Digital Intelligence Testing
In fact, the intelligence test is to examine the applicant's ability to solve problems under restrictions. This type of questions will appear in the recruitment interview of multinational corporations, which plays an extremely significant role in examining a person's way of thinking and the ability to change his or her way of thinking. According to some studies, such capabilities are often closely related to the strain and innovation status at work. Therefore, when answering these questions, we must break through the mindset and try to consider the problem from different perspectives, constantly Reverse Thinking and empathy, and associate the question with the scenario we are familiar, avoid confusion.
Intelligence Test on numbersInterview Example 1: Jeff and diamond like playing game of coins. one day they designed a new set of rules :( 1) totally 10 coins. (2) one can take away 1, 2, or 4 coins at one time by turns. (3) who takes the last loses. given these rules whether the winning status is pre-determined or not? (J and d like to play coin games. They set a set of rules one day: (1) a total of 10 coins; (2) 1, 2, and 4 coins each time; (3) whoever gets the last one will lose. Are you sure you will lose the game ?) Answer: (1) In the future, you must leave one to ensure that you can win. (2) Try to keep the other party 2, 3, and 5. (3) That is, leave, after obtaining the image. (4) If you leave 6 after the result is retrieved, and the other party obtains 2 results, which is in conflict with (3), 6 is excluded. (5) If you leave 8 after the result is retrieved, and the other party takes 4, the situation is the same as (3), so 8 is also excluded. (6) Likewise, 9 won't work. If I leave 9 for the other party and 2 for the other party, 7 for the other party will be in conflict with (3, therefore, it is also excluded. (7) Obviously, I can only extract 1, 4, and 7. (8) because only 1, 4, and 7 can be used to win the game, it is impossible for me to beat the game first. Obviously, only the other party can smoke the game first, that is, the first person to lose the game. Some basic methods to answer questions about the intelligence test with extended knowledge are as follows. (1) division first removes irrelevant problems. Problems that can be identified are first identified, so as to narrow down the scope of the unknown as much as possible, so as to facilitate problem analysis and resolution. This way of thinking is very useful in our work and life. (2) the progressive method analyzes the data layer by layer based on known conditions and ensures that each step is accurate. There may be several branches. We should first start with a simple branch based on the principle of first being easy, then difficult. (3) The reverse method starts from the final result of the question and continues step by step until the answer to the question is obtained. Some problems are easily solved using this method, but it is difficult to use other methods. (4) let's make one or more assumptions about the given problem, and then analyze it based on the given conditions. If there is a conflict with the conditions given by the question, it indicates that the hypothesis is incorrect, let's make another or another assumption. If there are only two possible results, the problem has been solved. In the history of science, "assumptions" have played a significant role. (5) Some Problems in the calculation method can be solved only after calculation. It should be noted that the questions in the quiz often contain implicit conditions, and sometimes the number given is useless. (6) analysis is the most basic method. Analysis is often used for various methods. It can be said that the level of analytical ability is a manifestation of a person's level of intelligence. The analytical ability is not only innate. To a large extent, it depends on the training of the day after tomorrow. It is necessary to develop a good habit of Analyzing Objective things. (7) Graph Method Based on the conditions known in the problem, use appropriate methods to draw a picture, which helps solve the problem. There are some problems. Before drawing a picture, you may feel that you have nowhere to start. After drawing the picture, you can see it clearly. (8) In fact, many problems must be solved using several different methods. The so-called comprehensive method is to combine various methods (including methods other than the aforementioned methods) to solve some problems. Where can I go for an interview example at USD? Three friends stayed in a hotel. At checkout, the total bill is USD. Each of the three friends shared $ and handed over the $ to the waiter, entrusting him to pay the account on behalf of the master station. However, at the time of settlement, the hotel offered a preferential price. The total station refunded $500 to the waiter and paid $500. The waiter deducted $500 from the refund of $200 and only refunded $300 to three guests. The three guests shared the USD 300 and each person retrieved the USD 100. In this way, each of the three guests actually pays 900 US dollars, a total of 700 US dollars, plus 200 US Dollars deducted by the waiter, a total of 900 US dollars, then where is the difference between the 100 US dollars? Answer: This question is purely a text game, but if your mind is not clear enough, you may be confused. The actual payment of $700 is equivalent to the actual settlement of $500 on the Total Station plus the deduction of $200 from the waiter. Adding $700 to the $200 is unreasonable, and adding the $700 to the $300 is justified, this is equivalent to the $ that the customer previously handed over to the waiter. Interview Example 3: click the mouse to start now! Contestants include Ralf, Willi, and Paul. Ralf was able to give 10 clicks in 10 seconds, Mike was able to give 20 clicks in 20 seconds, and Paul was able to give 5 clicks in 5 seconds. The time used by the above members is calculated as follows: from the first blow to the end of the last blow. Are they flat? If not, who first clicks 40 mouse clicks? Resolution:
NSecond click
NThe mouse clicks the first mouse to start timing.
N-1 requires
NSeconds, then if you press 40 mouse clicks, Ralf needs (40-1)/(9/10) = 39/0. 9 seconds, Wylie needs (40-1)/(19/20) = 39/0. 95 seconds, Paul needs (40-1)/(4/5) = 39/0. 8 seconds, so we hit it first. Answer: Willey finishes the attack first. Interview Example 4: Is the equation 8 + 8 = 91 correct with the first feeling? Describe the reason. Answer: Right. Example 5: My father called her daughter and asked her to buy some daily necessities for herself. She also told her that the money was put in an envelope on her desk. The daughter found the envelope and saw 98 on it. She thought there was 98 yuan in the envelope, so she took out the money and put the money into her schoolbag. In the store, she bought 90 yuan of things, and only found that she had not left 8 yuan, but 4 yuan. Back home, she told her father about it and suspected her father had made a mistake in money. His father said with a smile that he did not know what was wrong and was wrong with his daughter. Q: Where is my daughter wrong? Answer: Take it down. 86 is regarded as 98. Interview example: at, the three children turned their pockets and took out all their money to see how much they had. The result is 320 yen in total. Two coins are 100 yen, two are 50 yen, and two are 10 yen. Each child does not carry the same coin. In addition, children without 100 yen coins did not bring 10 yen coins, and children without 50 yen coins did not bring 100 yen coins. Can you figure out what coins the three Japanese children used to carry? Answer: first child:, 50, 10; Second Child:, 50; Third Child: 10. Example 7: There is a worm that splits every two seconds. After the split, two new bugs will be split after two seconds. If there is only one worm in a bottle at first, change it to 2 in 2 seconds, and then change it to 4 in 2 seconds ...... Two minutes later, a bottle of worms was full. Assume that two of these worms are placed in the bottle. Q: How long has it taken to become a full bottle? Answer: after 1 minute 58 seconds, it happens that a bottle is full. Because it takes only 2 seconds to change from a worm to two worms. When there is only one bug in the bottle, it will become 2 after 2 seconds. At this time, the situation is the same as that of the first two bugs in the bottle. The time difference between the two cases is 2 seconds. So, after 1 minute 58 seconds, it was just a full bottle. Interview Example 8: stylenx is a lion and female with wings in ancient Greek mythology. It is said that she was waiting for someone to guess the riddle near theobis. Once, she asked Prince thebs to guess: "There is an animal, four legs in the morning, two legs at noon, three legs at night. What is the animal ?" The wise prince said, "It's a person ." He guessed it. What kind of questions will you ask if you are a modern stylenx? For example, what symbols can be added between 1 and 0 to make the number larger than 0 and smaller than 1? Do you know? Answer: 0.1 interview Example 9: The company has 1 000 apple and 10 boxes. After loading 1 000 apple into 10 boxes in advance, ask: how can customers provide the whole box to customers regardless of how many apples they need? Answer: 1, 2, 4, 8, 16, 32, 64,128,256,489. The reason is simple.
NOnly the number of boxes that should be loaded is the first one
N-The sum of the quantity of 1 boxes plus 1. Interview Example 10: There is a 100-storey building with two identical glass pawns in your hands. Dropping a pawn from a certain floor of the building will break down. Use these two glass pawns in your hands to find out an optimal strategy to know the critical layer. Analysis: Set the total floor
H,
A(
N) (Such
A(1 ),
A(2 )......) Indicates the layer where each piece is thrown.
A(
N), You must try one by one along the layer where the last throwing is located, that is, you must throw at most
A(
N)-
A(
N-1)-1 +
NTimes. And so on, consider the average value, sum each time, and get the result after elimination :(
A(
N)-
N+ (1 +
N)*
N/2 )/
NBecause
A(
N) =
H, So the equation is
H/
N+
N/2 + 1/2. The minimum value occurs in
N= (
H* 2) ^ (1/2. Substitution
H= 100.
NAbout 14. That is, in the worst case, it takes about 14 times to complete. Answer: 14 times at most. For the first time from layer 14, if the layer is broken, the layer starts to be thrown, and layers are added until Layer 13. 14 times in total. If it is not broken, it will be thrown again at Layer 27. And so on, from layer 15 to layer 26 for a total of 12 times. The first 14 layers and 27 layers are added twice, so it is also 14 times. Run the following commands in sequence: 1414 + 13 = 2727 + 12 = 3939 + 11 = 5050 + 10 = 6060 + 9 = 6969 + 8 = 7777 + 7 = 8484 + 6 = 9090 + 5 = 9596979899 up to 14 times. The algorithm is as follows. # Include <iostream> using namespace STD; int kk [5]; int fmin2 [101]; int fmax2 [101] [101]; int F (int n ); int main () {memset (fmin2, 101, sizeof (fmin2); memset (fmax2, 0, sizeof (fmax2); fmin2 [0] = 0; fmin2 [1] = 1; cout <"100 floor:" <F (100) <Endl; return 0;} int F (int n) {If (fmin2 [N] <101) return fmin2 [N]; for (INT I = 1; I <= N; I ++) {int d = f (n-I); fmax2 [N] [I] = (I-1)> D? (I-1): D;} int min = 101; For (INT I = 1; I <= N; I ++) {min = min <fmax2 [N] [I]? Min: fmax2 [N] [I];} fmin2 [N] = 1 + min; return 1 + min;} example 11: There is a 27 cm fine wood pole, there is an ant in each of the 5 locations 3 cm, 7 cm, 11 cm, 17 cm. The wooden pole is very small and cannot pass through two ants at the same time. At the beginning, the head of the ant financial system is either left or right. They only move forward or turn their heads, but do not move back. When any two ants meet, they both turn their heads to the opposite direction. It is assumed that the ant Financial can walk 1 cm away every second. Write a program to find the minimum and maximum time for all ants to leave the wooden pole. Resolution: The Ghost algorithm can be used. The ghost means "passing energy" and "penetrating ". First, we should judge the ant that is closest to the center point. We can use the program to make a good decision, that is, the ant that is located at 11 cm points. In fact, the fastest time for this ant is the minimum time. Second, it depends on the outside ant. The program makes a good decision, that is, the 3 cm ant. In fact, the slowest time of this ant is the maximum time. Other ants can ignore it. Answer: The maximum time is 27-3 = 24, and the minimum time is 11-0 = 11. You can use the exhaustive method to verify the algorithm as follows. # Include <iostream> using namespace STD; // constant set const int APOs = 0; // a point position const int BPOs = 30; // B point position const int maxant = 5; // maximum number of ants const int speed = 1; // speed // global variable // known initial position (must be an odd number) int poslist [maxant] = {3, 7, 11, 17, 23}; // use a 5-bit binary number to indicate the starting direction of 5 ants: 00000 ~ 11111, a total of 32 types of Enum antflag {antflag1 = 0x1, antflag2 = 0x2, antflag3 = 0x4, antflag4 = 0x8, antflag5 = 0x10 // antflag6 = 0x20 // if there are more //...}; int antflist [] = {antflag1, antflag2, antflag3, antflag4, antflag5}; // calculate the start direction of the ant financial system based on the binary number int startdir (INT val1, int val2) {int IR = (antflist [val1] & val2 )? 1:-1; return IR;} class ant; // ant class ant {PRIVATE: int m_id; // ant ID (0 ~ 4) bool m_life; // life state, initially 1, 0 int m_pos after exit; // coordinates on the wooden pole (0 ~ 27) int m_dir; // direction (1,-1) int m_speed; // speed (1) int m_time; // crawling time (0 ~?) Public: static int count; // The number of existing ants. Static int antlist [maxant]; // stores the public: ant (); void Init (); // ant financial initializes int GETID () {return m_id;} // obtains the ID number int gettime () {return m_time;} // returns the time void setdir (INT Val) {m_dir = startdir (m_id, Val);} // void checklife () in the initial direction; // check the life status void changedir (); // void runpos () in the change direction (); // location after second void print () {cout <"ID:" <m_id <"POS:" <m_pos <"dir: "<m_dir <" time: "<m_time <Endl ;}}; // end antant: ant () {m_id = count; Init (); count ++;} void ant: Init () {m_pos = poslist [m_id]; m_speed = speed; m_life = 1; m_time = 0;} void ant: checklife () {If (m_life) {If (m_pos <APOs | m_pos> BPOs) {m_life = 0; count --;} elsem_time ++ ;}} void ant: changedir () {If (m_life) {m_dir * =-1 ;}} void ant: runpos () {If (m_life) m_pos + = m_dir * m_speed; antlist [m_id] = m_pos;} // a class funant {public: int lasttime; // The time when the last ant leaves ant ants [maxant]; // The ant object array contains 5 public: funant () {}// set the ant initial direction void funsetdir (INT d) {for (INT I = 0; I <maxant; I ++) Ants [I]. setdir (d) ;}// the screen outputs all the ant information void print () {for (INT I = 0; I <maxant; I ++) Ants [I]. print ();} // until the last ant leaves the pole, output the time void run () {While (ANTS [0]. count) {for (INT I = 0; I <maxant; I ++) {ants [I]. runpos (); // mobile ant location ants [I]. checklife (); // checks whether the ant is on the pole} changedir (); // checks if the ant encounters and changes the direction} lasttime = ants [0]. gettime (); For (INT I = 1; I <maxant; I ++) {bool B = lasttime <ants [I]. gettime (); If (B) {lasttime = ants [I]. gettime () ;}} print () ;}// checks the adjacent ant's position function. If the position is the same, call the void changedir () {for (INT I = 0; I <MAXANT-1; I ++) {If (ANTS [0]. antlist [I] = ants [0]. antlist [I + 1]) {ants [I]. changedir (); ants [I + 1]. changedir () ;}}}; int ant: antlist [] = {3, 7, 11, 17, 23}; int ant: Count = 0; //////////////////////////////////////// // int main () {int maxlist [32]; // store the 32 result Arrays for (INT I = 0; I <32; I ++) {ant: Count = 0; funant A1; a1.funsetdir (I); a1.run (); maxlist [I] = a1.lasttime; cout <"lasttime:" <a1.lasttime <Endl ;}int min, Max; min = max = maxlist [0]; for (INT I = 0; I <32; I ++) {If (max <maxlist [I]) max = maxlist [I]; If (min> maxlist [I]) min = maxlist [I];} cout <"Max: "<max <Endl <" Min: "<min <Endl; return 0;} // end main
