Hangover

时间限制: 1000ms 内存限制: 65536KB

问题描述

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 are 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.

输入描述

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.

输出描述

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.

样例输入

1.00
3.71
0.04
5.19
0.00

样例输出

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

来源

{Mid-Central USA 2001}

问题分析：（略）

这个问题和《POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover【数学计算】》是同一个问题，代码拿过来用就AC了。

程序说明：参见参考链接。

参考链接：POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover【数学计算】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++程序如下：

/* POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover */  #include <iostream>
#include <cstdio>  using namespace std;  const double one = 1.0;  int main()
{  double len, sum, d;  int i;  while((cin >> len) && len != 0.00) {  i = 1;  d = 2.0;  sum = one / d;  while(sum < len) {  d += 1.0;  sum += (one / d);  i++;  }  cout << i << " card(s)" << endl;  }  return 0;
}

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NUC1003 Hangover【数学计算+水题】

Hangover

时间限制: 1000ms 内存限制: 65536KB

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