Climbing Worm

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：输入三个数，分别是总路程、每分钟前进路程、每分钟下滑距离，
前进一分钟后下滑一分钟，求要多少时间到达。
问题分析：用循环判断到达是为几分钟。
程序说明：先判断是否能到达，再用for循环求几分钟时满足条件。
AC通过的C++程序如下：

include

using namespace std;
int main()
{
int n, u, d,s=0;
while (cin >> n >> u >> d)
{
if (u == 0 || u < d)
{
break;
}
else {
int work = 0;
for(s = n / u;work {
s++;
work = u (s/2) - (d((s/2) - 1));
}
}
s--;
cout << s << endl;
s = 0;
}
}