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HangOver

系統(tǒng) 2033 0

HangOver

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7693????Accepted Submission(s): 3129


Problem Description

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.

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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.

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.

?

?

Sample Input

1.00 3.71 0.04 5.19 0.00

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?

Sample Output

3 card(s) 61 card(s) 1 card(s) 273 card(s)

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?

Source

Mid-Central USA 2001

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題目沒什么難度,分明就是某年NOIP的級數(shù)求和,不過題目里如果不說,我還真不一定能想到,這個(gè)結(jié)論得記一下.

      #include<stdio.h>
      
        

#include
      
      <
      
        string
      
      .h>


      
        int
      
       f[
      
        1024
      
      
        ];


      
      
        double
      
       s[
      
        1024
      
      
        ];


      
      
        void
      
      
         getprepared()

{

    memset(f,
      
      
        0
      
      ,
      
        sizeof
      
      
        (f));

    memset(s,
      
      
        0
      
      ,
      
        sizeof
      
      
        (s));

    s[
      
      
        1
      
      ]=
      
        0.5
      
      
        ;

    
      
      
        for
      
       (
      
        int
      
       i=
      
        2
      
      ;i<=
      
        1000
      
      ;i++) s[i]=s[i-
      
        1
      
      ]+
      
        1.0
      
      /(i+
      
        1
      
      
        );

    
      
      
        for
      
       (
      
        int
      
       i=
      
        1
      
      ;i<=
      
        520
      
      ;i++
      
        )

    {

        
      
      
        double
      
       x=i/
      
        100.0
      
      
        ;

        
      
      
        for
      
       (
      
        int
      
       j=
      
        1
      
      ;j<=
      
        1000
      
      ;j++
      
        )

        
      
      
        if
      
       (s[j]>=
      
        x)

        {

            f[i]
      
      =
      
        j;

            
      
      
        break
      
      
        ;

        }

    }

}


      
      
        int
      
      
         main()

{

    getprepared();

    
      
      
        double
      
      
         ss;

    
      
      
        while
      
       (scanf(
      
        "
      
      
        %lf
      
      
        "
      
      ,&ss)!=
      
        EOF)

    {

        
      
      
        if
      
       (ss==
      
        0
      
      ) 
      
        return
      
      
        0
      
      
        ;

        
      
      
        int
      
       x=
      
        100
      
      *
      
        ss;

        printf(
      
      
        "
      
      
        %d card(s)\n
      
      
        "
      
      
        ,f[x]);

    }

    
      
      
        return
      
      
        0
      
      
        ;

}
      
    

?

HangOver


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