hdu 1049 Climbing Worm

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1049

 

题目描述：

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9278    Accepted Submission(s): 6052

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
   
   
   
   

 

题意：

墙高n，虫的速度是u，虫爬一分钟要休息一分钟，在休息的一分钟里虫下降d。问给出n，u，d，求出虫到达n高度的最短分钟数。

 

题解：这里将爬一分钟和休息一分钟合并成一个动作，这样的一个动作就是一个周期。这里最短指的是虫第一次到达这个高度，而接下来肯定是休息时间，这样肯定要用周期来无限接近于n-u，然后虫子下一分钟的上爬u就刚好到达n的高度，所以抽象成这样的表达式t=(n-u)/(u-d)，(u-d)就是虫子一个周期内上爬的高度，这里如果t*(u-d)<(n-u)的话要让t++。即使整数t歌周期必须要>=(n-u).然后再往上爬一分钟就到了n的高度。由于一个周期是两分钟这样答案就是2*t+1。

 

代码：

/*
acm
hdu:Climbing Worm

*/
#include<stdio.h>
#include<stdlib.h>

int n=0,u=0,d=0,t=0;
int ans=0;

/*for test*/
int test()
{
	return(0);
}

/*main process*/
int  MainProc()
{
	while(scanf("%d%d%d",&n,&u,&d)!=EOF&&n>0)
	{
		t=(n-u)/(u-d);
		if(t*(u-d)<n-u)
		{
			t++;
		}
		ans=t*2+1;
		printf("%d\n",ans);
	}
	return(0);
}

int main()
{
	MainProc();
	return(0);
}