本文主要是介绍NUC1003 Hangover【水题】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Hangover
时间限制: 1000ms 内存限制: 65536KB
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.
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.
1.00 3.71 0.04 5.19 0.00
3 card(s) 61 card(s) 1 card(s) 273 card(s)
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;
}
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