hdu -1049 Climbing Worm

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14400    Accepted Submission(s): 9719

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
  

 

Source

East Central North America 2002

 

//非常水的一道题，题意就是洞总长n英尺，每一分钟爬u英尺，然后休息一分钟，掉落d英尺；求出洞所用时间？

代码如下：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

int main()
{
    int n,u,d;
    while(scanf("%d%d%d",&n,&u,&d)!=EOF,n)
    {
        int c=0;
        while(n>0)//判断爬虫是否出界
        {
            if(n>0)//计算所用时间
            {
               n-=u;
               c++;
               if(n==0)
                break;
            }
           // cout<<n<<endl;//测试数据
            if(n>0)
            {
                n+=d;
                c++;
            }
            //cout<<n<<endl;
        }
        cout<<c<<endl;//输出结果
    }
    return 0;
}