Problem:
Score: 100 points

Problem Statement

In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019.

Integers M_1, D_1, M_2, and D_2 will be given as input.

It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1.

Determine whether the date 2019-M_1-D_1 is the last day of a month.

Constraints

Both 2019-M_1-D_1 and 2019-M_2-D_2 are valid dates in the Gregorian calendar.

The date 2019-M_2-D_2 follows 2019-M_1-D_1.

Input

Input is given from Standard Input in the following format:

M_1 D_1
M_2 D_2

Output

If the date 2019-M_1-D_1 is the last day of a month, print 1; otherwise, print 0.

Sample Input 1

11 16
11 17

Sample Output 1

0

November 16 is not the last day of a month.

Sample Input 2

11 30
12 1

Sample Output 2

1

November 30 is the last day of November.

Solution in PHP7 (7.0.15) (judge verdict: Wrong Answer)

  //標準入力
  $fh = trim(fgets(STDIN));
  $print = [];

//配列に格納
  if ($fh) {
    while (!feof($fh)) {
      $stdinLine = fgets($fh);
      $print[] = explode(" ",$stdinLine);
    }
    fclose($fh);
  }

  //日付1
  $day1 = "2019-".$print[0][0]."-".$print[0][1];
  $day10 = trim($day1);
  $year1 = "2019-".$print[0][0];

  //日付2
  $day2 = "2019-".$print[1][0]."-".$print[1][1];
  $day20 = trim($day2);
  $year2 = "2019-".$print[1][0];

  //指定してくれる関数
$firstDate = date('Y-m-d', strtotime('first day of ' . $year1));
$lastDate = date('Y-m-d', strtotime('last day of ' . $year1));


  if ($day10 == $lastDate ) {
  print("1");
  } else {
  print("0");
}