NUC1003 Hangover

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

1.003.710.045.190.00

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：

这个题与《POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover》完全相同，代码直接拿过来用就AC了。

程序说明：

参见参考链接。

参考链接：POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover

题记：

程序写多了，似曾相识的也就多了。

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;  }