Climbing Worm

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探讨了一只虫子如何在特定条件下从深井中爬出的问题，通过编程实现了一个算法来计算虫子完全爬出井所需的时间，考虑了虫子的爬升速度、休息时的滑落距离及井的深度。

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Climbing Worm
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23393 Accepted Submission(s): 16011

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

问题简述： 一只虫子在深n英寸的井里，每分钟它能往上爬u英寸，每爬一分钟要休息一分钟，休息时，虫子会往下滑d英寸，求虫子爬多久能爬出这口井。
问题分析：
1）由于爬一分钟休息一分钟，则当分钟对2取余等于1时，则加上u，否则就减去d；
2)循环进行到最后一步，时间i还会自增，所以要输出i-1；
3）限制每一次输入的n，u，d都不等于0。

已AC的代码：

#include<iostream>
using namespace std;
int main()
{
    int n, u, d;
    while (cin >> n >> u >> d&&n!=0&&u!=0&&d!=0)
    {
        int i, h = 0;
        for (i =1;h<n;i++)
        {
            if (i % 2 == 1)
                h += u;
            else
                h = h - d;
        }
        cout << i-1<<endl;
    }
}