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;
}
}
