# HackerRank Day of the Programmer problem solution

In this Day of the Programmer problem you have Given a year, y, find the date of the 256th day of that year according to the official Russian calendar during that year. Then print it in the format dd.mm.yyyy, where dd is the two-digit day, mm is the two-digit month, and yyyy is y.

## Problem solution in Python programming.

```#!/bin/python3

import sys

y = int(input().strip())
count=0
if(y>1918 and y<=2700):
if(y%400==0 or (y%4==0 and y%100!=0)):
count=1
if(count==1):
print('12.09.'+str(y))
else:
print('13.09.'+str(y))
if(y<1918 and y>=1700):
if(y%4==0):
count=1
if(count==1):
print('12.09.'+str(y))
else:
print('13.09.'+str(y))
if(y==1918):
print('26.09.1918')```

## Problem solution in Java Programming.

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static int daysInFeb(int year){
if(year == 1918){
return 15;
} else if (year < 1918) {
if (year % 4 == 0) {
return 29;
} else{
return 28;
}
} else { //year > 1918
if (year % 400 == 0){
return 29;
} else if (year % 100 == 0){
return 28;
} else if (year % 4 == 0){
return 29;
} else {
return 28;
}
}
}

public static int daysInMonth(int month, int year){
if (month == 2) return daysInFeb(year);
else if (Arrays.asList(1,3,5,7,8,10,12).contains(month)){
return 31;
} else return 30;
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int year = in.nextInt();
int month = 0;
int dayssofar = 0;
while (dayssofar <= 256){
month++;
dayssofar += daysInMonth(month,year);
}
int day = daysInMonth(month,year);
if (month == 2 && year == 1918) day += 13;
while (dayssofar > 256) {
day--;
dayssofar--;
}

System.out.format("%02d.%02d.%04d",day,month,year);

}
}```

### Problem solution in C++ programming.

```#include <bits/stdc++.h>

using namespace std;

int main(){
int y;
cin >> y;
if(y<1918){
if(y%4==0)cout<<"12.09."<<y<<endl;
else cout<<"13.09."<<y<<endl;
}
else if(y==1918){
cout<<"26.09."<<y<<endl;
}
else{
if(y%400==0){
cout<<"12.09."<<y<<endl;
}
else if(y%4==0&&y%100!=0){
cout<<"12.09."<<y<<endl;
}
else cout<<"13.09."<<y<<endl;
}
return 0;
}```

### Problem solution in C programming.

```#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main()
{
int y,leap;
scanf("%d",&y);
leap=0;
if (y % 4 == 0) leap=1;
if (y > 1918)
{
if (y % 100 == 0) leap=0;
if (y % 400 == 0) leap=1;
}
if (y != 1918)
{
if (leap == 0) printf("13.09.%d\n",y);
else printf("12.09.%d\n",y);
}
else printf("26.09.%d\n",y);
return 0;
}```

### Problem solution in JavaScript programming.

```process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
input_stdin += data;
});

process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});

return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
var leap = 0; //make 1 if it is a leap year...remove a day
if(y < 1918){
if(y%4 === 0){leap = -1}
console.log(13 + leap + ".09." + y);
}else if(y > 1918){
if(y%4 === 0 && y%100 !== 0 || y%400 === 0){leap = -1};
console.log(13 + leap + ".09." + y);
}else{ //case for 1918
console.log("26.09.1918");
}
}```