NYOJ 156 Hangover

最新推荐文章于 2024-12-10 18:38:12 发布

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最新推荐文章于 2024-12-10 18:38:12 发布

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分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/qq523176585/article/details/19076471

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探讨如何通过堆叠方式使卡片最大限度地悬空于桌子之外。利用数学原理计算不同数量的卡片可以实现的最大悬空长度，并提供了一段C++代码示例来解决实际问题。

Hangover

时间限制：1000 ms | 内存限制：65535 KB

难度：1

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'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.

输入

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

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)

思路：略。

#include<iostream>
using namespace std;
int main()
{
	double d,sum,count;
	while(cin>>d){
		
		if(d==0)return 0;
		sum=0;count=1;
		int i;
		for(i=2;sum<d;i++)
			sum+=count/i;
		cout<<i-2<<" card(s)"<<endl;
	}
}