Library Fine

Problem Statement :

Your local library needs your help! Given the expected and actual return dates for a library book, create a program that calculates the fine (if any). The fee structure is as follows:

1. If the book is returned on or before the expected return date, no fine will be charged (i.e.: fine = 0).
2. If the book is returned after the expected return day but still within the same calendar month and year as the expected return date, 
fine = 15 Hackos * (the number of days late).
3. If the book is returned after the expected return month but still within the same calendar year as the expected return date, the 
fine = 500 Hackos * (the number of months late).
4. If the book is returned after the calendar year in which it was expected, there is a fixed fine of 10000 Hackos.

Charges are based only on the least precise measure of lateness. For example, whether a book is due January 1, 2017 or December 31, 2017, if it is returned January 1, 2018, that is a year late and the fine would be 10000 Hackos.

d1, m1, y1 = 14, 7, 2018
d2, m2, y2 = 5, 7, 2018

The first values are the return date and the second are the due date. The years are the same and the months are the same. The book is 14 - 5 = 9 days late. Return 9 * 5 = 135.

Function Description

Complete the libraryFine function in the editor below.

libraryFine has the following parameter(s):

d1, m1, y1: returned date day, month and year, each an integer
d2, m2, y2: due date day, month and year, each an integer

int: the amount of the fine or 0 if there is none

Input Format

The first line contains 3 space-separated integers, d1, m1, y1, denoting the respective day, month  and year on which the book was returned.
The second line contains 3 space-separated integers, d2, m2, y2, denoting the respective day, month  and year on which the book was due to be returned.

1 <= d1, d2 <= 31
1 <= m1, m2 <=12
1 <= y1, y2 <= 3000
It is guaranteed that the dates will be valid Gregorian calender dates.

Solution :


                            Solution in C :

C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {    
    int edd, emm, eyyyy, add, amm, ayyyy;
    int fine =0;
    scanf("%d%d%d%d%d%d", &add, &amm, &ayyyy, &edd, &emm, &eyyyy);
    if(ayyyy < eyyyy){
        fine = 0;    
    else if(ayyyy > eyyyy){
        fine = 10000;
    else if(ayyyy == eyyyy){
        if(amm < emm){
            fine = 0;
        else if(amm > emm){
            fine = 500 * (amm - emm);
        else if(amm == emm){
            if(add <= edd){
                fine = 0;
            else if(add > edd){
                fine = 15 * (add - edd);
    printf("%d", fine);
    return 0;

C ++  :

using namespace std;
int main()
    int actual[3],expected[3],i,j;
    else if(actual[2]-expected[2]>0)
    else if(actual[1]-expected[1]<0)
    else if(actual[1]-expected[1]>0)
    else if(actual[0]-expected[0]>0)


Java  :

import java.util.Scanner;

class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(;
        int d2 = sc.nextInt(), m2 = sc.nextInt(), y2 = sc.nextInt();
        int d1 = sc.nextInt(), m1 = sc.nextInt(), y1 = sc.nextInt();
        if (y2 > y1) {
        } else if (y2 < y1 || m2 < m1 || m2 == m1 && d2 <= d1) {
        } else if (m1 == m2) {
            System.out.println(15 * (d2 - d1));
        } else {
            System.out.println(500 * (m2 - m1));

python 3  :


def main(N):
	sum0 = 0
	D = N[0][0]-N[1][0]
	M = N[0][1]-N[1][1]
	Y = N[0][2]-N[1][2]

	if D > 0 and M == 0 and Y == 0:
		sum0 = 15*(D)
	elif M > 0 and Y == 0:
		sum0 = 500*(M)
	elif Y > 0:
		sum0 = 10000*(Y)


if __name__ == '__main__':
	N = [list(map(int,input().split(" "))) for i in [1,2]]

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