Climbing Worm

ClimbingWorm

Time Limit : 2000/1000ms (Java/Other)   MemoryLimit : 65536/32768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) :1

Font:Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

An inch worm is at the bottom ofa well n inches deep. It has enough energy toclimb u inches every minute, but then has to rest a minute before climbingagain. During the rest, it slips down d inches. The process of climbingand 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 justreaches the top of the well at the end of its climbing, we'll assume the wormmakes it out.

Input

There will be multiple problem instances. Each line willcontain 3 positive integers n, u and d. These give the values mentioned in theparagraph above. Furthermore, you may assume d < u and n < 100. A valueof n = 0 indicates end of output.

Output

Each input instance should generate a single integer on aline, indicating the number of minutes it takes for the worm to climb out ofthe well.

Sample Input

10 2 1

20 3 1

0 0 0

Sample Output

17

19

Source

East Central North America 2002

 

 参考代码：

#include <stdio.h>

int main()
{
	int n,u,d;
	int count, sum;
	scanf("%d%d%d",&n,&u,&d);
	while(n)
	{
		sum=count=0;
		while(sum<n)
		{
			sum=sum+u;
			count++;
			if(sum<n)
			{
				sum=sum-d;
				count++;
			}
		}
		printf("%d\n",count);
		scanf("%d%d%d",&n,&u,&d);
	}
	return 0;
}