开新坑——【poj】Hangover

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.Description

 

 

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)

又要开一个新坑= =，洛谷的还没怎么做，学校就要求在poj上刷题QWQ头大，明明省赛刚刚结束，好吧，回归poj，刚做完A+B，看第二题直接高精度........于是挑了一个比较简单的做为新坑代表吧

突然发现不能用#include<bits/stdc++.h>，好吧，不能偷懒了QWQ

#include<iostream>/*由于不能用万能头文件于是只能勤快点了QWQ*/
using namespace std;
int main()
{
	double m;
	while(cin >> m&&m!=0){
		int i;
		double sum=0;
		for(i=2;;i++){
			sum+=1.0/i;
			if(sum>=m){
				cout << i-1 << " "/*第一次写忘记空格了*/ << "card(s)" << endl;
				break;
			}
		}
	}
	
}