Everyone should have heard of this old topic: there are 1000 identical bottles, of which 999 are ordinary water and one bottle is poison. Any creature that drinks poison will die in a week. Now, you only have 10 mice and a week, how do you check out which bottle is poisonous?The answer to this question is also classic: numbering the bottles from 0 to 999, then converting them all to 10-bit binary numbers. Get the first rat to drink. 1 to 1000 all bina
Time limit per Test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output
The Dragon and the princess are arguing about what to do on the New Year's Eve. the dragon suggests flying to the mountains to watch fairies dancing in the moonlight, while the princess thinks they shoshould just go to bed early. they are desperate to come to an amicable agreement, so they decide to leave this up to chance.
They take turns drawing a mouse from a bag which initially containsWWhi
D. Bag Of MiceThe Dragon and the princess is arguing about-what-does on the New year's Eve. The dragon suggests flying to the mountains to watch fairies dancing in the Moonlight and while the Princess thinks they shou LD just go to bed early. They is desperate to come to a amicable agreement, so they decide to leave the to chance. they take turns drawing a mouse from a bag Which initially Contains w white and b black mice. The person was the first
Problem Description: There are 1000 identical bottles, of which 999 are ordinary water and 1 are poison.Any life that has been poisoned will die after a week. Now you only have 10 mice and 1 weeks of time, how to test out which bottle is poisonous medicine?Answer:According to 2^10=1024, so 10 mice can determine which of the 1000 bottles is poisonous. The specific implementation of 3
A few days ago to a foreign company in Shanghai to participate in the written test, because the examination is tight, some of the major problems can not be completed, is very depressed. Now we are going to discuss one of the questions in this article---the problem of cat eating mice. Write this article just want to communicate with you and learn, inevitably there will be mistakes and deficiencies, hope to get everyone's criticism, in this we are grate
1000 bottles of liquid 10 white mice interview questions answer, 1000 bottles of questions answer
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Really, I dreamed of it last night.
A mouse ran to a friend's house, and my friend refused to kill it and said it was cute. The result quickly became a problem.Then I woke up, hey! I don't know what preference this dream is. I haven't found Zhou gongjie on the Internet for a long time.However, I woke up after my dream was completed. I thought of a program question, which I had come up:
There were 30 mice in one household, which were so rampant tha
The questions are as follows:
There are 1000 Identical bottles in the lab, but one of them is toxic. You can use lab mice to test which bottle is a poison. If the mouse drinks the poison, it will die in a week. The potion in other bottles has no side effects. Could you tell me the minimum number of mice that can be used to find out which bottle is poison within a week:A.9B. 10C. 32D. 999.E. None of the abo
1056. Mice and rice (25) Time Limit 30 ms memory limit 32000 kb code length limit 16000 B discriminant program standard
Mice and riceIs the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. the goal of each mouse is to eat as much rice as possible in order to become a fatmouse.
First the playing order is randomly decided for
Title Link: Http://codeforces.com/problemset/problem/148/DThe main topic: a bag of W White mouse, b black Mouse; A and b take the mouse in turn (not put back), who first caught microscope, who win; because B rude, every time a mouse, will run out of a first;The probability of seeking a win;Topic Analysis:This kind of probability DP state is fixed, DP (i, j) indicates the probability of the current state awin;The probability of 1:dp[i][0],a win is 1;dp[0][j] probability is 0;2:DP[I][J] The follo
the behavior of the Press Bar to disappear. This is not the case.
The method of intermittent rewards, but instead, made these mice feel like drug addiction, and continue to press on, whether or not they can be rewarded. Jenkins then compares a fixed frequency (such as a press bar getting food four times) with a random intermittent reward, and finds that in a random reward situation, eliminating the mouse Press Bar takes the longest time.
The di
Mice and Holes CodeForces, holescodeforces
Mice and Holes CodeForces-797F
There are n rats and m holes on a digital axis. The coordinates of the mouse are x [1],..., x [n], the coordinate of the hole is p [1],..., p [m], each hole can accommodate c [1],..., c [m]: Ask the minimum value of the sum of the distances that every mouse runs after hiding in the hole.
Analysis:
If the sequence of holes and
I. Description of the problem
The existing n mice in a circle, a cat from any place to eat mice, every time a mouse to eat, please give the last mouse number? The title requirement is to give the mouse number n, output the cat last eat the mouse number.
Second, problem solving
We assume that there are N mice, the sequence number is 1,2,3,......, n-1,n, and the
Codeforces Round #105 D. Bag of mice probability dp,
Http://codeforces.com/contest/148/problem/D
The question is that dragon and Princess take turns pumping mice from the bag, W white teachers in the bag, and D black teachers. The Princess smokes first, and the first pulls the white mouse to win, the dragon will randomly run out of a mouse each time it is pumped. Give W and D for you to find the probability
Probability DPThe 9th question in Kuangbin summaryAh ... The data given by the topic are only the number of rats and black rat, so we can only do this (GAO) (P).Obviously can use the number of two kinds of mice as state = =My WA Approach:Make F[i][j] said from (w,b) take the mouse one until (i,j) "This time round process take" two people have been dead heat probability, it is obvious (I,J) This state win probability is I/(I+J), then the probability of
DP [x] [Y]: Now there are X white mice, y black mice, and the probability that the princess will win.
So:
Assume that the princess gets the white mouse directly. The probability is X/(x + y), and the princess wins.
Assuming that the princess gets the black mouse, the probability is Y/(x + y), then the princess assumes that to win, the dragon must get the black mouse, the probability is (Y-1)/(x + Y-1 );
The
Presumably many people have more than one computer, such as a desktop + notebook, and many times we will open them at the same time. But have you found that if there are more than one computer set of keyboard and mouse on the table, it is very tiring to switch back and forth. If you can control all your computers at the same time by using just a set of mouse buttons, will you be excited about it?SynergyIt's a good tool for this! It allows you to share a set of keys on multiple computers and even
Requirements: 1. To have linkage, the behavior of the mouse and the master is passive.2. Considering extensibility, cat calls can cause other linkage effects.Point: 1. Linkage effect, run code as long as the cat.cryed () method is executed. 2. Abstract the mouse and the masterRating Standard: Public InterfaceObserver {voidResponse ();//The response of the Observer as if the mouse had seen the cat's reaction } Public InterfaceSubject {voidAimat (Observer obs);//for which observers, this
ProgramDESIGN: when the cat shouted, all the mice started to escape and the host was awakened. (C # language)Requirements:
1. The behavior of rats and Masters Should Be passive.2. Considering scalability, the call of a cat may cause other association effects.
Key points: 1. linkage effect, runCodeRun the cat. cryed () method. 2. abstract the mouse and the hostScoring standard: Constructs cat, mouse, and master classes, and enables the program to r
Probability, $DP $.$DP [i][j][0]$ says there is still $i$ a white cat, $j $ A black cat, the probability of the princess reaching the target State in case of a shot.$DP [i][j][1]$ says there is still $i$ a white cat, $j $ A black cat, the probability of reaching the target State in case of a dragon shot.At first $dp[i][0][0]$ are $1$, the answer is $dp[w][b][0]$. The recursive type is easy to write:$DP [i][j][0]=i/(I+J) +j/(i+j) *dp[i][j-1][1]$$DP [i][j][1]=j/(I+J) *i/(i+j-1) *dp[i-1][j-1][0]+j*
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