Climbing Worm

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本文探讨了一种经典的爬虫井问题，即一只虫子每分钟向上爬升一定距离后会休息并下滑，直到最终爬出井口。文章通过具体实例展示了如何计算虫子完全爬出井的时间，并提供了一个简洁的 C 语言程序实现。

Climbing Worm

时间限制:1000女士内存限制:32768KB64位输入输出格式:%I64d & %I64u

提交状态

描述

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.

输入

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.

输出

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.

样例输入

样例输出

     
      17
19

#include<stdio.h>
int main()
{
    int n,u,d;
    while(scanf("%d%d%d",&n,&u,&d)&&n||u||d)
    {
        int a=0;
        while(n)
        {
            n=n-u;
            a++;
            if(n<=0)
                break;
            a++;
            n=n+d;
        }
        printf("%d\n",a);
    }
    return 0;
}