poj ID1003

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.

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)


Submission:

#include<iostream>
using namespace std;
int main()
{
    int i=2;
    float n;
    float sum=0;
    while(cin>>n&&n!=0.00)
    {      
       i=2; 
       sum=0;                        
       while(sum<=n)
       {
          sum=sum+1.00/i;
          i++;
        }
        cout<<i-2<<" card(s)"<<endl;
    }   
       return 0;
}

这次没有那么那么顺利，稍微有点细节上的问题，sum每次循环之后都应该重新赋值为0，否则的话，只会一次一次的累加，结果就会出问题。另外输出的那一行结果，card前面是有一个空格的，少了的话，也是不能ac的，细节，还是细节!