Intelligence logic question (1)

[1] Suppose there is a pond with an infinite amount of water in it. There are two empty kettles with 5-litre and 6-litre capacities respectively. The question is how to get 3 litre water from the pond with only the two kettles.

From full 6 to empty 5 inverted, 1 litre left, continue to full 5 inverted, this 1 litre again empty 5, then 6 full, inverted 5 inside, because there is 1 litre of water in 5, 6 can only pour 4 litre of water to 5, then the remaining 2 litre of 6, pour 5 into the empty 5, then fill 6 to 5 to pour 3 litre, 3 litre remaining.

 

[2] three young men fell in love with a girl at the same time. They decided to fight with a gun to decide who would marry the girl. Xiao Li's hit rate is 30%, Xiao Huang is better than him, the hit rate is 50%, and the best gunner is Xiao Lin. He never makes a mistake and the hit rate is 100%. Because of this obvious fact, for the sake of fairness, they decided to follow the order: Xiao Li shot the gun first, Xiao Huang second, and Xiao Lin finally. Then it repeats until they have only one person left. Who has the greatest chance to survive these three individuals? What policies should they adopt?

Kobayashi will kill porn when it's his turn and Xiao Huang is not dead. Then he will make a single pick with cainiao Li.

Therefore, Huang will beat the forest if the forest is not dead, or else he will die.

After computation and comparison (the process is omitted), Mr. Li will decide to beat Mr. Lin first.

As a result, Xiao Li has the vitality of 873/2600 ≈ 33.6%;

Xiao Huang has the vitality of 109/260 ≈ 41.9%;

Kobayashi has 24.5% of its vitality.

Oh, in this way, the first shot of Xiao Li will go to the ground. Then, of course, he will beat the enemy and who will beat the enemy;

As always, Xiao Huang beat the forest first, but Xiao Lin killed Huang first, and his family was narrow!

Finally, Li, Huang and Lin had a survival rate of about 38: 27: 35;

Cainiao have a high chance of returning their beauty to their lives.

Li Xianzhi shot an empty gun (if the partner works for zhonglin, he will suffer the most). Huang will select Lin to take a shot (if Lin is not hit, he must have finished playing first) lin will select Huang to take a shot (after all, it has a high hit rate) Li Huang showdown 0.3: 0.280.4 possibility Li Lin showdown 0.3: 0.60.6 probability success rate 0.73

Li and Huang beat Lin and Li Huang showdown 0.3: 0.40.7*0.4 possibility Li Lin showdown 0.3: 0.7*0.6*0.70.7*0.6 probability success rate 0.64

 

[3] place n round coins of the same size on the desktop of a rectangle. Some of these coins may be incomplete on the desktop, or some may overlap with each other. When another coin is placed and its center is inside the desktop, the new coin must overlap with some of the original coins. Prove that the entire desktop can be completely covered with 4 n coins.

To prevent the new coin from overlapping the original coin, the center distance between the two coins must be greater than the diameter. That is to say, for any point on the desktop, the distance to the nearest center is less than 2. Therefore, the entire desktop can be covered with n coins with a radius of 2.

Scale down the desktop and coin by a factor. Then, the length and width of each small desktop half of the original desktop can be covered with n coins with a radius of 1. Then, divide the original table into four equal small tables, so each small table can be covered with n coins with a radius of 1. Therefore, the entire desktop can be covered with 4 n coins with a radius of 1.

 

[4] five Yuan coins of the same size. What should we do if we need to contact each other?

Put 1 at the bottom, 2 3 on the top of 1, and 4 5 on the top of 1.

(5) Mr. S, Mr. P, and Mr. Q knew that there were 16 cards in the drawer of the table: red peach A, Q, 4 black peach J, 8, 4, 2, 7, 3 K, Q, 5, 4, 6 blocks A, 5. Professor John picked a card from the 16 cards and told Mr. P about the number of cards and the color of the card to Mr. Q. Professor John asked Mr. P and Mr. Q: Can you tell from the known points or colors what the card is? So Mr. S heard the following conversation: 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. Mr. Q: I know. After hearing the above conversation, Mr. s thought about it and correctly launched the card. Excuse me: What is this card?

Box 5

 

[6] A professor of logic has three students, all of whom are very smart! One day, the professor gave them a question. The professor pasted a piece of paper on each person's head and told them that each person's paper had a positive integer, and the sum of two numbers is equal to the third one! (Each person can see the other two numbers, but not his/her own.) The professor asked the first student: Can you guess your own number? Answer: No. Ask the second question. No. The third question. No. ask the first question. No. The second question. No. The third question. I guess it is 144! The professor smiled with satisfaction. Can you guess the number of the other two people?

After the first round, it indicates that any two numbers are different. In the second round, the first two people did not guess, indicating that none of them is twice that of others. Now we have the following conditions: 1. Each number is greater than 02. It ranges from two to two. 3. Any number is not twice the other number. Each number may be the sum or difference of the other two. The third person can guess 144, and one of the three conditions must be excluded. Assume that the difference between two numbers is x-y = 144. At this time, 1 (X, Y> 0) and 2 (X! = Y). Therefore, to deny x + y, we must make 3 unsatisfied, that is, x + y = 2y, and obtain x = y, not true (otherwise the first round can be guessed), so it is not the difference between two numbers. Therefore, it is the sum of two numbers, that is, x + y = 144. Similarly, if both 1 and 2 are satisfied, 3 is not satisfied, that is, x-y = 2y. If the two equations are the same, x = 108, y = 36.

The order of the two rounds of guesses is as follows: first round (1, 2), second round (3, 1, 2 ). In this way, the information obtained at the end of each round is the same (that is, the preceding three conditions ).

Let's assume we are C. Let's take a look at how C is made: C sees 36 of A and 108 of B. Because of the condition, the sum of two numbers is the third, then you can either 72 or 144 (guess this is because 108 is the sum of 36 and 72, and 144 is the sum of 108 and 36. Raise your hand if you cannot understand this sentence ):

If your (c) is 72, then B can see it in the second round. below is the idea of C being 72 and B: in this case, B sees 36 of A and 72 of C, so he can guess himself, 36 or 108 (guess this is because 36, if 36 and 36 are equal to 72,108, the sum of 36 and 108 is ):

If your (B) header is 36, then C can be seen in the first round. below is the idea of B's 36 and C: in this case, C sees 36 of A and 36 of B, so he can guess himself, 72 or 0 (this is not explained ):

If your (c) header is 0, then a can see it in the first round. The following figure shows how C is 0 and A: in this case, A sees B's 36 and C's 0, so he can guess himself, is 36 or 36 (this is not explained), then he can report his 36. (Then, reverse push and reverse push) Now, A does not report its own 36 in the first round. C (in B's imagination) can know that it is not 0 on its head, if other ideas are the same as those of B (that is, B's head is 36), C can report its own 72 in the first round. Now C does not report his 36 in the first round. B (in C's imagination) can know that his head is not 36, if other ideas are the same as those of C (that is, C's head is 72), B can report its 108 in the second round. Now, if B does not report his 108 in the second round, C can know that his head is not 72, and the only thing on C is 144.

 

[7] There are 2n people queuing into the cinema, the fare is 50 cents. Among the 2n, N are only 50 cents, and n are $1 (paper ). When a stupid cinema starts selling tickets, there is no 1 cent. Q: How many queuing methods are there to make a 50 cent change to a cinema every time you buy a ticket for $1?

Note: One dollar = 100 cents for a person who owns one dollar, has a banknote, and cannot be broken into two 50 cents.

Recursion is available in this question.AlgorithmHowever, if the time complexity is 2, you can use the dynamic programming method. The time complexity is the square of N, which is much simpler to implement. But the most convenient thing is to directly use the formula: number of queues = (2n )! /[N! (N + 1)!].

If you don't consider whether the cinema can find money, there will be a total of (2n )! /[N! N!] Queuing method (that is, extracting the number of combinations of n persons from 2n persons). For each queuing method, if the cinema cannot find money, it is called unqualified, this queuing method is (2n )! /[(N-1 )! (N + 1)!] (N-1-1 combination of individuals from 2n), so the number of qualified queuing types is (2n )! /[N! N!] -(2n )! /[(N-1 )! (N + 1)!] = (2n )! /[N! (N + 1)!]. Why is the unqualified number (2n )! /[(N-1 )! (N + 1)!], It's too complicated to say, so I won't talk about it here.

 

[8] A diamond is placed at the door of each elevator on the first floor to the tenth floor, with different diamond sizes. When you take the elevator from the first floor to the tenth floor, the elevator door on each floor will open once. You can only get a diamond once. How can you get the largest one?

First, take the diamond on the first floor, and then compare the diamond in the handle on each floor with the diamond on the first floor, if the diamond on the first floor is larger than the diamond in the hand, replace the diamond in the hand with the diamond on that floor.

 

[9] the U2 choir will arrive at the concert venue within 17 minutes. On the way, it is necessary to cross a bridge. Four people will depart from the same end of the bridge. You have to help them reach the other end, the sky is dark, and they only have one flashlight. At the same time, a maximum of two people can bridge the bridge together, while a flashlight must be held when crossing the bridge, so someone has to bring the flashlight back and forth to both ends of the bridge. A flashlight cannot be lost. The walking speed varies between the four people. If the two people walk with each other, the speed of the slow person prevails. Bono takes 1 minute to bridge the bridge, edge takes 2 minutes to bridge the bridge, Adam takes 5 minutes to bridge the bridge, and Larry takes 10 minutes to bridge the bridge. How do they cross the bridge in 17 minutes?

2 + 1 first 2

Then 1. Return and send the flashlight 1.

5 + 10 + 10

2. Send a flashlight 2.

2 + 1 past 2

Total 2 + 1 + 10 + 2 + 2 = 17 minutes

 

[10] How can I divide 140 grams of salt into 50 and 90 grams of salt by using 7 grams and 2 grams of weight for each day?

140-> 70 + 70 70-> 35 + 35

35 + 70 = 105

105-> 50 + 7 + 55 + 2

55 + 35 = 90