ACM学习-POJ-1003-Hangover

灰太狼 2022-10-13 10:28 144阅读 0赞

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Hangover

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;92150</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;44638</td> 
  </tr> 
 </tbody> 
</table>

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][]

**问题要求：已知c=1/2+1/3+1/4+....1/(n+1).现给出一个值m，求n的值使得c刚好超过m。**

**问题分析：问题很简单，就是遍历，直到找到满足条件的那个n。 但是要注意运算的时候进行类型转换。（最早做的时候卡在这里了，一时粗心没注意）**

**下面给出AC代码**

#include <stdio.h>
    
    int main()
    {
        double sum_;
        double result_;
        int n;
        
        while ((~scanf("%lf", &sum_)) && sum_ != 0.00   )
        {
            result_ = 0.00;
            for (n=2;result_ <= sum_ ; n++)
            {
                result_ += 1.00/(double)n;//注意类型的转换
            }
            printf("%d card(s)\n", n-2);
        }
        return 0;
    }

[Mid-Central USA 2001]: http://poj.org/searchproblem?field=source&key=Mid-Central+USA+2001