POJ 1003 Hangover

最新推荐文章于 2024-06-06 09:06:09 发布

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本博客介绍了一个计算使用卡片堆叠实现的最大偏移量的问题，通过数学公式和编程实现来解决实际问题。

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Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 94952		Accepted: 45999

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)

解题思路：本题是很简单的，但是要注意的是“/”如果是两个整数相除的话，其实是整除的，所以要用一个小数点标识一下啦。

代码如下：

#include<stdio.h>

int main(void){
    float sum=0.0f;
    float c;
    int n;
    while((scanf("%f",&c),c!=0.00)){
	sum=0.0f;
	n=2;
	while(sum<c){
	    sum+=1.0/n++;
	}
    	printf("%d card(s)\n",n-2);
    }
    return 0;
}