poj 1003 Hangover

Hangover

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)

宿醉

描述

你能让一堆卡片悬在桌子上方多远?如果你有一张卡片，你可以创造出半张卡片长度的最大悬垂。(我们假设纸牌必须垂直于桌子。)用两张卡片，你可以使顶部的卡片悬在底部的卡片长度的一半，底部的卡片悬在桌子的卡片长度的三分之一，总的最大悬在1/2 + 1/3 = 5/6卡片长度。一般来说，你可以让n张卡片以1/2 + 1/3 + 1/4 +…+ 1/(n + 1)牌长，上面的牌比第二张高出1/2，第二张比第三张高出1/3，第三张比第四张高出1/4，等等，下面的牌比桌子高出1/(n + 1)，如下图所示。

输入

输入由一个或多个测试用例组成，后面跟着一行包含数字0.00，表示输入的结束。每个测试用例都是包含一个浮点数c的单行，其值至少为0.01，最多为5.20;C正好包含三个数字。

输出

对于每个测试用例，输出所需的最小卡数，以达到至少c卡长度的悬垂。使用示例中所示的输出格式。

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)


题意：求最小的n让1+1/2+1/3+...+1/n大于给的一个实数。
思路：找一个数sum记录结果，写一个循环不断累加 sum+=1.0/n，
直到sum大于c，输出n-2.

代码：
#include<stdio.h>
int main()
{
    double c,sum;
    int n,card;
    while(scanf("%lf",&c)!=EOF&&c)
    {
        if((int)(c*100)==0)
	    break;
        sum=0;
        n=2;
        while(sum<=c)//循环模拟。直到超过
	{
            sum+=1.0/n;
            n++;
        }
        printf("%d card(s)\n",n-2);
    }
    return 0;
}