Codeforces Solution | 1311A: Add Odd or Subtract Even

Problem Description:

We have 2 numbers a and b as input. We can obtain b from a. In each move we can add any odd number or subtract any even number. We have to count the minimum move required to obtain b from a.

Ideological Analysis:

Case 1: If \(a = b\) then, no move is required. So, answer will be ‘0’.

Case 2: If \(a \le b\) then,

• If the difference of a and b is odd then one move is required. Because we can add the difference with a \( (a + x) \) and obtain b.
  For example, \(a = 2\) and \(b = 5\) , \(difference = b – a = 3\) \(a + 3 = b\). So, the output will be ‘1’.
• If the difference is even, then two moves are required. Because we can have even number by adding 2 odd numbers.
  For example, \(a = 2\) and \(b = 6\), \(difference = b – a = 4\) \(a + 3 + 1 = b\). \( (a + x + x) \) So, the output will be ‘2’.

Case 3: If \(a \ge b\) then, (opposite of Case 2)

• if the difference of a and b is odd then two moves are required. \( (a + odd number – even number) \)
  For example, \(a = 5\) and \(b = 2\) , \(difference = a – b = 3\) \(a + 1 - 4 = b\). \( (a – x + y) \) So, the output will be ‘2’.
  
• If the difference is even, then one move is required. Because we can subtract the difference from a \( (a – y) \) and obtain b.
  For example, \(a = 6\) and \(b = 2\), \(difference = a – b = 4\) \(a - 4 = b\). \( ( a – y ) \) So, the output will be ‘1’.

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