hdoj 1045 Climbing Worm

爬井小虫模拟算法

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本文介绍了一道经典的模拟算法题——爬井小虫。题目描述了一只小虫如何通过不断攀爬和休息来最终爬出井口的过程，并探讨了在不同参数下小虫爬出井口所需的时间。

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8979 Accepted Submission(s): 5849

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

题目意思明确，这是一道模拟算法题，找到worm爬出well的时间，根据不同的时间点依次累加时间即可。

#include <iostream>  
  
using namespace std;  
  
int main()  
{  
 int d, u, n;  
 int time,len;  
 while (cin >> n >> u >> d && (u + d + n)) {  
  len = time = 0;  
  while (len < n) {  
   time ++;  
   if (time%2 == 1) len += u;  
   else len -= d;  
  }  
  cout << time << endl;  
 }  
 return 0;  
}