Climbing Worm(蜗牛爬井问题，递推)

最新推荐文章于 2024-07-25 16:30:00 发布

张志来

最新推荐文章于 2024-07-25 16:30:00 发布

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分类专栏： ACM

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探讨一只寸虫如何在不断爬升与滑落的过程中，最终成功爬出特定深度的井，通过数学建模的方式计算出所需的具体时间。

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Climbing Worm

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

解析：题意是一只蜗牛从井底往上爬，每分钟往上爬u厘米，然后休息一分钟，在休息的时候会往下滑d厘米。问要多少分钟爬出n厘米的井。

注意的是：在最后的一分钟如果刚好跑到井口也算爬出来了。就不会出错了。

#include"stdio.h"
int main()
{
	int n,u,d,i,j,sum;
	while(scanf("%d %d %d",&n,&u,&d)!=EOF)
	{
		sum=0;
		if(n==0&&u==0&&d==0)
		break;
		i=1;
		j=0;
	    while(sum<n)
	    {
    		sum=u+(i-1)*(u-d);
    		i++;
    		j=j+2; 
    	}
    	printf("%d\n",j-1);
	}
	return 0;
}