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  • [POJ1003]Hangover

    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 111811   Accepted: 54608

    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.


    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.

    Sample Input

    1.00
    3.71
    0.04
    5.19
    0.00
    

    Sample Output

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

    Source

    思路

      级数求和

    var n,k:real;
    procedure main;
    var i:longint;
    begin
        i:=1;k:=0;
        while k<n do
            begin
                inc(i);
                k:=k+(1/i);
            end;
        writeln(i-1,' card(s)');
    end;
    begin
        while true do
            begin
                read(n);
                if n=0 then halt;
                main;
            end;
    end.
    View Code
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  • 原文地址:https://www.cnblogs.com/yangqingli/p/4907556.html
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