answer
A
The meaning of problems:
Given a set of points and a plane of the original flat interrogation point set, and only a vertex of the right triangle in the number of interrogation points are concentrated statistically plane. Two point set size of not more than \ (2000 \) .
Problem solution:
Day 2 of the original title was it?
Even I would, sort polar angle, two-hand sweep over statistics answer, pay attention to accuracy.
Experience:
mouth Hu, playing the title of princess or to rely on a solid foundation.
D
Meaning of the questions:
Given a positive integer size appropriate, ask it in the reciprocal of the decimal is not a finite decimal.
Experience:
do not say that readers will think I am silly.
E
Meaning of the questions:
are complex, try to summarize.
Given a trellis, there are two lattice A can, vice versa.
The first line of FIG grid has several robots, the robot will try to move down until it meets a converter to change direction.
There are four such converters, they can come from a robotic transfer in both directions to go in another direction, but there is no converter to support a U-turn, the robot will continue to move after the turn, until next met a converter.
Converter may be only on the grid can pass, and can not be placed on the same square.
FIG bottom of the grid a number of outlets, each outlet of a robot considered only escaped arrival trellis diagram, the robot outlet do not have the determined corresponding relation, but not receiving a plurality of outlet robots.
Any time a robot can be in the same grid without the risk of conflict (in addition to export out).
Seeking a discharge converter program, so that the robot can escape from all of the grid of FIG.
Solution:
find each grid is the most horizontal / vertical forward through time, because the path of the robot could not repeat (otherwise they will come to the same exit), it is impossible inverted repeat (this will only make the robot go back to retrograde is the starting point of retrograde person).
Because it is the equivalent of between robots, so we engage in a network stream, the source side even to the robot, each grid is connected to the surrounding side, the export side even to the meeting point, all traffic side is 1.
There will be a lot easier to find illegal situation (but not including the case of retrograde path and repeat).
- The first is a grid -> left once, has been under -> Right again.
This is not easy to find excellent, because if these two paths to exist, they must also have to pay elsewhere (either themselves and their intersection, or there are two points of intersection between each other), we can directly answer here optimized away.
- The first is a grid -> left once, and was the right -> the next time.
Between the robot are equivalent. This situation is equivalent to not put converter.
There are two cases are symmetric, not to say.
finished.
Experience:
I thought, to have any feelings on the test?
F
Meaning of the questions:
Given a cactus forest, and asked how many Border Erase program to make it into a tree forest.Control of Desertification
Solution:
That goes without saying? Direct running ring, ring side at least a delete, delete does not delete the side of the tree does not matter.
Others
Gugu Gu, the waiting up files.
Overall experience
Marser is not the first day, I think she (fog).