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  • poj1003

                                                                                                              Hangover
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 99369   Accepted: 48162

    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

    Mid-Central USA 2001
     1 #include<iostream>
     2 using namespace std;
     3 
     4 int main()
     5 {
     6     double sum = 0.5;
     7     double num[301]={0,0.5};
     8 
     9     for(int i=2;sum<=5.20;i++)
    10     {
    11         sum = sum + 1.0/(i+1) ;
    12         num[i] = sum;
    13     }
    14     double n;
    15     while(cin>>n)
    16     {
    17         if(n==0.00)
    18             break;
    19         for(int i=1;i<301;i++)
    20         {
    21             if(num[i]>=n)
    22             {
    23                 cout<<i<<" card(s)"<<endl;
    24                 break;
    25             }
    26         }
    27     }
    28     return 0;
    29 }
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  • 原文地址:https://www.cnblogs.com/jhldreams/p/3750656.html
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