pku oj overhang叠加卡片求最少的卡片数

这个估计是里面第二简单的了，因为第一简单的是求a+b

哈哈，一submit就ac了

题目如下：

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)

简单的归结就是一个级数求和的问题

#include <iostream>
using namespace std;

class Overhang {
private:
    double a;
public:
    int getCardsNum();
    void getValue() { cin >> a; }
    double geta() { return a; }
};

int Overhang::getCardsNum() {
    int num=1;
    double sum=0.0;
    while(true) {
        sum += 1.0/(num+1);
        num++;
        if (sum >= a)
            return --num;
    }
}
int main(void) {
    Overhang t;
    while(true) {
        t.getValue();
        if ((t.geta() -  0.0) < 0.0001)
            return 0;
        cout << t.getCardsNum() << " card(s)" << endl;
    }
}