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

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;

    }

}

巴特西

pku oj overhang叠加卡片求最少的卡片数

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