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FAVDICE - Favorite Dice |
BuggyD loves to carry his favorite die around. Perhaps you wonder why it's his favorite? Well, his die is magical and can be transformed into an N-sided unbiased die with the push of a button. Now BuggyD wants to learn more about his die, so he raises a question:
What is the expected number of throws of his die while it has N sides so that each number is rolled at least once?
Input
The first line of the input contains an integer t, the number of test cases. t test cases follow.
Each test case consists of a single line containing a single integer N (1 <= N <= 1000) - the number of sides on BuggyD's die.
Output
For each test case, print one line containing the expected number of times BuggyD needs to throw his N-sided die so that each number appears at least once. The expected number must be accurate to 2 decimal digits.
Example
Input: 2 1 12 Output: 1.00 37.24
Added by: | Matthew Reeder |
Date: | 2006-10-29 |
Time limit: | 1s |
Source limit: | 30000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | Al-Khawarizm 2006 |
hide comments
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2017-03-16 07:37:01
Coupon collecter problem, very useful piece of knowledge. Great problem. |
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2017-03-14 12:59:14
Geometric random variable>>>>AC... |
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2017-03-07 21:02:58
Coupon Collector problem, helped! |
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2017-03-02 21:23:28
Mathematics always screw me... learned something new... HINT : expected number of trials until success = 1/p given p is the probability of success in every trial. |
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2016-09-11 19:08:23
Is there any new approach? Last edit: 2016-09-11 19:09:45 |
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2016-09-09 22:26:51
coupon collector application |
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2016-08-21 04:54:46 Pratik S
Why is this argument wrong? Let E(n) be the desired expectation. Now Choose some n-1 of these n numbers. The number of turns required for getting these at least once is E(n-1). Now the exp. number of throws to get that last number is n, So total number of exp throws is E(n-1) + n. Since we could have chosen any number as the last number in an equally likely way, the expectation recurrence should be:- E(n) = n*(E(n-1)+n)/n = E(n-1)+n Blows my mind ... can't find the fault!!! |
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2016-06-04 21:44:28
Hints.. in comment section. :( |
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2016-02-19 20:33:18
Easy after learning "Coupon Collector Problem"!!!! |
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2016-02-09 21:39:37
after reading coupon collector problem the question becomes too easy to solve .only need to take care of data type conversion. |