HDOJ 1049-Climbing Worm【数学】



Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15450    Accepted Submission(s): 10488

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

 

解题思路

有一个小虫在一个深为N的井里,他每一分钟就会向上跳u英尺，跳完一下就会休息一分钟，这一分钟就会向下滑d英尺，问青蛙需要多长时间才能跳出井。

#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
	int n,u,d;
	while(scanf("%d%d%d",&n,&u,&d)&&(n||u||d))
	{
		int t;
		t=0;
		int wie=0;
		while(1)
		{
			if(wie>=n)
			break;
			wie+=u;
			t++;
			if(wie>=n)
			break;
			wie-=d;
			t++; 
		}
		printf("%d\n",t);
	}
	return 0;
}