E - Hangover

E - Hangover
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu
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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)

此题如果直接来求会不精确如果数极大答案就会趋向一个固定值

#include<stdio.h>double delta=1e-8;int zero(double x)//在精度detla范围内，若x是小于0的负实数，则返回-1；若是大于0的正实数则返回1；若x为0，则返回0    {        if(x<-delta)            return -1;        return x>delta;    }int main(){    double a;    double len[999];    int i;    while(scanf("%lf",&a)&&a)    {        len[0]=0.0;        for( i=1;zero(len[i-1]-a)<0;i++)        {            len[i]=len[i-1]+1.0/double(i+1);        }        printf("%d card(s)\n",i-1);    }}