<< problem 18 - Maximum path sum I Factorial digit sum - problem 20 >>

# Problem 19: Counting Sundays

You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

# My Algorithm

There is a sweet algorithm called "Zeller's congruence' that determines the weekday of any given date.
Its Wikipedia page has a detailled explanation and comes with a few examples, too: en.wikipedia.org/wiki/Zeller's congruence)
You will even find instructions on how to convert that algorithm to source code - my implementation is inside the function getWeekday.

We are only interested in Sundays, that means only an algorithm's result of 1 is of importance.
My program calls getWeekday for every first of every month and increments a counter sum if getWeekday returned 1.

## Alternative Approaches

The original problem can be easily solved by brute-force, too.
An overview of other algorithms: en.wikipedia.org/wiki/Determination of the day of the week

## Modifications by HackerRank

Arbitrary input dates that are ridiculously far in the future ... but not too far apart from each other.
Therefore the "2800 years optimization" isn't actually needed to achieve full score.

## Note

The weekday pattern repeats every 2800 years and contains 4816 Sundays on the first of a month:
- 7 days a week x 400 years in the leap year cycle → 2800
- processing any such cycle produces 4816 matches

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 1900 1 1 1910 1 1" | ./19

Output:

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>

const unsigned int Sunday = 1;
// based on Zeller's congruence
// January = 1, February = 2, ..., December = 12
// returns 0 => Saturday, 1 => Sunday, 2 => Monday, ... 6 => Friday
unsigned int getWeekday(unsigned long long year, unsigned int month, unsigned int day)
{

// January and February are counted as month 13 and 14 of the previous year
if (month <= 2)
{
month += 12;
year--;
}

// Wikipedia provides an altered formula better suited for software implementation
return (day +
13 * (month + 1) / 5 +
year + year / 4 - year / 100 + year / 400)
% 7;
}

int main()
{
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned long long year1, year2;
unsigned int month1, month2, day1, day2;
std::cin >> year1 >> month1 >> day1; // from
std::cin >> year2 >> month2 >> day2; // to

// wrong input order ?
if (year2 < year1 || (year2 == year1 && month2 < month1))
{
std::swap(year1,  year2);
std::swap(month1, month2);
}

// jump forward to the first day of the month
unsigned long long currentYear  = year1;
unsigned int       currentMonth = month1;
if (day1 > 1)
{
currentMonth++;

// from December to January of next year
if (currentMonth > 12)
{
currentMonth -= 12;
currentYear++;
}
}

// number of relevant Sundays
unsigned int sum = 0;

// same patterns every 2800 years
while (currentYear + 2800 < year2)
{
currentYear += 2800;
sum += 4816;         // there are 4816 Sundays on the first of a month in 2800 years
}
// note: a constant-time approach would be to use MOD ...
//       but the while-loop is probably easier to understand

// simple scan through all months
while (currentMonth < month2 || currentYear < year2) // days already match, they are both 1
{
// count Sundays
if (getWeekday(currentYear, currentMonth, 1) == Sunday)
sum++;

currentMonth++;

// from December to January of next year
if (currentMonth > 12)
{
currentMonth -= 12;
currentYear++;
}
}
// check last month, too
if (getWeekday(currentYear, currentMonth, 1) == Sunday)
sum++;

std::cout << sum << std::endl;
}
return 0;
}


This solution contains 13 empty lines, 16 comments and 1 preprocessor command.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

February 24, 2017 submitted solution

# Hackerrank

My code solves 8 out of 8 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too

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The 270 solved problems (level 10) had an average difficulty of 31.3% at Project Euler and
I scored 13,386 points (out of 15600 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 18 - Maximum path sum I Factorial digit sum - problem 20 >>
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