MINCOUNT - Move To Invert
A triangle made of coins of height h is as follows:
- It has h coins at the base and h-1 coins one level above base and so on. (Coins are placed as shown in the figure below.)
- At the top-most level there will be only one coin.
Now given h, the task is to invert this triangle by moving the minimum number of coins.
For example when h=4 triangle is:
For h=4 at least 3 coins must be moved to invert it.
Input
In the first line N will be given and then N lines follow with each line having a integer which is the height of triangle in that test case. (0 ≤ h < 1010.)
Output
For each test case output in a separate line the minimum number of moves required to invert the triangle. Output fits in long long data type.
Example
Input: 1 3 Output: 2
hide comments
Gopinath:
2013-03-19 19:40:30
Same Algo in Python and C++ , TLE and in C , AC.. :) |
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@looser@:
2013-03-05 20:26:29
same algo in python TLE and in c AC |
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Ouditchya Sinha:
2013-03-03 09:23:18
It's weird, my AC solution gives 1294379940242040320 for 10^10... looks like test cases are weak. |
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Po:
2013-03-03 07:08:04
Answer for 100 is 1683....HOW..Could somebody pls explain? |
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bigBOSS:
2012-12-27 05:14:39
any tricky case plzzz. getting WA |
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secret22:
2012-07-27 08:49:32
how to minimise the time?
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Pranshul Agarwal:
2012-06-22 03:27:23
test cases are weak.... output for 10^10 4368837285860298922 still got AC |
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eliminator:
2012-06-12 18:48:28
:)AC gives=>
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eliminator:
2012-06-12 18:47:52
top row = 1, second row = 2 3, third row = 4 5 6, fourth row = 7 8 9 10 Solution: move 7 to the left of 2, move 10 to the right of 3, move 1 below 8&9 |
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The Mundane Programmer:
2012-06-03 17:56:18
not getting how it is inverted by moving 3 coins....... |
Added by: | Abhilash I |
Date: | 2006-12-16 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO |
Resource: | IIIT Hyderabad Local Programming Contest |