POJ_Hangover

Hangover
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 95164 Accepted: 46128
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)

翻译：

　　若将一叠卡片放在一张桌子的边缘，你能放多远？如果你有一张卡片，你最远能达到卡片长度的一半。（我们假定卡片都正放在桌 子上。）如果你有两张卡片，你能使最上的一张卡片覆盖下面那张的1/2，底下的那张可以伸出桌面1/3的长度，即最远能达到 1/2 + 1/3 = 5/6 的卡片长度。一般地，如果你有n张卡片，你可以伸出 1/2 + 1/3 + 1/4 + … + 1/(n + 1) 的卡片长度，也就是最上的一张卡片覆盖第二张1/2，第二张超出第三张1/3，第三张超出第四张1/4，依此类推，最底的一张卡片超出桌面1/(n + 1)。下面有个图形的例子：

　　现在给定伸出长度C（0.00至5.20之间），输出至少需要多少张卡片。

解决思路

/*输入一个浮点数，记为c，找出累加式1/2 + 1/3 + 1/4 + ... + 1/(n + 1)大于等于c的最小的n，0.01<=c<=5.20，输入0.00代表输入结束。*/#include "stdio.h"void main(){    double c,sum;    int n=1;    scanf("%lf",&c);    while(c!=0){        sum=0;        n=1;        //printf("Next PUT:%d Card\n",n);        while(sum<c){            sum+=(1.0/(++n));            //printf("Next PUT:%d Card\n",n);        }        printf("%d card(s)\n",n-1);        scanf("%lf",&c);    }}