HDU 1056 HangOver(数学题)

                                                                       HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11757    Accepted Submission(s): 5102

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)
  
  

 

题意：一直 1/2  +  1/3  +  1/4  +  ...  +  1/( n   +  1)叠加，直到>m. （和poj一模一样的题）

AC代码：

#include<cstdio>
int main()
{
	float a=0.0;
	while(1)
	{
	    scanf("%f",&a);
		float sum=0;
		float i=0;
		int count=0;
		if(a==0.00)
		{
			break;
		}
		for(i=2.0;sum<a;i++)
		{
			sum=sum+1/i;
			count++;
		}
		printf("%d card(s)\n",count);
	}
	return 0;
}