HDU 1056

/*
Problem 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.



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.

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)
*/
刚开始也吓到了，以为是图一类题，后来仔细一看下面这个式子，额有点眉目，意淫之下，结果还是出来了
下面是直接法
//大意是 1 / 2 + 1 / 3 + ···+ 1 / (n+1)
//需要多少步运算可以大于等于这个数，输出步数。注意是大于等于
#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    double a,sum;
    int i,cou;
    while(scanf("%lf",&a),a!=0.00){
    i=2;
    sum=0;
    while(sum<a){//此处<=应该为<
        sum+=1.0/i;
        cou=i;
        i++;
    }

    printf("%d card(s)\n",(cou-1));
  }
    return 0;
}