1045 HangOver

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HangOver

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Time limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 7773   Accepted Submit: 4111  

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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.

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.

Example input:

1.00
3.71
0.04
5.19
0.00

Example output:

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

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Problem Source: Mid-Central USA 2001

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http://acm.zju.edu.cn/show_problem.php?pid=1045

 

#include  < iostream >
using   namespace  std;

int  solve( float  x);

int  main()
... {
    float len;
    cin >> len;
    
    while ( len != 0.00 )
    ...{
        cout << solve(len) << " card(s)" << endl;
        
        cin >> len;
    }
    
    return 0;
}

int  solve( float  x)
... {
    int n = 1;
    float sum = 0.00;
    
    while ( 1 )
    ...{
        sum += (float)(1/(float)(n+1));
        if ( sum >= x )
        ...{
            break;
        }
        n++;
    }
    
    return n;
}