Hangover问题

How far can you make a stack of cardsoverhang a table? If you have one card, you can create a maximum overhang ofhalf a card length. (We're assuming that the cards must be perpendicular to thetable.) With two cards you can make the top card overhang the bottom one byhalf a card length, and the bottom one overhang the table by a third of a cardlength, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. Ingeneral you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) cardlengths, where the top card overhangs the second by 1/2, the second overhangstha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottomcard overhangs the table by 1/(n + 1). This is illustrated in the figure below.

Input

The input consists of one or more testcases, followed by a line containing the number 0.00 that signals the end ofthe input. Each test case is a single line containing a positive floating-pointnumber c whose value is at least 0.01 and at most 5.20; c will contain exactly threedigits.

Output

For each test case, output the minimumnumber 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)     61card(s)    1 card(s)     273 card(s)

程序代码：

#include<stdio.h>

double sum(int a);       //定义一个double型的求和函数，用来求放置a个卡片后露出桌面的长度和

int main()

{

    int n;        //n表示卡片的数量

    float C;      //C表示输入的长度（浮点型）

    scanf("%f",&C);      //输入

    while(C!=0)       //当输入的C不等于0时就进入循环进行判断和计算

    {

       for(n=1;;n++) //因为最少是一块，所以n赋初值为1，每次增加一块

       {

           if(C<=sum(n))     //如果卡片露出桌面的长度和大于等于输入的长度C时，跳出循环

           break;

       }

       printf("%d card(s)\n",n);   //输出

       scanf("%f",&C);   //再次输入，进入下一次循环，直到输入0时停止

    }

    return 0;

}

double sum(int a)    //累加求和函数

{

    int i;

    double s=0;

    for(i=1;i<=a;i++)

    {

       s=s+1/(i+1.0);

    }

    return s;

}