Poj1003------Hangover

题目

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.

题目大意就是，我给你一个长度，作为目标长度，看需要多少张牌叠放在一起，其露出桌子的长度能够到这个目标长度。叠放的规则是，一张牌至少要保证有1/2的长度有支撑，所以为了达到最大化，最上面那张牌是露出1/2，下一张是1/3，以此类推，直到1/n，露出的部分到达长度c

问题

主要就是读题，还有就是，循环会用到一个1/i，这个i很可能随习惯就设成int类型了，按照题目的翻译，的确从1/2到1/n分母都是整型，但是！！你累加的结果s要求是浮点型，为了避免c++取整性，（所有c++计算题涉及得出精确浮点结果都需要避免系统陷阱！！)所以要么把i设成double，要不写成1.0/i，

代码

#include <iostream>
using namespace std;
int main()
{double n,tmp,s=0;while(cin>>n&&n!=0){for(double i=2;;i++)//注意1除i变成整数 ！！！！！ {s=s+1/i;//double i  or  1.0/i if(s>=n) {cout<<i-1<<" card(s)"<<endl;s=0;break;}}}return 0;
}