Codeforces Round #200 (Div. 2)

I'm going to start from 200 to the present thanks for the idea.

A: Not the same as the last input, ans++

B: Find two maximum matches at a time

C: Assuming that the resistor is now connected to the X/y (x+y)/y parallel to the y/(X+y) can be seen also consumes 1 resistors formed by the new values reciprocal to each other this conclusion allows us to change the numerator denominator of the resistance value at random (without affecting the total number of resistors used) the apparent resistance value >1 when we take a series  The resistor of the whole number can therefore be obtained the method resistor >1 when the series <1 is inverted and >1 ... This is a very similar process.

D: Use the stack to simulate the same character with the top of the stack on the stack or the last stack to determine whether the stack is empty

E: Two-part answer greedy judgment this answer can be set up for a head x first to read it to the left is not read to find out the need for time L then there are two strategies one is to read L and then retrace as far as possible to the right to read the other is to read the right but to ensure that can be retraced to read L Two strategies greedy choice can

Codeforces Round #200 (Div. 2)