Plus you need to check ((x2 — x1) / x + (y2 — y1) / y ) % 2 == 0 because there are no rules that move only x or y.

remember that we must change a character. first compare s[0.. n/2-1] and s[n-1.. n-1-(n/2-1)]

YES : only one difference || ( n % 2 == 1 && every character is equal )

NO : (Actually using 'else') ( n % 2 == 0 && every character is equal ) || more than 1 differences

How to solve C?

I can get gcd using Euclidean Algorithm(if you know any better idea, can I ask you to tell any idea).

Just Programming skills not a specific algorithm (sorry about confusion)

This contest is good for me to overcome my laziness about studying algorithm.

I just find min, max value of div1, div2. if there is no div2 ==>> Infinity. if a max value of div2 is bigger than a min value of div1 ==>> Impossible. Finally, there is solution = 1899 + position — maxValueDiv2 to find maximum value. Drawing range will be helpful (Sorry Bad English)

Contest always end up with hacking :(

My solution C simulating statement is accepted. is there any solution without simulating?

I agree with you. I had the same experience before.

I have a question. I wonder how people can solve Div2 B, though they are confused about it... Is it a gap between me and high ranks?

there are two copies.

Div2 B, they said "It is guaranteed that the puzzle pieces are one 4-connected piece." and didn't accept rotate, flip and overlap. ==> only move. Because there are two copies of puzzle, only rectangular input is accepted. no stairs shaped, window shaped...

.XXXX XXXX. XXXX. or .XX XX. XX. .XX XX. XX. It can't be rectangle

I hope this will be helpful :) It seems like binary search.

ll recur(int n, ll k) { ll temp = pow(2, (double)n — 1); if (k == temp) return n; else if (k > temp) { k -= temp; } return recur(n — 1, k); }

Hack Festival!@#(@($*!@(#!@

Although I missed a mistake as you mentioned, pretest was permitted. I noticed it, corrected immediately and got -50 points :( But I used the information for hacking! and got +100 points :)

I can feel the power of Alyona's mother