ATM CodeChef Solution

Problem – ATM CodeChef Solution

Pooja would like to withdraw X $US from an ATM. The cash machine will only accept the transaction if X is a multiple of 5, and Pooja’s account balance has enough cash to perform the withdrawal transaction (including bank charges). For each successful withdrawal the bank charges 0.50 $US.

Calculate Pooja’s account balance after an attempted transaction.

Input

Positive integer 0 < X <= 2000 – the amount of cash which Pooja wishes to withdraw.

Nonnegative number 0<= Y <= 2000 with two digits of precision – Pooja’s initial account balance.

Output

Output the account balance after the attempted transaction, given as a number with two digits of precision. If there is not enough money in the account to complete the transaction, output the current bank balance.

Example – Successful Transaction

Input:
30 120.00
Output:
89.50

Example – Incorrect Withdrawal Amount (not multiple of 5)

Input:
42 120.00
Output:
120.00

Example – Insufficient Funds

Input:
300 120.00
Output:
120.00

ATM CodeChef Solution in C++17

#include <bits/stdc++.h>
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.precision(2);

    int a;
    float b;
    cin >> a >> b;
    
    if ((a%5==0) && (b-a-0.5>=0))
    {
        cout << fixed << b-a-0.5 << endl;
    }
    else
    {
        cout << fixed << b << endl;
    }

    return 0;
}

ATM CodeChef Solution in Pyth 3

n,atm=map(float,input().split())
n=int(n)
if (n+0.5<=atm and n%5==0):
    print(float(atm-n-0.5))
else:
    print(float(atm))

ATM CodeChef Solution in Java

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

class Solution{
    public static void main(String[] args) throws Exception{
        InputStreamReader i = new InputStreamReader(System.in);
        BufferedReader bf = new BufferedReader(i);
        String[] in = bf.readLine().split(" ");
        float n = Float.parseFloat(in[0]);
        float f = Float.parseFloat(in[1]);
        
        if(n%5==0 && f>=n+0.5f)
        System.out.println(f-n-0.5f);
        
        else{
            System.out.println(f);
        }
        
    }
}

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