POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover【数学计算+水题】

Hangover Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 125708 Accepted: 61333

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

Regionals 2001 >> North America - Mid-Central USA

问题链接：POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover。

题意简述：已知c=1/2+1/3+1/4+....1/(n+1)，输入m（m是浮点数），求正整数n使得c>=m。

问题分析：从小到大试探n即可。

程序说明：（略）

AC的C语言程序如下：

/* 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; }