HANGOVER - Hangover
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.
Input
The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.
Output
For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.
Input: 1.00 3.71 0.04 5.19 0.00 Output: 3 card(s) 61 card(s) 1 card(s) 273 card(s)
hide comments
John Doe:
2015-02-12 12:08:59
Daniel, I'm not disputing that you can get AC with the given formula.
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Daniel Carvalho:
2015-02-12 02:20:45
Extremelly easy, but bad formulated. John Doe, your formulation is wrong. Use the given general formula. |
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John Doe:
2015-02-11 16:43:04
Does the problem specification reflect reality?
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D:
2015-02-07 17:02:59
can't see the image.
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shreya goyal:
2015-02-05 21:39:15
AC IN FIRST GO
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Added by: | Wanderley Guimarăes |
Date: | 2006-06-09 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | ACM Mid Central Regionals 2001 |