Hangover问题

在acm上看到一个十分有意思的问题http://poj.org/problem?id=1003
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)

Source
Mid-Central USA 2001
尝试了一下

#include 

using namespace std;

int main()
{
     
int i,n;
class hanover
{
     
    public:
    float a;//card length

    int m;//card number
};

float sum;
hanover a[100];//定义数组出错！一定用具体的数长度
i=0;
for(i=0;scanf("%f",&a[i].a),a[i].a!=0.00;i++);
i=n;//到这里成功输入，m作为元素个数
sum=0;

for(i=0;i<n;i++)
{
     
for(float j=2 ;sum<a[i].a;j++)//一开始这里的n,m混淆
    {
     sum=sum+1/j;
    a[i].m=j-1;}
printf("%d",a[i].m);
printf("\n");
sum=0;
}

/*for(i=0;*/
return 0;

}