NYOJ 156 Hangover

最新推荐文章于 2019-04-29 16:27:38 发布

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最新推荐文章于 2019-04-29 16:27:38 发布

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分类专栏：语言入门文章标签： C语言基础 ACM ACM入门基础 NYOJ 156 Hangover

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本文详细解析NYOJ第156题Hangover，适合ACM初学者，通过实例探讨问题解决思路，涵盖C语言基础知识点。

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

1.00

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

POJ

上传者

iphxer

/*
**类型：入门级题目
**题目来源：NYOJ  156
**时间：2017/7/22
**解决方案：
*/
#include<stdio.h>
int main()
{
	double c;
	while(scanf("%lf",&c)!=EOF&&c!=0.00)   //%lf 
	{
		double sum = 0.0;
		int count = 0;
		if(c <= 0.5)
		{
			printf("1 card(s)\n");
		}else{
			for(int i = 2; ; i++)
			{
				sum = sum + 1.0/i;
				count ++;
				if(sum >= c)
				{
					printf("%d card(s)\n",count);
					break;
				}
			}
		}
	}	 
	return 0;
}
 //再次看一下%f  %lf  以及思考最大值5.20，可以表示最大数为？