hdu Climbing Worm

最新推荐文章于 2018-09-11 17:29:33 发布

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本文探讨了一个关于爬虫攀爬井壁的问题，通过设定特定参数，如井深、每分钟爬升及下滑距离，来计算爬虫完全爬出井口所需的时间。文章提供了完整的代码实现，并解释了判断条件。

点击打开链接密码：syuct

Climbing Worm

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

Total Submission(s) : 4 Accepted Submission(s) : 3

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Problem Description

An inch worm is at the bottom of a well n inches deep. It has enough energy to climb u inches every minute, but then has to rest a minute before climbing again. During the rest, it slips down d inches. The process of climbing and 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 just reaches the top of the well at the end of its climbing, we'll assume the worm makes it out.

Input

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

Output

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

Sample Input

10 2 1
20 3 1
0 0 0

Sample Output

17
19

好像做过，水题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int n,u,d;
	while(scanf("%d%d%d",&n,&u,&d),n+u+d)
	{
		int m=0,now=0;
		while(1)
		{
			now+=u;
			m++;
			if(now>=n)
				break;
			now-=d;
			m++;
			if(now>=n)
				break;
		}
		printf("%d\n",m);
	}
	return 0;
}