Climbing Worm

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本文探讨了一个经典的爬虫问题，即一只爬虫如何在有限的能量下爬出特定深度的井。文章详细介绍了爬虫每分钟爬升和休息时下滑的具体情况，并提供了一段C语言代码作为解决方案，帮助读者理解爬虫最终成功爬出井所需的总时间。

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ClimbingWorm

Time Limit : 2000/1000ms (Java/Other) MemoryLimit : 65536/32768K (Java/Other)

Total Submission(s) : 1 Accepted Submission(s) :1

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Problem Description

An inch worm is at the bottom ofa well n inches deep. It has enough energy toclimb u inches every minute, but then has to rest a minute before climbingagain. During the rest, it slips down d inches. The process of climbingand resting then repeats. How long before the worm climbs out of the well?We'll always count a portion of a minute as a whole minute and if the worm justreaches the top of the well at the end of its climbing, we'll assume the wormmakes it out.

Input

There will be multiple problem instances. Each line willcontain 3 positive integers n, u and d. These give the values mentioned in theparagraph above. Furthermore, you may assume d < u and n < 100. A valueof n = 0 indicates end of output.

Output

Each input instance should generate a single integer on aline, indicating the number of minutes it takes for the worm to climb out ofthe well.

Sample Input

10 2 1

20 3 1

0 0 0

Sample Output

Source

East Central North America 2002

参考代码：

#include <stdio.h>

int main()
{
	int n,u,d;
	int count, sum;
	scanf("%d%d%d",&n,&u,&d);
	while(n)
	{
		sum=count=0;
		while(sum<n)
		{
			sum=sum+u;
			count++;
			if(sum<n)
			{
				sum=sum-d;
				count++;
			}
		}
		printf("%d\n",count);
		scanf("%d%d%d",&n,&u,&d);
	}
	return 0;
}