Hangover

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48443		Accepted: 22764

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)





这个题总体来说十分简单，下面是源代码：

//Hangover

//Time Limit: 1000MS		Memory Limit: 10000K



#include<stdio.h>



int main()

{

	float a;

	int i;

	float b=0;

	while(scanf("%f",&a))

	{

		if(a<0.01){ printf("这个数太小了/n");

			continue;}

		else if(a>5.20){ printf("这个数太大了/n");

			continue;}

		else 

		{	b=0;

			for(i=2;;i++)                        

			{

				b=b+(float)1/(float)i;

				if (b>a) 

				{

					printf("%d card(s)/n",i-1);

					break;

				}

			}

		}



	}



}



刚开始没有写else下b=0；这句话，认为每次计算结束后b会重新复制，观察才发现，程序一直没有退出while循环，所以b的值是保存着的，所以为了

结果的每次都正确需要每次都给b清零！！