It is 2012, and it's a leap year. So there is a "February 29" in this year, which is called leap day. Interesting thing is the infant who will born in this February 29, will get his/her birthday again in 2016, which is another leap year. So February 29 only exists in leap years. Does leap year comes in every 4 years? Years that are divisible by 4 are leap years, but years that are divisible by 100 are not leap years, unless they are divisible by 400 in which case they are leap years.

In this problem, you will be given two different date. You have to find the number of leap days in between them.

Input

Input starts with an integer T (≤ 550), denoting the number of test cases.

Each of the test cases will have two lines. First line represents the first date and second line represents the second date. Note that, the second date will not represent a date which arrives earlier than the first date. The dates will be in this format - "month day, year", See sample input for exact format. You are guaranteed that dates will be valid and the year will be in between 2 * 10³ to 2 * 10⁹. For your convenience, the month list and the number of days per months are given below. You can assume that all the given dates will be a valid date.

Output

For each case, print the case number and the number of leap days in between two given dates (inclusive).

Sample

Sample Input	Sample Output
Click to copy 4 January 12, 2012 March 19, 2012 August 12, 2899 August 12, 2901 August 12, 2000 August 12, 2005 February 29, 2004 February 29, 2012	Click to copy Case 1: 1 Case 2: 0 Case 3: 1 Case 4: 3

Notes

The names of the months are {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"}.
The numbers of days for the months are {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31} respectively in a non-leap year.
In a leap year, number of days for February is 29 days; others are same as shown in previous line.

Problem Name: February 29

LightOj-1414

Problem Idea:

In this problem, you will be given two different date which format is (Month Date, Year). You have to find the number of leap days in between them.
Note : leap day means February 29.

Solution Approach:

We know that leap year is a year which divisible by 4 but not divisible by 100 or which divisible by 400.
For example:
year= 2020
First condition is, year%4==0 and year%100 !=0.
Second condition is, year%400==0.
Now, if any year accepts one or both of these conditions then we called this year is a leap year otherwise NOT a leap year.

In this problem , count the number of leap days between the range .

Hint

At first count the total number of leap year between of the range .then
If Starting Year of the range is a leap year but the Month already cross February so neglecting this leap year of the total number of leap year .
If Ending Year of the range is a leap year but the Month and date are less than February 29 so neglecting this leap year of the total number of leap year.

You know that a Leap comes after 4 Years later. So, find the number of Multiples of 4 between two ranges Dividing by 4. Then As you did for Checking Leap Year find the Multiples of 100. Further, find the Multiples of 400. Then apply the Rule given Below -

Number of Leap Years = Multiples of 4 - Multiples of 100 + Multiples of 400.
Where,

Multiples of 4 = ( Ending Year - (Starting Year -1) ) /4
Multiples of 100 = ( Ending Year -( Starting Year - 1 ) ) /100
Multiples of 400 = ( Ending Year - (Starting Year -1 ) ) /400

Time Complexity: O(1) per test_case;

Cpp Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{

    int test_case;
    cin >> test_case;
    string start_month, end_month;
    char comma;
    int start_date, end_date, start_year, end_year;
    for (int i = 1; i <= test_case; i++)
    {
        cin >> start_month >> start_date >> comma >> start_year;
        cin >> end_month >> end_date >> comma >> end_year;
        if (start_month == "January" or start_month == "February") start_year = start_year;
        else start_year++;

        if (end_month == "January" or (end_month == "February" and end_date < 29)) end_year--;
        else end_year = end_year;

        int mul_of_4 = end_year / 4 - (start_year - 1) / 4;
        int mul_of_400 = end_year / 400 - (start_year - 1) / 400;
        int mul_of_100 = end_year / 100 - (start_year - 1) / 100;

        int total = mul_of_4 + mul_of_400 - mul_of_100;
        cout << "Case "<<i<<": "<<total << endl;
    }
}

Python Code:


import calendar
month={"January":0, "February":1, "March":2, "April":3, "May":4, "June":5, "July":6, "August":7, "September":8, "October":9, "November":10, "December":11}
for i in range(1,int(input())+1):
    start=list(input().split())
    end=list(input().split())
    year=calendar.leapdays(int(start[2]),int(end[2])+1)
    end_date=end[1][:-1]
    if calendar.isleap(int(start[2]))==True and (month[start[0]]>1):
        year-=1
    if calendar.isleap(int(end[2]))==True and (month[end[0]]<1 or (month[end[0]]==1 and int(end_date)<=28)):
        year-=1
    print("Case {}:".format(i),year)

Happy Coding!

Written by: Md. Rasel Meya

Tutorial source