HDU 1049 Climbing Worm

最新推荐文章于 2021-03-07 16:48:10 发布

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本文探讨了一款名为ClimbingWorm的算法问题，该问题涉及爬虫的爬升速度、休息时间及洞深之间的关系。通过输入洞深、爬升速度和下滑速度，算法计算出爬虫爬出洞口所需的时间。详细解析了问题求解过程，并提供了代码实现，旨在帮助理解算法逻辑和实际应用。

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8181 Accepted Submission(s): 5273

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

Sample Output

  
   17
19

Source

East Central North America 2002

分析：小学奥数题。。只要注意计算时要防止出现已经爬出洞口再滑下来的情况，所以虫子最后一次一定是爬到洞口，不会再下滑

代码：

#include<stdio.h>

int main()
{
    int n,u,d;
    while(scanf("%d%d%d",&n,&u,&d))
    {
        if(n==0&&u==0&&d==0)
            break;
        int i;
        for(i=0;;i++)
        {
            if(i*(u-d)+u>=n)
                break;
        }
        i=i*2+1;
        printf("%d\n",i);
    }
    return 0;
}