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10. a classic question is, to a 3 * 3 * 3 cube cube into 27 units, if each is allowed to re-place across the board after the location of each piece, at least can Jidao? The answer is 6 knives, because the middle of the unit cube of six faces are cut out later, so how need 6 knife. Consider this question: If the A n * n * n unit cube into a cube, take at least Jidao?
Answer: In fact, starting from a stronger proposition but make the problem easier. For an a * b * c of the rectangular, we need f (a) + f (b) + f (c) knife, which f ( x) =? log (x) / log (2)?. Need only note that in any step of the process, the current maximum block after they cut the required number is the same knife the whole process needs a few knives, as other small knives need not exceed the maximum number of blocks of the be the number of knives, they can work together with the largest block of parallel processing. This shows that our optimal decision is the biggest block to the current as small as possible, that should block as much as possible the current equivalent to the largest cut in half. Using mathematical induction, we can quickly get the conclusion of the beginning of the paragraph.
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10 an exciting intellectual issue, see you do a few road?

2. how to use a coin equal probability to generate a 1 to 3 random integers? If the coin is unfair to it?
Answer: If it is a fair coin, then threw twice, \
If it is unfair coins, notes that a \ 3, the remaining cases again. Another way is even better, throwing three coins, \

3, the rest of the situation again.

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5. Initially, two pockets have a ball. The back of the n-2 balls turn into the pocket, into which pocket the probability of the ball with the pocket has been proportional to the number. This way, the ball pocket that fewer average expected number of balls?
answer: first consider how a seemingly unrelated issues 1 to n,[link widoczny dla zalogowanych], generate a random order. First, write down on paper the numbers 1; then, the two wrote in a left or right; then, the three wrote in the left, far right, or inserted between 1 and 2 ... ... In short, the number i, etc. probability to the previous i-1 into the number of generated (including the far left and far right,[link widoczny dla zalogowanych], including a) a total of i-space one. This is obviously generated by a completely random order.
we change the topic description of the process point of view: an imaginary ball with a rope tied the two together, this label is a rope. Next, put a small ball which split into two balls, two small balls attached to rope labeled 2. In short, the \ Lenovo our previous discussion, the label is in fact a string of random permutation, that is the beginning position of the rope 1 may be the last such appearance in each place. That is, it both sides of the ball number (1, n-1), (2, n-2), (3, n-3), ... ..., (n-1, 1) that n-1 种 conditions, which may occur. Therefore, the pouch is about the number of balls n / 4 个. Precisely, when n is odd, the ball pouch number (n +1) / 4; when n is even, the ball pouch number of n ^ 2 / (4n-4).

8. arbitrarily to an 8 * 8 01 matrix, you can only choose a 3 * 3 or 4 * 4 sub-matrix and all the elements inside the inversion. Is there a way to all numbers in the matrix all into 1?
answer: no. Large matrix in the 36 small matrix 3 * 3 4 * 4 and 25 small matrix, so a total of 61 possible operations. Obviously, given a sequence of operations, the sequence of these operations is irrelevant; In addition, the use of an operating sequence of two or more of the same operation is useless. Therefore, the real sequence of different operations only 2 ^ 61. However, the 01 8 * 8 matrix of a total of 2 ^ 64 species, so not every situation has a way to achieve their goals.
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1. to a blind man of 52 playing cards, and told him that there are exactly 10 cards face up to. Requirements of the cards into two piles of this blind, so that each stack of cards face up card in the number of sheets as much. The blind how to do?
answer: the cards into two piles, a pile of 10, a pile 42. Then, all of the small pile of cards which all turned over.
3. 30 枚 face value of the same coin incomplete in a row, A, B and two one turn to one end of a row of coins, and took away the outside of the rocket coins. If you first take a coin , can guarantee the money will not be less than the opponent it?
answer: who can allow himself to be always taken an odd position to take coins, or even always taken the position of the coins. Count the position of the face value is odd or even position more than the sum of the face value of the sum of many, and always take these positions on the coins on it.
6. Consider an n * n chessboard, to a common edge of two adjacent lattice is called lattice. Initially, some cells yard has a virus. After every second, so long as at least two adjacent lattice grid infected with a virus , then he will also be infected. To make all the boxes have been infected, initially requires at least a few lattice with a virus? Gives a solution and prove optimality.
Answer: at least n, such as na diagonal grid. n-grid is also required. When a new grid is infected, all were infected with the graphics grid consisting of the perimeter will be reduced by 0, 2 or 4 units (the number depends on the specific reduction of the lattice around it by the number of infections). Also because when all the lattice have been infected, graphics, perimeter is 4n, so initially at least n-infected cells.
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7. in an m * n chessboard, a k-cell to have placed a piece. Whether all the pieces were always stained red and blue two, so that each row and each column of red and blue pieces of a pawn up poor?
Answer: Yes. Build a bipartite graph G (X, Y), where X has m vertices represent the board of the m-line, Y has n vertices representing the board of the n columns. Ji -row pieces are listed in X (i) and Y (j) an edge between the connected. First find out the graph G where all the ring (as is the bipartite graph of the loop length must be even), to ring in the side of alternating red and blue coloring. Did not stain the rest of the plan must be a number of trees. Tree recursively for each operation: remove a leaf node and the corresponding side, the rest of the tree, two red and blue stained legitimate , then just remove the vertices and edges added back to this side of the appropriate color to meet the requirements.

4. a circular orbit with n gas stations, fuel filling stations the sum of all the car just enough to run around. Proved that one can always find a gas station, making the initial fuel tank is empty when the car from here, smooth ring around back to the beginning.
Answer: there is always a gas station, the oil only enough to run it under a gas station (or all the stations of the oil will not add up to full. ) To the next station in advance to move all the oil to the gas station, and the oil had been removed out of the gas station ignored. In the rest of the filling stations continue to search for oil enough to reach the place where the next gas station, gas stations continue to merge until only one station so far. Obviously from here will be able to successfully finish the course.
Another proof: let the car tank installed more than enough oil, gas station from which the starting random test lap. Every car at a gas station, the record at this time the rest of the fuel tank, and then the gas stations all equipped with the oil. Finish lap test, check the gas station just on the road to which the least amount of oil left, then empty the tank from there obviously must be able to run the entire distance.

9. five holes in a row, one for possession of a fox hole. Each night, the fox will jump to an adjacent hole; every day, you are only allowed to check one of the holes. How can we ensure that the fox will eventually be caught?
Answer: According to the 2, 3, 4, 2, 3, 4 in order to seize the inspection can guarantee a fox hole fox. To illustrate this scheme is feasible, with a collection of F, said the fox may appear, initially F = {1, 2, 3, 4, 5}. If it is not No. 2 hole, then the next day, the fox had run to the F = {2 , 3, 4, 5}. If the No. 3 hole at this time it is not, then the third day of the fox must go to the F = {1, 3, 4, 5}. At this point it is not 4, if the hole, then another after a night F = {2, 4}. If the No. 2 hole this time it is not, then another day F = {3, 5}. If it is not in the 3rd hole at this time, another day it will go to No. 4 hole.
program is not the only, the following options are possible:
2, 3, 4, 4, 3, 2
4, 3 , 2, 2, 3, 4
4,[link widoczny dla zalogowanych], 3, 2, 4, 3, 2
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