A. Axis of Symmetry
B. Binary Tree
- n parity to determine the winner.
C. Constructing Ranches
- Right point on the path is greater than, twice the maximum value, which is a point on a polygonal path could spell Sufficient Conditions.
- Tree divide and conquer, a path statistics can be.
- Please explain how there F0_0H next card often.
D. Defining Labels
Attendance
E. Erasing Numbers
- Consider the first \ (i \) bit of the answer, because we only care about the elements to be deleted and \ (a [i] \) the size of the relationship, the ratio \ (a [i] \) small element is set to 0, than \ (a [i] \) large elements is set to 1.
- First of prefixes, suffixes were eliminated. Consider a sequence of results can be divided into three categories eliminate whole [0] [1] [Full 01] have
F. Falling Objects
G. Game Design
- \ (f (u) \) to consider \ (u \) for the sub-tree root optimum program number.
- \ (f (u) = f (lson) * f (rson) \) or \ (f (lson) * f (rson) +1 \) or \ (1 \)
- According to \ (K \) shows a structure to binary.
H. Hold the Line
I. Incoming Asteroids
idea
- For type 1, a total of y, there are k planet. Points about the pot, each planet needs \ ([\ frac {y} {k}] \) target.
- When a planet to complete the goal, the rest of the job and then shared equally go on, we found that the total amount of work the last minute pot is 2/3
- With a heap to maintain the pot, and then write a BU, BU sample cycle of death [].
J. Junior Mathematician
- Decisions from high to low, the recording module \ (m \) based at, \ (F (X) the -X-\) and the sum of each digit, digits DP.