poj 1003 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)
这个题就是一个大水题！！！！翻译过来就是1/2+1/3+1/4+……+1/（n+1）什么时候>c,
#include<cstdio>
int main()
{
    float c;
    while(scanf("%f",&c)==1&&c!=0.00)
    {
        float s=0;//不是int
        int sum=0;
        float t=2;//这个也不能是int
        while(s<c)
        {
            s=s+1/t;
            t++;
            sum++;
        }
        printf("%d card(s)\n",sum);
    }
    return 0;
}
一开始我就是这么写的，但是输入后没有输出，我就看了一下其他人写的，但是是前天晚上看的，我们昨天休息，所以也没看，今天早上到了之后看的，就想着先自己改改，但是还没改对，我又想看别人写的代码了，
但是我又看了一下题，再 改了一下，我就没抱希望自己会对，但是运行没错误，就输入数据，这个我也没抱希望，结果！！！第一个数据对了，激动！！！就继续测试第二个，第三个，都对！！！然后就提交，通过！！！
然后反思了一下，感觉自己太依赖别人的代码了，一不会就上网找别人的代码，尤其是改不对的时候，我就认为自己笨，肯定做不出来，不会就没办法，然后就给自己找借口什么的，哎，以后呢，要相信自己啦
但是看别人的代码跟我的不一样，我的就是一看题就那么写，跟英语翻译一样，一个字对应一个词，而别人不是这么写的，他们会变通一下，他用的就是把每个相加后都存起来，然后用输入的数根存好的数比较，
这样我感觉对于输入比较大的数据来说会好一点，我这个就是一个输入就循环一次，数太多就不好弄