杭电1056——HangOver（水题）

最新推荐文章于 2024-07-27 08:45:00 发布

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本文探讨如何通过巧妙地堆叠卡片来实现最大过挂长度，包括基本原理、数学公式以及实例演示。重点在于理解每张卡片如何依次增加过挂长度，最终达到目标长度。

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Problem 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.
这里写图片描述

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.

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)

题目比较简单，但是刚开始还是WA了好几次，后来把数据类型从float 换成double 就AC了，不知道为什么。。。。。

# include<stdio.h>

int main()
{
    double c,sum;
    int i;
    while(scanf("%lf",&c)!=EOF&&c!=0)
    {
        sum=0;
        for(i=2;sum<c;i++)
        {
            sum+=1.0/i;
        }
        printf("%d card(s)\n",i-2);
    }
    return 0;
}