2018.10.1 Logic Question Training

There are now a total of 100 horses with 100 stones, 3 species of horses, large horses, medium-sized horses and small horses. One horse can carry 3 pieces of stone at a time, the medium horse can carry 2 pieces, and the small horse 2 can carry a stone. Q. How many horses, medium and small horses are needed? (The crux of the problem is that it just has to run out of 100 horses).

解：设大型马有x匹，中型马有y匹，小型马有z匹，根据题意可得：    x+y+x=100?①    3x+2y+ 12z=100②， ②×2－①得：5x+3y=100，所以有y= 100-5x3 ，因为x、y必须是正整数，所以有：  x=17    y=5    z=78                  x=14   y=10  z=76 ，x=11  y=15  z=74 ，     x=8  y=20  z=72 ，     x=5  y=25  z=70 ，       x=2  y=30  z=68 ，答：可能有：大型马17匹、5匹中型马，78匹小型马；大型马14匹、10匹中型马，76匹小型马；大型马11匹、15匹中型马，74匹小型马；大型马8匹、20匹中型马，72匹小型马；大型马5匹、25匹中型马，70匹小型马；大型马2匹、30匹中型马，68匹小型马；

Suppose there is a pond with an infinite number of water. There are 2 empty water bottles, 5 litres and 6 liters respectively in volume. The question is how to get 3 liters of water from the pond using only these 2 kettles.

　　答：由满6向空5倒，剩1升，把这1升倒5里，然后6剩满，倒5里面，由于5里面有1升水，因此6只能向5倒4升水，然后将6剩余的2升，倒入空的5里面，再灌满6向5里倒3升，剩余3升。

Zhou Wen's mother is the analyst of Yu Lin Cement plant. One day, Zhou Wen came to the laboratory to do his homework. I want to go out to play after finish. "Wait, mother also want to test you a topic," she went on, "you see these 6 to do the test glass, the front 3 is filled with water, the back 3 is empty." Can you move only 1 glasses, handy the cups filled with water and the empty cups between them? "Zhou Wen, the school's famous" small smart ", she only thought for a while to do. What do you think of "Little smart", please?

　　答：设杯子编号为ABCDEF，ABC为满，DEF为空，把B中的水倒进E中即可。

Three young men fell in love with a girl, and in order to decide who could marry the girl, they decided to have a duel with a pistol. Xiao Li's hit rate is 30%, small yellow than he better, hit ratio is 50%, the most outstanding gunman is Kobayashi, he never made mistakes, hit ratio is 100%. Because of this obvious fact, for the sake of fairness, they decided to follow the order: Xiao Li first shot, Xiao Huang second, Kobayashi finally. Then cycle until they are left with only one person. So who has the greatest chance of surviving among the three? What kind of strategy should they take?

　　答：小林在轮到自己且小黄没死的条件下必杀黄，再跟菜鸟李单挑。所以黄在林没死的情况下必打林，否则自己必死。　　小李经过计算比较（过程略），会决定自己先打小林。于是经计算，小李有873/2600≈33.6%的生机；　　小黄有109/260≈41.9%的生机；小林有24.5%的生机。哦，这样，那小李的第一枪会朝天开，以后当然是打敌人，谁活着打谁；　　小黄一如既往先打林，小林还是先干掉黄，冤家路窄啊！最后李，黄，林存活率约38：27：35；　　菜鸟活下来抱得美人归的几率大。　　李先放一空枪（如果合伙干中林，自己最吃亏）黄会选林打一枪（如不打林，自己肯定先玩完了）林会选黄打一枪（毕竟它命中率高）李黄对决0.3:0.28 0.4可能性李林对决0.3:0.60.6可能性成功率0.73　　李和黄打林李黄对决0.3:0.40.7*0.4可能性李林对决0.3:0.7*0.6*0.70.7*0.6可能性成功率0.64

Two inmates were held in a cell. Every day, the prison provides a pot of soup for the cell, allowing the two inmates to divide themselves. At first, these two people often quarrel, because they always think that the other side of the soup more than their own. Then they found a way to make the most of the two worlds: one for the soup, and the other for the first choice. So the dispute was settled. Now, however, a new prisoner has been added to the cell, and it is now three people to divide the soup. A new approach must be found to maintain peace between them. What should we do? Note: Psychological problems are not logical.

　　答：让甲分汤，分好后由乙和丙按任意顺序给自己挑汤，剩余一碗留给甲。这样乙和丙两人的总和肯定是他们两人可拿到的最大。然后将他们两人的汤混合之后再按两人的方法再次分汤。

Five yuan coins of the same size. Require two-phase contact, how should be placed?

　　答：底下放一个1，然后2 3放在1上面，另外的4 5竖起来放在1的上面。

On a rectangular table, you put n a round coin of the same size. Some of these coins may not be entirely on the table, or they may overlap each other, and when one more coin is placed in the center of the table, the new coins must overlap with some of the original coins. Please prove that the entire desktop can be completely covered with 4n of coins.

　　答：要想让新放的硬币不与原先的硬币重叠，两个硬币的圆心距必须大于直径。也就是说，对于桌面上任意一点，到最近的圆心的距离都小于2，所以，整个桌面可以用n个半径为2的硬币覆盖。　　把桌面和硬币的尺度都缩小一倍，那么，长、宽各是原桌面一半的小桌面，就可以用n个半径为1的硬币覆盖。那么，把原来的桌子分割成相等的4块小桌子，那么每块小桌子都可以用n个半径为1的硬币覆 盖，因此，整个桌面就可以用4n个半径为1的硬币覆盖。

A ruler with a ball, a length of approximately 2/3 of the diameter of the ball. How do you measure the radius of the ball? A lot of ways to see who's more ingenious

Guess the question Mr. S, Mr. P, Mr. Q they know there are 16 cards in the drawer of the table: Hearts A, Q, 4 spades J, 8, 4, 2, 7, 3 grass flowers K, Q, 5, 4, 6 blocks a, 5. Professor John picked out a card from the 16 cards and told Mr. P the number of points in the card to tell Mr. Q about the color of the card. At this time, Professor John asked Mr. P and Mr. Q: Can you infer from the known points or suits what the card is? So Mr. S heard the following dialogue: Mr. P: I don't know this card. Mr. Q: I know you don't know this card. Mr. P: Now I know this card. Q: I know, too. After listening to the dialogue above, Mr. S has thought about it and then correctly introduced the card. Excuse me: What card is this card?

　　答：方块5

A professor of logic, there are three students, and three students are very clever! One day the professor gave them a question, and the Professor put a note on each person's forehead and told them that each person's note was written with a positive integer, and that the sum of some two numbers equals the third one! (Everyone can see another two numbers, but can't see their own) The professor asked the first student: can you guess your own number? Answer: No, ask the second, can't, third, can't, ask first, can't, second, can't, third: I guessed it, it's 144! The professor smiled with satisfaction. Could you guess the number of the other two people?

　　A: After the first round, it means that any two numbers are different. In the second round, the first two people did not guess, indicating that no number is twice times the number of other numbers. Now there are several conditions: 1. Each number is greater than 02.22 is not equal to 3. Any number is not twice times the number of other numbers. Each number may be the sum or the difference of another two, the third person can guess 144, inevitably according to the previous three conditions to exclude one of the possible. Hypothesis: Is the difference of two numbers, namely x-y=144. At this time 1 (x,y>0) and 2 (x! =y) are satisfied, so to deny X+y must make 3 not satisfied, that is, x+y=2y, solution x=y, not set up (otherwise the first round can be guessed), so not two of the difference. So is the sum of two numbers, namely x+y=144.　　Similarly, at this time all meet, must make 3 not satisfied, namely x-y=2y, two equations, can get x=108,y=36. These two rounds of guessing are in fact the same: the first round (number one, number second), the second round (number third, number one, second).　　This is the same information that everyone gets at the end of each round (that is, the previous three conditions). So suppose we are C, to see how C is done: C See a of 36 and B of 108, because the condition, two number and is the third, then oneself either 72 or 144 (guessed this is because 72, 108 is 36 and 72 and 144 is 108 and 36) and. This kind of sentence can not understand the hands of the hand: assuming that their (C) is 72, then B in the second round to be able to see, the following is if C is 72,b idea: In this case, b see is a of 36 and C 72, then he can guess himself, is 36 or 108 (guess this is because 36, 36 plus 36 equals 72,108 is 36 and 108): If you assume that you (B) head is 36, then C in the first round can be seen, the following is if B is 36,c idea: in this case, C saw a 36 and a 36 of B, then he can guess himself, is 72 or 0 (this is no longer explained): If you assume (C) the head is 0, then a in the first round can be seen, the following is if C is 0,a idea: In this case, a see is b 36 and C 0 , then he can guess himself, is 36 or 36 (this no longer explains), then he can quote his head of 36. (and then reverse push inverse push), now a in the first round did not report their own 36,c (in the imagination of B) can know that their head is not 0, if the other and b the same idea (refers to the B head is 36), then C in the first round can quote their own 72. Now c in the first round did not report their 36,b (in C's imagination) can know their head is not 36, if the other and C idea(72 on the head of c), B can report its 108 in the second round. Now b in the second round did not report their own 108,c can know that their head is not 72, then the only one on the C head may be 144.

A city happened a car hit people escape incident, the city only two colors of the car, blue 15% Green 85%, the incident when there is a person at the scene saw, he refers to the blue car, but according to the experts in the field analysis, then that condition can see the right probability is 80% then, the car is the probability of a blue car is how much?

　　答：15%*80%/（85％×20％＋15%*80%）

One man has 240 kilograms of water, and he wants to make money in dry areas. He carries a maximum of 60 kilograms each time, and consumes 1 kilograms of water per kilometer (evenly water consumption). Assuming that the price of water at the starting point of 0, later, and transport distance is proportional to (that is, at 10 km is 10 yuan/kg, 20 km is 20 yuan/kg ...) And suppose that he must return safely, how much money can he make, please?

　　答：f（x）=（60-2x）*x,当x=15时，有最大值450。450×4

1=5,2=15,3=215,4=2145 so 5 =?

　　答：因为1=5，所以5=1.

2n individuals lined up at the cinema, with a fare of 50 cents. Of these 2n individuals, n individuals were only 50 cents and the other n individuals had $1 (paper tickets). The Stupid movie theater started selling tickets at 1 cents. Q: How many queueing methods make it possible for a movie theater to have 50 cents every time a ticket is bought for $1. Note: $1 = 100 cents a person with 1 dollars, with a paper money, can't break into 2 50 cents

　　答：本题可用递归算法，但时间复杂度为2的n次方，也可以用动态规划法，时间复杂度为n的平方，实现起来相对要简单得多，但最方便的就是直接运用公式：排队的种数=（2n）！/[n!（n+1）！]。

2018.10.1 Logic Question Training