Hangover(poj 1003)

1 问题描述

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)

大致意思

若将一叠卡片放在一张桌子的边缘，你能放多远？如果你有一张卡片，你最远能达到卡片长度的一半。（我们假定卡片都正放在桌 子上。）如果你有两张卡片，你能使最上的一张卡片覆盖下面那张的1/2，底下的那张可以伸出桌面1/3的长度，即最远能达到 1/2 + 1/3 = 5/6 的卡片长度。一般地，如果你有n张卡片，你可以伸出 1/2 + 1/3 + 1/4 + … + 1/(n + 1) 的卡片长度，也就是最上的一张卡片覆盖第二张1/2，第二张超出第三张1/3，第三张超出第四张1/4，依此类推，最底的一张卡片超出桌面1/(n + 1)。现在给定伸出长度C（0.00至5.20之间），输出至少需要多少张卡片。

题目地址poj 1003

2 算法思路

非常简单的题目，第i次的长度为1i+1。还需要考虑的就是0的情况，因为浮点数无法直接判断相等，但是因为精度就小数点后2位，可以先乘以100在判断是否为0

3 算法实现

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
double val;
while(cin >> val) {
// double无法直接判断相等，可以乘100转成int判断是否是0
if((int)(val*100) == 0) return 0;
int i;
double len = 0.00;
for(i = 1; true; ++ i) {
len += 1.0/(i+1);
// 大于输入的值就结束
if(len > val) {
printf("%d card(s)\n", i);
break;
}
}
}
return 0;
}

4 数据测试