HDU 1049 Climbing Worm （简单模拟）

井底之虫的逃逸

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探讨了一道经典的数学问题——井底之虫如何在反复爬升与滑落中最终逃离深井。通过分析问题背景及输入输出要求，提供了一个简洁的C++代码解决方案。

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
问题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1049
问题分析：高中做过的蜗牛爬树问题，只要用一层循环相加行动的里程数（d-u）即可，再判断行动的里程数加d是否等于总里程数，是则+1退出循环。
AC通过的C++语言程序如下：

#include<iostream>
using namespace std;
int main()
{
	int n, u, d;
	while (cin >> n >> u >> d)
	{
		if (n != 0)
		{
			int sc = 0, time = 0;
			for (int i = 1; sc != n; i++)
			{
				if (sc + u >= n) { time++; break; }
				sc = i * (u - d);
				time += 2;
			}
			cout << time << endl;
		}
	}
	return 0;
}