HDOJ_1049Climbing Worm(爬)

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

 

问题描述：

一英寸的蠕虫在一英寸深的井的底部。它有足够的能量每分钟爬上你的脚，但是在爬之前必须休息一分钟。在余下的时间里，它下降到D英寸。攀登和休息的过程然后重复。虫子从井里爬出来还有多久？我们总是把一分钟的一分钟算为一分钟，如果虫子在爬完的时候到达了井顶，我们就假设虫子把它弄出来了。

输入：

将有多个问题实例。每一行包含3个正整数n，u和d。这些值给出上面段落中提到的值。此外，可以假设d＜u和n＜100。n＝0的值表示输出结束。

输出：

每一个输入实例都应该在一行上生成一个整数，表示蠕虫从井里爬出来所需的时间。

输入：

10 2 1 20 3 1 0 0 0

输出：

17 19

#include
int main()
{
    int n,u,d,time,wz;
    while(scanf("%d%d%d",&n,&u,&d)&&(n!=0&&u!=0&&d!=0))
    {
        time=0,wz=0;
        while(wz