the rule, vehicles would be fined as X Rs.

// Example 1:

// Input :

// 4 -> Value of N

// {5,2,3,7} -> a[], Elements a[0] to a[N-1], during input each element is separated by a new line

// 12 -> Value of D, i.e. date 

// 200 -> Value of x i.e. fine

// Output :

// 600  -> total fine collected 

// Explanation:

// Date D=12 means , only an even number of vehicles are allowed. 

// Find will be collected from 5,3 and 7 with an amount of 200 each.

// Hence, the output = 600.

// Example 2:

// Input :

// 5   -> Value of N 

// {2,5,1,6,8}  -> a[], elements a[0] to a[N-1], during input each element is separated by new line

// 3 -> Value of D i.e. date 

// 300 -> Value of X i.e. fine 

// Output :

// 900  -> total fine collected 

// Explanation:

// Date D=3 means only odd number vehicles with are allowed.

// Find will be collected from 2,6 and 8 with an amount of 300 each.

// Hence, the output = 900 

// Constraints:

// 0<N<=100
// 1<=a[i]<=9
// 1<=D <=30
// 100<=x<=5000 
// The input format for testing 

// The candidate has to write the code to accept 4 input(s).

// First input – Accept for N(Positive integer) values (a[]), where each value is separated by a new line.

// Third input – Accept value for D(Positive integer)

// Fourth input – Accept value for X(Positive integer )

// The output format for testing 

// The output should be a positive integer number (Check the output in Example 1, Example e) if no fine is collected then print ”0”.

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

int CalculateFine(vector<int>& arr, int D, int X){
    int fine = 0;
    for(auto x : arr){
        if(D % 2 == 0 && x % 2 != 0){
            fine += X;
        }else if(D % 2 != 0 && x % 2 == 0){
            fine += X;
        }
    }
    return fine;
}

int main(){
    int N;
    cin>>N;
    vector<int> arr(N);

    for(int i=0; i<N; i++){
        cin>>arr[i];
    }
    int date;
    cin>>date;

    int fine;
    cin>>fine;

    cout<<CalculateFine(arr,date,fine);

    return 0;
}