Hangover

Time Limit: 1000MS Memory Limit: 10000K

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)

思路：

只要能读懂题目就行了，也就是每次都加上1/x，然后当总和大于n时结束。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    double n;
    while (~scanf("%lfd", &n) != EOF && n) {
        int x = 2;
        double sum = 0;
        while (sum <= n) {
            sum += 1.0 / x;
            x++;
        }
        printf("%d card(s)\n", x - 2);
    }
    return 0;
}